# MATH32062 Notes. 1 Affine algebraic varieties. 1.1 Definition of affine algebraic varieties

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1 MATH32062 Notes 1 Affine algebraic varieties 1.1 Definition of affine algebraic varieties We want to define an algebraic variety as the solution set of a collection of polynomial equations, or equivalently, a set where every element of a set of polynomials vanishes. Then every combination of those polynomials vanishes on the set, so we can use the ideal generated by those polynomials to define the algebraic variety. This motivates the definition below. Using ideals instead of concrete polynomials also turns out to be more useful for proving most of the theorems about algebraic varieties. In the following, K denotes an arbitrary field, unless specified otherwise. Definition. An affine algebraic variety is a set of the form V(I) = {(a 1, a 2,..., a n ) K n f(a 1, a 2,..., a n ) = 0, f I}, where I is an ideal I K[x 1, x 2,..., x n ]. Examples: 1. If I = {0}, then V(I) = K n. 2. If I = K[x 1, x 2,..., x n ], then V(I) =. 3. Let f 1, f 2,..., f m be polynomials of degree 1 in x 1, x 2,... x n, and let I = f 1, f 2,... f m be the ideal generated by them. Then V(I) is the set of solutions of the system of linear equations f 1 = f 2 =... = f m = 0. Such a variety is called an affine subspace of K n. This, in particular, includes sets containing a single point. If P = (b 1, b 2,..., b n ) K n, then the ideal I = x 1 b 1, x 2 b 2,... x n b n defines the variety V(I) = {P }. 4. Let I = y 2 x 3 R[x, y], then V(I) is the cuspidal cubic curve, also called Neile s semicubical parabola after William Neile, who computed its arc length in It is defined by the equation y 2 = x 3 in R 2, shown below on the left. 1

2 Let I = y 2 x 3 x 2 R[x, y], then V(I) is the nodal cubic curve defined by the equation y 2 = x 3 + x 2 in R 2, shown above on the right. 6. Let I = x 2 + y 2 1 R[x, y], then V(I) is the circle of radius 1 centred at the origin defined by the equation x 2 + y 2 = 1 in R If I = x, y R[x, y], then V(I) = {(0, 0)} R If I = x 2, y 2 R[x, y], then V(I) = {(0, 0)} R If I = x 2 + y 2 + z 2 1, x 2 x + y 2 R[x, y, z], then V(I) R 3 is a figure of eight curve, shown below

3 10. The special linear group SL(n, R) consisting of all n n real matrices with determinant 1 is also an affine algebraic variety. The set of all n n real matrices is R n2, the determinant is a polynomial in the entries of the matrix, therefore SL(n, R) = V( det(a) 1 ). Note. Examples 7 and 8 show that different ideals can define the same variety. The exact nature of the correspondence between ideals and varieties will be investigated later in Section 1.3. Proposition 1.1 Let K be a field. Let f 1, f 2,..., f m K[x 1, x 2,..., x n ] and let I = f 1, f 2,..., f m. Then V(I) = {(a 1, a 2,..., a n ) K n f i (a 1, a 2,..., a n ) = 0, i, 1 i m}. Proof. f i I for each i, 1 i m, so if (a 1, a 2,..., a n ) V(I), then f i (a 1, a 2,..., a n ) = 0 for every i, 1 i m. Conversely, assume that f i (a 1, a 2,..., a n ) = 0 for every i, 1 i m. Any f I can be written as f = 1 i m, therefore f(a 1, a 2,..., a n ) = so (a 1, a 2,..., a n ) V(I). m i=1 f i g i for suitable g i K[x 1, x 2,..., x n ], m f i (a 1, a 2,..., a n )g i (a 1, a 2,..., a n ) = 0, i=1 Definition. A ring R is Noetherian if and only if all of its ideals can be generated by finitely many elements. (Equivalently, R is Noetherian if and only if for any increasing sequence of ideals I 1 I 2 I 3... I n I n+1... there exists an N such that I n = I N for all n N.) Theorem 1.2 (Hilbert Basis Theorem) (Not examinable) If K is a field, then the polynomial ring K[x 1, x 2,..., x n ] is Noetherian for any n 0. Note: The Hilbert Basis Theorem justifies why we only considered ideals generated by a finite set of polynomials in Proposition 1.1. The combination of Proposition 1.1 and Theorem 1.2 shows that any affine algebraic variety is the set of solutions of finitely many polynomial equations. The proposition below describes some methods which can be used to construct further affine algebraic varieties from known ones. Proposition 1.3 (i) Let V 1 = V(I 1 ), V 2 = V(I 2 ),..., V k = V(I k ) be affine algebraic varieties in K n. Then V 1 V 2... V k = V(I 1 I 2... I k ) = V(I 1 I 2... I k ) 3

4 is also an affine algebraic variety. (ii) Let V α = V(I α ), α A be affine algebraic varieties in K n. Then ( ) V α = V I α α A is also an affine algebraic variety. (iii) Let V 1 = V(I 1 ) K m, V 2 = V(I 2 ) K n be affine algebraic varieties. Then V 1 V 2 K m K n = K m+n is also an affine algebraic variety. Proof. (i) Let s assume first that k = 2. Let P V 1 V 2 and let f I 1 I 2. If P V 1, then f(p ) = 0 because f I 1, if P V 2, then f(p ) = 0 because f I 2. Therefore in either case we have f(p ) = 0, so P V(I 1 I 2 ), V 1 V 2 V(I 1 I 2 ). We also have V(I 1 I 2 ) V(I 1 I 2 ), therefore α A V 1 V 2 V(I 1 I 2 ) V(I 1 I 2 ) (1) (I 1 I 2 I 1 I 2, since all the elements of the form i 1 i 2, i 1 I 1, i 2 I 2 are contained in both I 1 and I 2 by the definition of an ideal, and if all the generators of I 1 I 2 are elements of I 1 I 2, then necessarily I 1 I 2 I 1 I 2. This means that the elements of V(I 1 I 2 ) have to satisfy all the polynomials in I 1 I 2, therefore V(I 1 I 2 ) V(I 1 I 2 ).) Let now P K n \ (V 1 V 2 ). Then P / V 1, so there exists f 1 I 1 such that f 1 (P ) 0. Similarly, P / V 2, so there exists f 2 I 2 such that f 2 (P ) 0. Then f 1 f 2 I 1 I 2 and (f 1 f 2 )(P ) = f 1 (P )f 2 (P ) 0, therefore P / V(I 1 I 2 ). This implies V(I 1 I 2 ) V 1 V 2. By combining this with (1), we obtain V 1 V 2 = V(I 1 I 2 ) = V(I 1 I 2 ). For k > 2, use induction on k. (ii) Assume that P V α. Then P V α = V(I α ) for every α A, so α A f α (P ) = 0 for every α A and f α I α. Any f α A I α can be written as f = f α with f α I α for every α A and f α = 0 for all but finitely α A many α. Hence f(p ) = f α (P ) = ( 0 = 0, so P V I α ). This α A α A α A implies ( V α V I α ). (2) α A Assume now that P K n \ α A α A V α. Then there exists α 0 A such that P / V α0, therefore there exists f I α0 such that f(p ) 0. Now f α A I α, 4

5 therefore P / V( α A I α ). This implies that ( V α A ) I α V α. α A By combining this with (2), we obtain ( ) V I α = V α. α A α A (iii) Let x 1, x 2,..., x m be co-ordinates on K m, and y 1, y 2,..., y n coordinates on K n. Let J 1 = I 1 K[x 1, x 2,..., x m, y 1, y 2,..., y n ]. (I 1 is an ideal in K[x 1,..., x m ], J 1 is the ideal generated by the elements of I 1 in the bigger ring K[x 1, x 2,..., x m, y 1, y 2,..., y n ].) We claim that V(J 1 ) = V 1 K n. Let P = (a 1, a 2,..., a m, b 1, b 2,..., b n ) V(J 1 ). f(p ) = 0 for every f I 1, since I 1 J 1. As f only involves the variables x 1, x 2,..., x m, 0 = f(p ) = f(a 1, a 2,..., a m, b 1, b 2,..., b n ) = f(a 1, a 2,..., a m ). Hence (a 1, a 2,..., a m ) V(I 1 ) = V 1, so P V 1 K n, therefore V(J 1 ) V 1 K n. Let now P = (a 1, a 2,..., a m, b 1, b 2,..., b n ) V 1 K n. Let f J 1. f can be written as f = r f i g i, where f i I 1 and g i K[x 1, x 2,..., x m, y 1, y 2,..., y n ]. i=1 Now f i (P ) = f i (a 1, a 2,..., a m, b 1, b 2,..., b n ) = f i (a 1, a 2,..., a m ) = 0. (The second equality holds because f i only involves on the variables x 1, x 2,..., x m, and the last equality holds because P V 1 K n.) Therefore f(p ) = r f i (P )g i (P ) = 0, so P V(J 1 ). This means V 1 K n V(J 1 ), by combining i=1 this with V(J 1 ) V 1 K n proved above, we obtain V(J 1 ) = V 1 K n. Similarly, if we define J 2 = I 2 K[x 1, x 2,..., x m, y 1, y 2,..., y n ], then V(J 2 ) = K m V 2. Hence by (ii). V 1 V 2 = (V 1 K n ) (K m V 2 ) = V(J 1 ) V(J 2 ) = V(J 1 + J 2 ) In practical calculations, the ideals defining the varieties are given by sets of generators. 5

6 To find a generating set for an ideal defining the union of the varieties, take all the possible products of the generators, one factor from each ideal. To find a generating set for an ideal defining the intersection of the varieties, take the union of the generating sets of the ideals. To find a generating set for an ideal defining the Cartesian of the varieties, take the union of the generating sets of the ideals, but remember that this ideal will be in a different ring. Examples: 1. Find an ideal I R[x, y] such that V(I) = {(1, 1), (2, 3)}. Let I 1 = x 1, y 1 and I 2 = x 2, y + 3. Then V(I 1 ) = {(1, 1)} and V(I 2 ) = {(2, 3)}. By Proposition 2.3 (i), I = I 1 I 2 = (x 1)(x 2), (x 1)(y + 3), (y 1)(x 2), (y 1)(y + 3) is a suitable ideal. 2. Let V 1 = V( x x ) R 2 and let V 2 = V( y 1 y 2 ) R 2. Then V 1 V 2 = V x x 2 2 1, y 1 y 2 R 4 is a cylinder. Definition. Let K be a field and let V K n be an affine algebraic variety. W is a subvariety of V if and only if W V and W is also an affine algebraic variety. Remark. By Proposition 1.3 (i) and (ii), affine algebraic varieties behave like closed sets in a metric space. This motivates the introduction of the Zariski topology on an affine algebraic variety, in which the closed sets are the subvarieties and the open sets are the complements of subvarieties. There are, however, certain differences from metric spaces, for example, any two non-empty Zariski open sets in R n have a non-empty intersection. 6

7 1.2 Affine spaces Affine spaces are the simplest algebraic varieties, they are the solutions sets of system of linear equations, i. e., they are defined by ideals generated by degree 1 polynomials. Definition. Let K be a field, and let V be a vector space over K. U V is an affine subspace of V iff U = or U = u 0 + W = {u 0 + w w W }, where u 0 U and W is a linear subspace of V. Examples: K n as a subspace of itself,, any set containing a single point are all affine subspaces of K n. The line y = 2x + 5 is an affine subspace of R 2, in this case we can take W = span{(1, 2)} and u 0 = (0, 5). Proposition 1.4 Let U = u 0 +W be a non-empty affine subspace in a vector space V. Then U = u + W for any u U, in other words, u 0 can be chosen to be an arbitrary element of U, but W is uniquely determined by U. Proof. Let u U, then u = u 0 + (u u 0 ), so u u 0 W. We have u + W = {u + w w W } = {u 0 + (u u 0 ) + w w W }. (u u 0 ) + w W as W is a vector space, therefore u + w u 0 + W for every w W, hence u + W u 0 + W. Similarly, u 0 + W = {u 0 + w w W } = {u + (u 0 u) + w w W } u + W, therefore u + W = u 0 + W = U. W can be obtained as U u = {v u v U} for any u U. Definition. The dimension of an affine space U = u 0 + W is defined to be dim W. (Sometimes it is convenient to define dim = 1.) Definition. Let K be a field. K n, considered as an affine subspace of itself is called an n-dimensional affine space over K, and is denoted by A n (K) or A n if the field is understood. Definition. Let V 1, V 2 be vector spaces. An affine map from V 1 to V 2 is defined to be a function Φ : V 1 V 2 which can be written in the form Φ(x) = T (x) + b, where T : V 1 V 2 is a linear transformation and b V 2. It is easily checked that the composition of affine maps is also an affine map. The invertible affine maps V V form a group. Within this group the translations form a normal subgroup and the quotient by this normal subgroup is the group of invertible linear transformations on V. 7

8 Proposition 1.5 Let V 1, V 2 be vector spaces and let Φ : V 1 V 2 be an affine map. (i) For any affine subspace U 1 V 1, the image Φ(U 1 ) is also an affine subspace of V 2. (ii) For any affine subspace U 2 V 2, the preimage Φ 1 (U 2 ) is also an affine subspace of V 1. Proof. Let x 1, x 2,..., x m be the co-ordinates on V 1, y 1, y 2,..., y n the co-ordinates on V 2. Let Φ(x) = T (x) + b, where T : V 1 V 2 is a linear transformation and b = (b 1, b 2,..., b n ) T V 2 is a vector. Let {t ij } 1 i n,1 j m be the matrix of T with respect to the co-ordinates on V 1 and V 2. (i) Let U 1 V 1 be an affine subspace. If U 1 =, Φ(U 1 ) =, too. Otherwise, we can write U 1 = u 1 + W 1, where u 1 U 1 and W 1 is a linear subspace of V 1. Then Φ(U 1 ) = {Φ(u 1 + w) w W 1 } = {T (u 1 + w) + b w W 1 } = {(T (u 1 ) + b) + T (w) w W } = (T (u 1 ) + b) + T (W ) As T (W ) is a linear subspace of W 2, it follows that Φ(U 1 ) is an affine subspace, as claimed. (ii) Let now U 2 be an affine subspace of V 2. There exist degree 1 polynomials f 1, f 2,..., f k in y 1, y 2,..., y n such that U 2 = {(y 1, y 2,..., y n ) V 2 f i (y 1, y 2,..., y n ) = 0, i, 1 i k}. Then Φ((x 1, x 2,..., x m ) T ) = ( b 1 + m t 1j x j, b 2 + j=1 m t 2j x j,..., b n + j=1 therefore Φ((x 1, x 2,..., x m ) T ) U 2 if and only if f i (b 1 + m t 1j x j, b 2 + j=1 m t 2j x j,..., b n + j=1 m ) T t nj x j, j=1 m ) t nj x j = 0 for every i, 1 i k. These are linear equations in the x i, 1 i m, therefore Φ 1 (U 2 ) is an affine subspace of V 1, as claimed. Definition. Let X, Y be subsets of a vector space V. X and Y are affine equivalent if and only if there exist mutually inverse affine maps Φ, Ψ : V V such that Φ(X) = Y and Ψ(Y ) = X. It is easy to verify that affine equivalence is an equivalence relation. j=1 8

9 Examples: 1. All affine subspaces of the same dimension are affine equivalent. This can be proved by showing that any k-dimensional affine subspace is affine equivalent to the linear subspace spanned by the first k standard unit vectors. 2. Any two triangles in R 2 are affine equivalent. An easy way to prove it is to show that any triangle is affine equivalent to the triangle with vertices (0, 0), (1, 0) and (0, 1). See ~gm/teaching/math32062/triangles.pdf for an example of calculating an explicit affine equivalence between two triangles. 3. Similarly, any two parallelograms in R 2 are because all parallelograms affine equivalent to the unit square. 4. Plane conics A conic (short for a conic section) is the set of points satisfying an equation of the form ax 2 + bxy + cy 2 + dx + ey + f = 0 with at least one of a, b, c not 0. By a sequence of change of co-ordinates corresponding to affine transformations and by multiplying the equation by a non-zero scalar, every such equation can be reduced to one of finitely many standard forms, which shows that there are finitely many affine equivalence classed of conics. The tranformation of the points is given inverse of the change of co-ordinates, for example, if the new co-ordinates are x = 2x and y = y + 5, then its effect on the points is the map (x, y) (x/2, y 5). The procedure consists of four steps. The co-ordinates after the ith step will be denote by x i and y i, the coefficients by a i, b i,..., f i. Step 1. If a 0, let x 1 = x, y 1 = y. If a = 0 but c 0, let x 1 = y, y 1 = x. If a = c = 0, then let x 1 = (x + y)/2, y 1 = (x y)/2. In this case b 0 and bxy = b(x 2 1 y 2 1). We have achieved that a 1, the coefficient of x 2 1 in the new equation is not 0. If necessary, we can multiply the equation by 1 to make a 1 positive. Step 2. Complete the square with respect to x 1, let x 2 = ( a 1 x 1 + b 1 y 1 + d ) 1 2a 1 2a 1 and y 2 = y 1. After this step, the equation will be of the form x c 2 y e 2 y 2 + f 2 = 0. Step 3. If c 2 0, complete the square with respect to y 2, let x 3 = x 2 and y 3 = c 2 (y 2 +e 2 /(2c 2 )), then the equation will have the form x 2 3±y 2 3+f 3 = 0. If c 2 = 0 but e 2 0, let x 3 = x 2 and y 3 = e 2 y 2 + f 2, so the equation becomes 9

10 x y 3 = 0. If c 2 = e 2 = 0, then let x 3 = x 2 and y 3 = y 2, so the equation will have the form x f 3 = 0. Step 4. If f 3 0, then let x 4 = x 3 / f 3, y 4 = y 3 / f 3 and after the change of variable divide the whole equation by f 3. After this, the form of the equation will be the same, but f 3 will be ±1. If f 3 = 0, then let x 4 = x 3, y 4 = y 3. At the end of this step, either the equation is x y 4 = 0, or it contains x 2 4, possibly ±y 2 4 and possibly ±1 and no other term. x 2 4 y = 0 can be tranformed into x 2 4 y = 0 by swapping x 4 and y 4, so these two are affine equivalent. Therefore any variety in R 2 defined by a degree 2 equation is affine equivalent to one of the following: Case 1. x 2 + y 2 1 = 0 (ellipses) Case 2. x 2 + y 2 = 0 (single point) Case 3. x 2 + y = 0 ( ) Case 4. x 2 y 2 1 = 0 (hyperbolas) Case 5. x 2 y 2 = 0 (two intersecting lines) Case 6. x 2 + y = 0 (parabolas) Case 7. x 2 1 = 0 (two parallel lines) Case 8. x 2 = 0 ( double line ) Case 9. x = 0 ( ) It is also true, but requires a separate proof, that these cases, apart from the two empty sets are not affine equivalent. See ac.uk/~gm/teaching/math32062/conic.pdf for examples of transforming the equation of a conic to one of the above forms. Over C we can use essentially the same method to reduce the equation of a conic to one of the above forms. The difference is that we do not need to multiply the equation to ensure a 1 > 0 in Step 1 and there is no need to take the modulus of c 2 and f 3 in Steps 2 and 3, resp. If a term occurs in the final equation, it will always have coefficient +1, so we will end up with one of Cases 3, 6, 8 or 9. Over C, the change of co-ordinates x = x, y = iy converts Case 1 into Case 4 and vice versa, so all ellipses and hyperbolas are affine equivalent over C. By a similar argument, Cases 1, 3 and 4 are affine equivalent over C, similarly Cases 2 and 5, and Cases 7 and 9. 10

11 1.3 The correspondence between varieties and ideals Definition. Let X A n. The ideal of X is defined as I(X)={f K[x 1, x 2,..., x n ] f(a 1, a 2,..., a n ) = 0, (a 1, a 2,..., a n ) X}. I(X) is an indeed ideal of K[x 1, x 2,..., x n ]. It is clear from the definition that X 1 X 2 implies I(X 1 ) I(X 2 ). Proposition 1.6 (i) If J K[x 1, x 2,..., x n ], then J I(V(J)). (ii) X V(I(X)) for any X A n (K). Equality holds if and only if X is an affine algebraic variety. Proof. (i) If f J and P V(J) then f(p ) = 0 by the definition of V, therefore f I(V(J)) by the definition of I, so J I(V(J)) as claimed. (ii) It is also clear from the definitions that X V(I(X)) for any set X A n and that if equality holds then X is an affine algebraic variety. The only thing that remains to be proved that if X is an algebraic variety, then X = V(I(X)). Assume that X = V(J) for some ideal J K[x 1, x 2,..., x n ]. By (i), J I(V(J)), therefore X = V(J) V(I(V(J))) = V(I(X)). We have already seen that X V(I(X)), therefore X = V(I(X)), as required. Definition. A field K is algebraically closed if and only if every polynomial of degree at least 1 in K[x] has a root in K. Examples: C and Q (the field of algebraic numbers in C) are algebraically closed. Q, R and the finite fields are not algebraically closed. Remark. For any field K, there exists a minimal algebraically closed field containing it, which is unique up to isomorphism, this field is called the algebraic closure of K. For example, the algebraic closure of R is C and the algebraic closure of Q is Q Definition. Let R be a (commutative) ring. Let I R be an ideal. The radical of I, denoted by I or rad I, is I = {x R n such that x n I}. I is called a radical ideal if and only if I = I. Algebraic facts: For any ideal I, I is also an ideal, I I and I = I, i. e., I is a radical ideal. 11

12 Theorem 1.7 (Hilbert s Nullstellensatz) Let K be an algebraically closed field. (i) Every maximal ideal of K[x 1, x 2,..., x n ] is of the form x 1 a 1, x 2 a 2,..., x n a n, where a i K, 1 i n. (ii) If J is an ideal of K[x 1, x 2,..., x n ], J K[x 1, x 2,..., x n ], then V(J). (iii) I(V(J)) = J for any J K[x 1, x 2,..., x n ]. Proof. (i) Not proved here. Note that x 1 a 1, x 2 a 2,..., x n a n is a maximal ideal of K[x 1, x 2,..., x n ] for any field K. We can define a ring homomorphism K[x 1, x 2,..., x n ] K, f f(a 1, a 2,..., a n ), whose kernel is exactly x 1 a 1, x 2 a 2,..., x n a n. Since the image is a field, the kernel is a maximal ideal. The substance of this part is that over an algebraically closed field the converse is also true. (ii) As J K[x 1, x 2,..., x n ], there exists a maximal ideal m containing J. By (i), m = x 1 a 1, x 2 a 2,..., x n a n for some a i K, 1 i n. Therefore {(a 1, a 2,..., a n )} = V(m) V(J), so V(J). (iii) If f J, then f k J for some k. Hence f k (P ) = (f(p )) k = 0 for every P V(J), therefore f(p ) = 0 for every P V(J), i. e., f I(V(J)). This means that I I(V(J)) without assuming that K is algebraically closed. We need to prove the other inclusion, I(V(J)) I. Let f I(V(J)). Let s introduce a new variable y. Let f 1, f 2,..., f r be a set of generators for J, and let J 1 = f 1, f 2,..., f r, fy 1 K[x 1, x 2,..., x n, y]. We claim that V(J 1 ) =. Assume that P = (a 1, a 2,..., a n, b) V(J 1 ) A n+1. As f i J 1 for every i, 1 i r, we have f i (a 1, a 2,..., a n ) = 0 for every i, 1 i r, so (a 1, a 2,..., a n ) V(J). Therefore f(a 1, a 2,..., a n ) = 0, too, but then (fy 1)(P ) = 1, so P / V(J 1 ) after all. Therefore V(J 1 ) =, as claimed. By (ii), this implies that J 1 = K[x 1, x 2,..., x n, y], in particular 1 J 1. This means that we can write 1 = (fy 1)g 0 + r g i f i for suitable g 0, g 1,..., g r K[x 1, x 2,..., x n, y]. Let y N be the highest power of y occurring in any of g 0, g 1,..., g r. Let s multiply the above equality by f N. Whenever there is a power y, say y k, occurring in g i for some i, 0 i r, we can write y k f N as (fy) k f N k, so for each i, 0 i r, f N g i (x 1, x 2,..., x n, y) = G i (x 1, x 2,..., x n, fy) for a suitable polynomial G i. i=1 12

13 Therefore f N = (fy 1)G 0 (x 1, x 2,..., x n, fy) + r f i G i (x 1, x 2,..., x n, fy). i=1 By substituting y = 1/f into the above equality, we obtain f N = r f i G i (x 1, x 2,..., x n, 1). i=1 The right-hand side is an element of J, therefore f J as required. The function V from ideals of K[x 1, x 2,..., x n ] to affine algebraic varieties in A n is not injective, and the function I from subsets of A n to ideals of K[x 1, x 2,..., x n ] is neither injective nor surjective. However, the combination of Proposition 1.6 and the Nullstellensatz implies that V and I create a bijection between radical ideals of K[x 1, x 2,..., x n ] and affine algebraic varieties in A n. 13

14 1.4 Irreducibility Definition. An affine algebraic variety V is reducible if and only if it can be written as V = V 1 V 2, where V 1, V 2 are also affine algebraic varieties, V 1 V V 2. If V is not reducible, it is called irreducible. (The irreducibility of V is equivalent to saying that if V = V 1 V 2, where V 1, V 2 are also affine algebraic varieties, then V 1 = V or V 2 = V.) Examples: 1.The variety V( x 2 y 2 ) R 2 is reducible, since it is the union of V( x y ) and V( x + y ), which are the lines y = x and y = x. 2. Any variety consisting of a single point is obviously irreducible. 3. Warning: Graphs can be useful in finding the irreducible components, but they can also be misleading. The graph below shows the the curve y 2 + (x 2 4)(x 2 1) = 0. Topologically it consists of two disjoint connected components, but it is irreducible, the two parts cannot be separated algebraically Theorem 1.8 Every affine algebraic variety V can be decomposed into a union V = V 1 V 2... V k such that every V i, 1 i k, is an irreducible affine algebraic variety and V i V j for i j. The decomposition is unique up to the ordering of the components. Definition. The V i, 1 i k, in the above theorem are called the irreducible components of V. Proof. Let s call a variety good if it can be written as the union of finitely many irreducible affine algebraic varieties and bad otherwise. We shall prove first that all affine algebraic varieties are good. Assume that V is a bad affine algebraic variety. V cannot be irreducible, so V = U 1 W 1 for some proper subvarieties. Furthermore, at least one of U 1 and W 1 must also be bad, since if both of them were good, their union would also be good. We may assume that U 1 is bad. Then by a similar argument, 14

15 U 1 = U 2 W 2, where U 2 and W 2 are proper subvarieties of U 1, and at least one of them must be bad, we may assume it is U 2. Continuing like this, we obtain an infinite strictly decreasing sequence U 1 U 2 U 3... of bad varieties. U i U i+1 implies I(U i ) I(U i+1 ) by the remark after the definition of I. By Proposition 1.6 (ii), V(I(U i )) = U i and V(I(U i+1 )) = U i+1, therefore I(U i ) I(U i+1 ). Thus I(U 1 ) I(U 2 ) I(U 3 )... I(U n ) I(U n+1 )... is a strictly increasing sequence of ideals, but this contradicts the Noetherian property of the polynomial ring (Theorem 1.2), so the assumption that V is bad was incorrect. Therefore any affine algebraic variety can be written as the union of finitely many irreducible affine algebraic varieties, and by discarding varieties which are contained in others, we can write V = V 1 V 2... V k such that every V i, 1 i k, is an irreducible affine algebraic variety and V i V j for i j. This proves the existence of the decomposition. Let V = V 1 V 2... V l be another decomposition of V into irreducible varieties with V i V j for i j. Then V 1 = V 1 V = (V 1 V 1) (V 1 V 2)... (V 1 V l ). As V 1 is irreducible, there exists i such that V 1 = V 1 V i, i. e., V 1 V i. By applying this argument to V i, we obtain that there exists j such that V i V j, therefore V 1 V j. But different components do not contain each other, therefore j = 1 and V 1 = V i. By applying this to each V r (1 r k) and V j (1 r l), we obtain a bijection between {V 1, V 2,..., V k } and {V 1, V 2,..., V l }, so the decompositions are the same apart from possibly the order of the components. Definition. Let R be a commutative ring. An ideal I R is called a prime ideal if and only if ab I implies a I or b I. Proposition 1.9 An affine algebraic variety W is irreducible if and only if I(W ) is prime. Proof. We shall prove the contrapositive, namely that W is reducible if and only if I(W ) is not prime. Assume that W is reducible, then W can be written as W = W 1 W 2, where W 1 and W 2 are affine algebraic varieties and W W 1, W W 2. W 1 W implies I(W 1 ) I(W ), but V(I(W )) = W and V(I(W 1 )) = W 1 by Proposition 1.6 (ii), therefore I(W 1 ) I(W ). Let f 1 I(W 1 ) \ I(W ). Similarly, let f 2 I(W 2 ) \ I(W ). If P W 1, then (f 1 f 2 )(P ) = 0 because 15

16 f 1 (P ) = 0, if P W 2, then (f 1 f 2 )(P ) = 0 because f 2 (P ) = 0, therefore (f 1 f 2 )(P ) = 0 for every P W, so f 1 f 2 I(W ). Neither f 1, nor f 2 is an element of I(W ), therefore I(W ) is not prime. This shows that if I(W ) is prime, then W is irreducible. Assume now that I(W ) is not prime. Let f 1, f 2 be polynomials such that f 1, f 2 / I(W ), but f 1 f 2 I(W ). Let W 1 = V( I(W ), f 1 ), W 2 = V( I(W ), f 2 ). W 1, W 2 are affine algebraic varieties. Clearly W 1 W, W 2 W, but W 1 W since f 1 / I(W ) and similarly W 2 W since f 2 / I(W ). For any P W, 0 = (f 1 f 2 )(P ) = f 1 (P )(f 2 )(P ), so f 1 (P ) = 0 or f 2 (P ) = 0. In the first case, P W 1, in the second P W 2, therefore W = W 1 W 2 and W is reducible. This shows that if W is irreducible, then I(W ) is prime. Remark. The criterion involves I(W ), so if W = V(J) for some ideal J, the primality or otherwise of J is not sufficient in general to tell whether W is irreducible or not. If the field is algebraically closed, then I(V(J)) = J by the Nullstellensatz. If J is prime, then it is automatically radical, so I(V(J)) = J and V(J) is irreducible. If J is radical but not prime, then V(J) is reducible. The functions V and I create a bijection between prime ideals of K[x 1, x 2,..., x n ] and irreducible affine algebraic varieties in A n. Properties of irreducible varieties and examples 1. Irreducibility is preserved under affine equivalence. If X and Y are affine equivalent affine algebraic varieties, then X is irreducible if and only if Y is. Let Φ : X Y be an affine equivalence and assume that X can be written as X = X 1 X 2, where X 1, X 2 are also affine algebraic varieties and X 1 X X 2, then Y = Φ(X 1 ) Φ(X 2 ), Φ(X 1 ), Φ(X 2 ) are also affine algebraic varieties and Φ(X 1 ) Y Φ(X 2 ). 2. Affine subspaces over an infinite field K are irreducible. (Note that algebraically closed fields are infinite.) Every affine subspace is equivalent to one of the form x 1 = x 2 =... = x k = 0 for some k. It can be proved that I(V( x 1, x 2,..., x k )) = x 1, x 2,..., x k if K is infinite and that x 1, x 2,..., x k K[x 1, x 2,..., x n ] is a prime ideal directly from the definition. 3. Let K be an algebraically closed field. Let H A n (K) be a hypersurface, that is, an algebraic variety defined by an ideal generated by a single polynomial f K[x 1, x 2,..., x n ], H = V( f ). f can be factorised as f = f α 1 1 f α 2 2 fr αr, where the f i, 1 i r, are irreducible polynomials such that f i is not a scalar multiple of f j if i j, and α i, 1 i r, are positive integers. This factorisation is unique up to scalar factors. 16

17 We shall now show that f = f 1 f 2 f r. Let g f. g can be factorised as g = g β 1 1 g β 2 2 gs βs, where the g i, 1 i s, are irreducible polynomial such that g i is not a scalar multiple of g j if i j, and β i, 1 i s, are positive integers. g f means that there exists a positive integer m such that g m = g mβ 1 1 g mβ 2 2 gs mβs f, therefore for each i, 1 i r, f i g mβ 1 1 g mβ 2 2 gs mβs. As f i is irreducible, it must divide g j for some j, 1 j s, so f i g. The f i are pairwise coprime, therefore it follows that f 1 f 2... f r g, g f 1 f 2 f r. Conversely, if g f 1 f 2 f r, then g max{α 1,α 2,...,α r} f. This shows that I(H) = f = f 1 f 2 f r. By Proposition 1.3 (i), r i=1 V( f i ) = V f 1 f 2 f r = H. By the argument used previously, f i K[x 1, x 2,..., x n ] is a prime ideal, therefore it is radical, so I(V( f i )) = f i by the Nullstellensatz, by using the fact that f i is prime again, it follows that V( f i ) is irreducible. V( f i ) does not contain V( f j ) if i j, because f j is not a multiple of f i. This shows that the irreducible components are V( f i ), 1 i r. In particular, H is irreducible if and only if r = 1, i. e., f is the power of a single irreducible polynomial. Lemma 1.10 Let n 2 be an integer, let K be a field and let p K[x 1, x 2,..., x n ] be a polynomial which does not involve the variable x i for some i, 1 i n, and which is not a square of a polynomial K[x 1, x 2,..., x n ]. Then x 2 i p is irreducible in K[x 1, x 2,..., x n ]. Proof. We can assume for simplicity that i = n. Let s assume that x 2 n p can be factorised. By considering it as a polynomial in x n with coefficients in K[x 1, x 2,..., x n 1 ], there are two possibilities: either one factor has degree 2 in x n and the other degree 0, or both factors have degree 1. In the first case the factors are ax 2 n + bx n + c and d, where a, b, c, d K[x 1, x 2,..., x n 1 ]. From the coefficient of x 2 n we get ad = 1, therefore a and d must be constants, so we do not get a proper factorisation. In the second case the factors are ax n + b and cx n + d with a, b, c, d K[x 1, x 2,..., x n 1 ]. Similarly to the previous case, from the coefficient of x 2 n we get ac = 1, therefore a and c must be constants. By dividing the first factor by a and multiplying the second by a, we can assume that a = c = 1. The x 2 n p = (x n + b)(x n + d) = x 2 n + (b + d)x + bd, so by comparing the coefficient of x n and the degree 0 term in x n we get b + d = 0 and p = bd. Hence d = b, so p = b 2, p = b 2, but p is not a square, this is a contradiction, therefore x 2 n p is irreducible as claimed.. 17

18 This simple lemma has lots of useful consequences. 4. Ellipses, hyperbolas, parabolas are irreducible over C and R. x 2 + y 2 1 = x 2 (1 y 2 ) and 1 y 2 is not a square, therefore x 2 + y 2 1 is irreducible in C[x, y] by Lemma 1.10 and x 2 + y 2 1 C[x, y] is a prime ideal. Prime ideals are radical, so by the Nullstellensatz (Theorem 1.7), I(V( x 2 + y 2 1 )) = x 2 + y 2 1. By Proposition 1.9, V( x 2 + y 2 1 ) is irreducible over C as x 2 + y 2 1 C[x, y] is a prime ideal. It can also be proved that I(V( x 2 + y 2 1 )) = x 2 + y 2 1 over R, too. The main idea is that if x 2 + y 2 1 does not divide f R[x, y], then f(x, y) = x 2 + y 2 1 = 0 only has finitely many solutions, but the unit circle x 2 + y 2 1 = 0 has infinitely many points over R, therefore f / I(V( x 2 + y 2 1 )). The rest of the proof is the same as over C. Any ellipse is affine equivalent to V( x 2 +y 2 1 ) over either C or R, therefore all ellipses are irreducible. The proof for hyperbolas and parabolas is similar. 5. If p(x) C[x] is a polynomial of odd degree, then V( y 2 p(x) ) is irreducible over C, as p(x) cannot be a square. This includes as special cases the nodal cubic curve, V( x 3 + x 2 y 2 ), the cuspidal cubic curve V( x 3 y 2 ) and elliptic curves (to be studied later in the course) Nodal cubic 4 Cuspidal cubic 6. Let C = V( x 2 y 2 +y 4 x 2 2y 2 +1 ) A 2 (C). This polynomial factorises 18

19 as x 2 y 2 + y 4 x 2 2y = (y 2 1) 2 + x 2 (y 2 1) = (x 2 + y 2 1)(y 2 1) = (x 2 + y 2 1)(y 1)(y + 1) y 1 and y + 1 have degree 1, so they are irreducible. We have already showed that x 2 + y 2 1 is irreducible. Therefore the irreducible components of C are the lines V( y 1 ), V( y + 1 ) and V( x 2 + y 2 1 ) Warning. The correspondence between the irreducible factors of the generator and the irreducible components described above only applies to hypersurfaces. An ideal generated by irreducible polynomials need not be prime. Factorisation is still a useful tool in decomposition of varieties into irreducible factors, but all elements of the ideal have to be considered, not just the generators. 7. The unit sphere V( x 2 + y 2 + z 2 1 ) A 3 (C) is irreducible, since x 2 + y 2 + z 2 1 = x 2 (1 y 2 z 2 ) and 1 y 2 z 2 is not a square, therefore x 2 + y 2 + z 2 1 is irreducible in C[x, y, z]. It can be proved that I(V( x 2 + y 2 + z 2 1 )) = x 2 + y 2 + z 2 1 over R similarly to the case of the circle, therefore the unit sphere is also irreducible over R. Any ellipsoid is affine equivalent to the unit sphere, therefore ellipsoids are irreducible over R or C. 8. Let I = x 2 + y 2 + z 2 1, 3x 2 + y 2 z 2 1 C[x, y, z] and let W = V(I) A 3 be the variety defined by I. The graph below shows this variety, the two surfaces are the sphere defined the equation x 2 + y 2 + z 2 1 = 0 and the hyperboloid of one sheet defined by 3x 2 + y 2 z 2 1 = 0, W is their intersection, the black curve. 19

20 The generators themselves are irreducible, but x 2 z 2 = (3x2 + y 2 z 2 1) (x 2 + y 2 + z 2 1) 2 I, too, and x 2 z 2 = (x + z)(x z). Therefore by the argument of Proposition 1.9, W can be decomposed into the union of W 1 = V( x 2 + y 2 + z 2 1, 3x 2 + y 2 z 2 1, x + z ) and W 2 = V( x 2 + y 2 + z 2 1, 3x 2 + y 2 z 2 1, x z ). We shall prove that these varieties are irreducible. Let s consider W 1 = V( x 2 +y 2 +z 2 1, 3x 2 +y 2 z 2 1, x+z ). The difference of x 2 +y 2 +z 2 1 and 3x 2 +y 2 z 2 1 is divisible by x+z so we can omit one of them from the set of generators, therefore W 1 = V( x 2 +y 2 +z 2 1, x+z ). The graph below show the intersection of the plane x+z = 0 with the sphere and the hyperboloid. 20

21 The plane x + z = 0 intersects both in the same curve, shown in black, this is W 1. The intersection of a sphere with a plane is a circle. By rotating W 1 through π/4 about the y-axis, we can transform it to the unit circle x 2 + y 2 1 = 0 in the xy-plane. We have already proved that the unit circle is irreducible, the rotation is a Euclidean transformation, which is a special case of an affine equivalence, therefore W 1 is irreducible. The irreducibility of W 2 = V( x 2 + y 2 + z 2 1, 3x 2 + y 2 z 2 1, x z ) = V( x 2 + y 2 + z 2 1, x z ) can be proved similarly. It is also clear that neither of W 1 and W 2 contains the other, for example, because (1/ 2, 0, 1/ 2) W 1, but (1/ 2, 0, 1/ 2) / W 2, while (1/ 2, 0, 1/ 2) W 2, but (1/ 2, 0, 1/ 2) / W 1. Therefore they are the irreducible components of W. 21

22 2 Morphisms, co-ordinate rings, rational maps, and function fields 2.1 Morphisms and co-ordinate rings The obvious functions to consider on an affine algebraic variety V A n (K) are the polynomials. Two polynomials f, g define the same function V K if and only if f g I(V ), this motivates the following definition. Definition. Let V A n (K) be an affine algebraic variety. Its co-ordinate ring, denoted by K[V ], is defined to be K[V ] = K[x 1, x 2,..., x n ]/I(V ). Remark. K means the field over which the variety is defined, so if the field is C, then the notation is C[V ], if the field is called L, then L[V ]. Definition. Let V A m (K) and W A n (K) be affine algebraic varieties. A morphism ϕ : V W is a function of the form ϕ(p ) = (ϕ 1 (P ), ϕ 2 (P ),..., ϕ n (P )) with ϕ i K[V ] for each i, 1 i n. Elements of K[V ] can be considered as morphisms V A 1. The composition of two morphisms ϕ : V W and ψ : W X is a morphism ψ ϕ : V X. Definition. The morphism ϕ : V W is called an isomorphism if and only if it has an inverse morphism, i. e., if there exists a morphism ψ : W V such that ψ ϕ = id V and ϕ ψ = id W. V and W are called isomorphic if and only if there exists an isomorphism between them. Isomorphic affine algebraic varieties are the same for most purposes. One of the goals of algebraic geometry is to classify algebraic varieties up to isomorphism and to decide whether two varieties are isomorphic. One of the tools is to find properties of algebraic varieties which are preserved under isomorphism. Examples: 1. Any affine map is a morphism. Invertible affine maps are isomorphisms. Affine equivalent affine algebraic varieties are isomorphic. 2. Let K be an arbitrary infinite field. Let t be the co-ordinate on A 1 and x, y the co-ordinates on A 2. Let V = A 1 and W A 2 be the parabola defined by the equation y x 2 = 0. Then I(V ) = {0}, I(W ) = y x 2 (this is where we need K to be infinite), so K[V ] = K[t] and K[W ] = K[x, y]/ y x 2. 22

23 ϕ : V W, t (t, t 2 ) is a morphism. t and t 2 are elements of K[V ] and x = t, y = t 2 satisfy y x 2 = 0 for any t. (This shows that ϕ(t) W for every t V = A 1, so ϕ is a morphism V W, not just V A 2.) ψ : W V, (x, y) x is also a morphism. x K[W ] (technically it should be x + I(W ), but this distinction is usually not made in practice) and there are no equations to check. (ψ ϕ)(t) = ψ(t, t 2 ) = t and (ϕ ψ)(x, y) = ϕ(x) = (x, x 2 ) = (x, y) for (x, y) W, therefore ϕ and ψ are inverses of each other, so V and W are isomorphic. 3. Let K be an infinite field of characteristic other than 2, for example K = R or C. Let V = A 1 and let W A 2 be the nodal cubic curve defined by the equation x 3 + x 2 y 2 = ϕ : V W, t (t 2 1, t(t 2 1)) is a morphism. t 2 1 and t(t 2 1) are elements of K[V ] and x = t 2 1, y = t(t 2 1) satisfy x 3 + x 2 y 2 = 0 for any t, so ϕ is a function V W. ϕ is not an isomorphism, since it is not injective, ϕ(1) = ϕ( 1) = (0, 0). (We need the characteristic to be different from 2 to ensure that 1 1.) In examples 4 5, K can be an arbitrary infinite field. 4. Let H A 2 be the hyperbola defined by xy 1 = 0, and let ϕ : H A 1 be the morphism (x, y) x. ϕ is not an isomorphism, since it is not surjective because 0 / im ϕ. Warning: The image of a morphism is not necessarily an algebraic variety. In this example im ϕ = A 1 \{0}, and this set is not a variety, since any proper subvariety of A 1 is finite. 5. Let V = A 1 and let W A 2 be the cuspidal cubic curve defined by the equation y 2 x 3 = 0. ϕ : V W, t (t 2, t 3 ) is a morphism. t 2 and t 3 are 23

24 elements of K[V ] and x = t 2, y = t 3 satisfy y 2 x 3 = 0 for any t, so ϕ is a function V W ϕ is bijective as a function, but it is not { an isomorphim. It has an inverse y/x, if x 0 as a function ψ : W V, ψ(x, y) =, but this does not 0, if x = 0 appear to be a morphism from that way it is defined. Let s assume that ϑ : W V is a morphism, which is the inverse of ϕ. The ϑ is represented by a polynomial f K[x, y] such that f(t 2, t 3 ) = t, but this is impossible. We shall prove later (see the example after Corollary 2.2) that V and W are not even isomorphic, not just that this particular morphism is not an isomorphism. Warning. Any isomorphism is bijective, since it has an inverse as a function, but as the above example shows, a bijective morphism is not necessarily an isomorphism. 6. Let x, y be the co-ordinates on A 2 (C) and u, v, w the co-ordinates on A 3 (C). Let V = A 2 and W A 3 be the variety defined by the equation uw v 2 = 0, it is the double cone shown below. 24

25 ϕ : V W, (x, y) (x 2, xy, y 2 )) is a morphism. x 2, xy, y 2 are elements of K[V ] and u = x 2, v = xy and w = y 2 satisfy uw v 2 = x 2 y 2 (xy) 2 = 0 for all x, y, therefore ϕ(v ) W, so ϕ is indeed a morphism V W, not just V A 3. ϕ is not an isomorphism because it is not injective, ϕ(x, y) = ϕ( x, y) for any (x, y) A 2, but (x, y) ( x, y) unless (x, y) = (0, 0). (This argument works as long as the characteristic of the field is not 2. In characteristic 2, (x, y) = ( x, y) and ϕ is bijective if the field is algebraically closed, but it is still not an isomorphism.) Theorem 2.1 Let V A m (K) and W A n (K) be affine algebraic varieties. Let y 1, y 2,..., y n be co-ordinates on A n. For a morphism ϕ : V W, define ϕ : K[W ] K[V ], ϕ (f) = f ϕ. : ϕ ϕ is a bijection between morphisms of affine algebraic varieties ϕ : V W and ring homomorphisms α : K[W ] K[V ] preserving K. Proof. Let f K[W ]. f = F + I(W ) for some F K[y 1, y 2,..., y n ]. Now ϕ (f) = f ϕ = F ϕ = F (ϕ 1, ϕ 2,..., ϕ n ), where ϕ 1, ϕ 2,..., ϕ n K[V ] are the components of ϕ. As F is a polynomial, ϕ (f) = F (ϕ 1, ϕ 2,..., ϕ n ) K[V ], too. If G K[y 1, y 2,..., y n ] is another polynomial such that f = G + I(W ), then F G I(W ), so F (ϕ(p )) = G(ϕ(P )) for every P V and therefore ϕ (f) is a well-defined element of K[V ]. This shows that ϕ is indeed a function K[W ] K[V ]. ϕ (λ) = λ for any λ K and ϕ (f +g) = ϕ (f)+ϕ (g), ϕ (fg) = ϕ (f)ϕ (g) for all f, g K[W ] by the definition of ϕ, therefore ϕ is a homomorphism of rings which preserves K. Given a homomorphism α : K[W ] K[V ] with α(λ) = λ for every λ K, 25

26 let ϕ i = α(y i + I(W )) K[V ] for i = 1, 2,..., n. Then ϕ = (ϕ 1, ϕ 2,..., ϕ n ) is clearly a morphism V A n. We need to prove that ϕ(v ) W so that ϕ is a morphism V W. Let P V an arbitrary point. In order to prove ϕ(p ) W, we need to show g(ϕ(p )) = 0 for any g I(W ). Let q : K[y 1, y 2,..., y n ] K[W ] = K[y 1, y 2,..., y n ]/I(W ) be the quotient map q(f) = f + I(W ). Then ϕ i = α(q(y i )), so g(ϕ(p )) = g(α(q(y 1 ))(P ), α(q(y 2 ))(P ),... α(q(y n ))(P )). g is a polynomial, composed of its variables, scalars in K, addition and multiplication, while α and q are ring homomorphisms preserving K, therefore g(α(q(y 1 )), α(q(y 2 )),..., α(q(y n ))) = α(q(g(y 1, y 2,..., y n ))) = α(0) = 0 since g I(W ) = ker q. If we substitute the co-ordinates of P, we obtain g(ϕ(p )) = 0. As this holds for every g I(W ), we have ϕ(p ) W, so ϕ is indeed a morphism V W. If ϕ = (ϕ 1, ϕ 2,..., ϕ n ) is a morphism of algebraic varieties V W, then ϕ i = ϕ (y i + I(W )), so the ring homomorphism constructed from ϕ is exactly ϕ. Conversely, if α is a ring homomorphism α : K[W ] K[V ] and ϕ : V W is the morphism of algebraic varieties constructed from α, then α = ϕ. This shows constructions associating a ring homomorphism to a homomorphism of algebraic varieties and a homomorphism of algebraic varieties to a ring homomorphism are inverses of each other, therefore : ϕ ϕ is a bijection. Note. ϕ ϕ is contravariant, i. e., if ϕ : V W and ψ : W X are morphisms, then (ψ ϕ) = ϕ ψ : K[X] K[V ]. The definition of the homomorphism ϕ may seem complicated, but it can be written down very easily. In Example 6 before the theorem, V = A 2, W = V( uw v 2 ) A 3 and ϕ : V W is the morphism ϕ(x, y) = (x 2, xy, y 2 ). ϕ : K[W ] K[V ] is defined by ϕ (u + I(W )) = x 2, ϕ (v + I(W )) = xy, ϕ (w + I(W )) = y 2 and since it is ring homomorphism that preserves K, ϕ (f) can be calculated for any f K[W ]. Corollary 2.2 ϕ : V W is an isomorphism of affine algebraic varieties if and only if ϕ : K[W ] K[V ] is an isomorphism of rings. V and W are isomorphic if and only if there is an isomorphism between K[V ] and K[W ] which preserves K. 26

27 Examples: In these examples K can be an arbitrary infinite field. 1. Let H = V( xy 1 ) A 2 be a hyperbola, we shall prove that it is not isomorphic to A 1. It can be proved by direct calculation that I(H) = xy 1. (Over an algebraically closed field this follows from the Nullstellensatz, as xy 1 is an irreducible polynomial.) In K[H], (x + I(H))(y + I(H)) = 1 + I(H), i. e., there exist invertible elements which are not scalars, whereas in K[A 1 ] the only invertible elements are the scalars. This shows that K[H] and K[A 1 ] are not isomorphic, therefore H is not isomorphic to A 1 either. 2. We shall show that the cuspidal cubic (Example 5 before Theorem 2.1) is not isomorphic to A 1. Let t be the co-ordinate on A 1 and x, y the co-ordinates on A 2. Let V = A 1 and let W A 2 be the cuspidal cubic curve defined by y 2 x 3 = 0 and let ϕ : V W be the morphism t (t 2, t 3 ). We already proved that ϕ is bijective, but it is not an isomorphism. Now we shall prove that W is not isomorphic to V by showing that K[W ] is not isomorphic to K[V ] = K[t]. It can be proved by direct calculation that I(W ) = x 3 y 2. (Over an algebraically closed field this follows from the Nullstellensatz, as x 3 y 2 is an irreducible polynomial.) In order to simplify notation, let f = f + I(W ) for f K[x, y]. ϕ (x) = t 2 and ϕ (y) = t 3, so ϕ (x a y b ) = t 2a+3b. This way we can obtain all nonnegative powers of t except t itself. As the x a y b (a, b Z 0 ) generate K[W ] as a vector space over K, the image of ϕ is the subring S of K[t] consisting of polynomials with no degree 1 term. This gives another proof of the fact ϕ is not an isomorphism, since ϕ is not an isomorphism, because it is not surjective. We shall prove that K[W ] is not isomorphic to K[V ] = K[t] by showing that K[W ] cannot be generated by a single element and K as a ring. If there existed an element in z K[W ] which together with K generated K[W ] as a ring, then ϕ (z) and K would generate S, the image of K[W ] under ϕ. (In fact, ϕ is an isomorphism between K[W ] and S.) We shall prove that S cannot be generated by a single element and K as a ring. Let s assume to the contrary that there exists such an element u S. We can write u as u = a 0 + a 2 t 2 + a 3 t a k t k for some k 0 and for suitable coefficients a 0, a 2, a 3,..., a k K. u and u a 0 generate the same ring, so we may assume a 0 = 0, so u = a 2 t 2 + a 3 t a k t k. The elements of the ring generated by u and K are of the form b 0 + b 1 u + b 2 u b n u n, for some n 0 and b 0, b 1, b 2,..., b n K. If S is generated by K and u, then all elements of S, in particular t 2 and t 3 27

28 haver to be able to written in this form with suitable coefficients b 0, b 1, b 2,..., b n. The degree 2 term in b 0 + b 1 u + b 2 u b n u n is b 1 a 2 t 2, the degree 3 term is b 1 a 3 t 3. In order to be able to get t 2 with a non-zero coefficient, we need to have a 2 0. Similarly, in order to get t 3 with a non-zero coefficient, we need a 3 0. This, however, means that if b 1 = 0 then both t 2 and t 3 have coefficient 0 in b 0 + b 1 u + b 2 u b n u n, while if b 1 0 then both t 2 and t 3 have a non-zero coefficient, we are not able to obtain just t 2 or just t 3. This contradiction shows that S cannot be generated by u and K. K[t] = K[V ] is clearly generated by t and K, so we can conclude that K[V ] = K[W ] and therefore V and W are not isomorphic either. The correspondence between varieties and ideal can be re-written in terms of varieties and co-ordinate rings. Definition. An element r in a ring is called nilpotent if and only if there exists a positive integer n such that r n = 0. Algebraic fact: An ideal I in a ring R is radical if and only if R/I has no nilpotent elements other than 0. Co-ordinate rings of affine algebraic varieties over K are rings which contain K, have no nilpotent elements other than 0 and are generated as rings by K and finitely many other elements (the images of the co-ordinate functions). Conversely, if we have such a ring R, we can take choose a set of elements x 1, x 2,..., x n / K which together with K generate R. The fact that they and K generate R means that there is a surjective ring homomorphism K[x 1, x 2,..., x n ] R, let J be its kernel, then R = K[x 1, x 2,..., x n ]/J by the First Ring Isomorphism Theorem. I is radical since quotient ring K[x 1, x 2,..., x n ]/J has no nilpotent elements other than 0. If K is an algebraically closed field, then J = I(V(J)) by the Nullstellensatz, so R = K[V(J)]. If K is algebraically closed, then V K[V ] gives a bijection between isomorphism classes of affine algebraic varieties and isomorphism classes of rings which which contain K, have no nilpotent elements other than 0 and which can be generated by K and finitely many other elements. An affine algebraic variety V is irreducible if and only if I(V ) is a prime ideal, which is equivalent to K[V ] being an integral domain, therefore in the above correspondence irreducible varieties correspond to integral domains which contain K and which can be generated by K and finitely many other elements. 28

29 If a property of affine algebraic varieties can be expressed purely in terms of algebraic properties of co-ordinate rings, then that property is preserved under isomorphism. Being an integral domain an algebraic property preserved by isomorphism, therefore irreducibility is preserved by isomorphism of affine algebraic varieties. 29

30 2.2 Rational functions and maps, function fields Definition. Let V be an irreducible affine algebraic variety. The function field of V, denoted by K(V ), is the set of fractions f/g f, g K[V ], g 0, considering two such fractions f 1 /g 1 and f 2 /g 2 equal if and only if f 1 g 2 = f 2 g 1 in K[V ]. (K(V ) is indeed a field with the natural operations.) The elements of K(V ) are called rational functions on V. Just like in the case of co-ordinate rings, K means the field over which the variety is defined, so if the field is C, then the notation is C(V ). A rational function ϕ on V is defined at P V if and only if it can be represented in the form f/g such that g(p ) 0, and in that case the value ϕ(p ) is defined to be ϕ(p ) = f(p )/g(p ), and this value is independent of the choice of f and g. Example: If V = A n, then K(A n ) = K(x 1, x 2,..., x n ), the field of rational functions in n variables. Lemma 2.3 Let V be an irreducible affine algebraic variety. Let ϕ = f/g f, g K[V ], g 0. If for some point P V, f(p ) 0 and g(p ) = 0, then ϕ is not defined at P. Proof. Let ϕ = f 1 /g 1 f 1, g 1 K[V ], g 1 0 be another representation of ϕ. By the definition of equality of rational functions, f 1 g = fg 1 in K[V ], so f 1 (P )g(p ) = f(p )g 1 (P ). g(p ) = 0, so the left-hand side is 0, but then f(p ) 0 implies that on the right-hand side g 1 (P ) = 0, because K[V ] is an integral domain. Remarks. 1. In general, all possible representations of a rational function ϕ need to be considered when deciding whether it is defined at a point. For example, let x, y, z, w be the co-ordinates on A 4. Let V = V( xy zw ) A 4, and let ϕ(x, y, z, w) = x/z. x/z not defined at P = (0, 1, 0, 0), but x/z = w/y, since xy = zw K[V ]. w/y is defined at P, so ϕ(p ) = 0/1 = However, if K[V ] is a unique factorisation domain, then ϕ can be written as ϕ = f/g with f, g coprime. In this case it is sufficient to consider only this one representation of ϕ, because all others are of the form ϕ = (fh)/(gh) for some h K[V ], therefore ϕ is defined at a point P if and only if g(p ) 0. In particular, this applies if V = A n for some n. Definition. Let V A m (K) and W A n (K) be affine algebraic varieties and let V be irreducible. A rational map ϕ : V W is a function defined on a non-empty subset of V given by ϕ(p ) = (ϕ 1 (P ), ϕ 2 (P ),..., ϕ n (P )) with ϕ i K(V ) for each i, 1 i n, and such that if ϕ(p ) is defined at P, i. e., 30

31 if ϕ i (P ) is defined for each i, 1 i n, then ϕ(p ) W. Rational functions and rational maps are generally not functions on V, only partial functions defined on a non-empty subset of V. The set where a rational function or a rational map is not defined is a subvariety, the locus where the denominators of all the representations of the rational function or of the components of the rational map vanish. The set where a rational function or a rational map is defined is the complement of the previous set, therefore it is a Zariski open subset. Definition. A rational map ϕ : V W is called dominant if and only if its image is not contained in any proper subvariety of W. Remark. Let V, W be an irreducible affine algebraic varieties, and let ϕ : V W and ψ : W X be rational maps. If the composite ψ ϕ : V X is defined, it is also a rational map. If ϕ is dominant, ψ ϕ is always defined. Definition. A rational map ϕ : V W is called a birational equivalence if and only if there exists a rational map ψ : W V such that ψ ϕ = id V and ϕ ψ = id W. (In this case ϕ and ψ are necessarily dominant.) V and W are birationally equivalent if and only if there exists a birational equivalence between them. A variety is rational if and only if it is birationally equivalent to A n for some n. Theorem 2.4 Let V A m (K) and W A n (K) be irreducible affine algebraic varieties. There exists a bijection between dominant rational maps ϕ : V W and field homomorphisms α : K(W ) K(V ) preserving K given by the following constructions: for a rational map ϕ : V W, define ϕ : K(W ) K(V ), ϕ (f) = f ϕ, and for a homomorphism α : K(W ) K(V ), define a rational map by (α(y 1 ), α(y 2 ),..., α(y n )), where y 1, y 2,..., y n are the co-ordinates on A n. Sketch of Proof: (Not examinable) We introduce an alternative view of rational maps. Let g K[V ], g 0. Let x 1, x 2,..., x m be the co-ordinates on A m. Let us consider A m+1 = A m A 1 with a new co-ordinate z on A 1 and let V g = V (I(V ), zg 1) A m+1. (V g is the graph of z = 1/g over V.) Let K[V ][1/g] be the subring of K(V ) generated by K[V ] and 1/g, it consists of the elements of K(V ) that can be written as f/g k for some f K[V ] and some non-negative integer k. There is an isomorphism between ι g : K[V ][1/g] K[V g ] given by ι g (f/g k ) = fz k. β g : V V g, (x 1, x 2,... x m ) (x 1, x 2,... x m, 1/g(x 1, x 2,..., x m )) is a rational map, while its inverse γ g : V g V, (x 1, x 2,... x m, z) (x 1, x 2,... x m ) is a morphism. For ρ K[V ][1/g], ι g (ρ) = ρ γ g and for σ K[V g ], ι 1 g (σ) = ρ β g. β g and 31

32 γ g give a bijection between the Zariski open subset {P V g(p ) 0} in V and V g. Let now ϕ : V W be a dominant rational map. We can assume that ϕ = (f 1 /g, f 2 /g,..., f n /g) for some g K[V ], if necessary we can multiply the numerator and denominator of each component by a suitable factor to get ϕ into this form. Now ϕ γ g : V g W is a rational map which is defined at every point of V g, therefore it is a morphism (problem sheet 4, question 5). By Theorem 2.1, there exists a corresponding homomorphism of rings (ϕ γ g ) : K[W ] K[V g ]. ϕ γ g is dominant as ϕ and γ are, therefore (ϕ γ g ) is injective (problem sheet 4, question 4 (a)). By composing it with the isomorphism ι 1 g, we obtain an injective morphism ι 1 g (ϕ γ g ) : K[W ] K[V ][1/g]. As it is injective, it can be extended to a morphism of function fields ϕ : K(W ) K(V ), and it can be checked that it has the property ϕ (y i ) = f i /g = y i ϕ for each i, 1 i n, therefore ϕ (f) = f ϕ for any f K(W ). Conversely, let α : K(W ) K(V ) be a morphism of fields preserving K. Let g K[V ] be such that α(y i ) = f i /g for for a suitable f i K[V ] for each i, 1 i n. Then ϕ(k[w ]) K[V ][1/g], so by Theorem 2.8, ι g α : K[W ] K[V g ] determines a morphism ψ : V g W. Moreover, ψ is dominant since ι g α is injective (problem sheet 4, question 3 (a)), because α is injective, as it is a homorphism of fields. Therefore ϕ = ψ β g is a rational map V W and ϕ = α. Corollary 2.5 ϕ : V W is a birational equvalence if and only if ϕ : K(W ) K(V ) is an isomorphism of fields. V and W are birationally equivalent if and only if K(V ) and K(W ) are isomorphic as extensions of K. A variety V is rational if and only if K(V ) = K(t 1, t 2,..., t k ) for some k. Birational equivalence is an equivalence relation on affine algebraic varieties, which is weaker than isomorphism. Birational equivalence classes of varieties are in bijection with the isomorphism classes of finitely generated extensions of K. Birational equivalence of V and W means that there exist g K[V ] and h K[W ] such that V g = Wh, so in a sense, V and W have isomorphic non-empty Zariski open subsets. Examples: Worked examples can be found in the separate handout at http: // 32

33 3 Tangent spaces, dimension and singularities Let H A n (K) be a hypersurface defined by the equation f(x 1, x 2,..., x n ) = 0, where f K[x 1, x 2,..., x n ], and and let P H. One of the possible ways of defining that the line {P + tv t K} (v = (v 1, v 2,..., v n ) K n \ {0}) is tangent to H at P is to require that f(p + tv), now simply a function of t, has a stationary point at t = 0. The Taylor expansion of f(p + tv) about t = 0 is d f(p + tv) f(p + tv) = f(p ) + t d t P +(terms of order 2 in t) n f = f(p ) + t P v j + (terms of order 2 in t). x j j=1 (Derivatives of polynomials can be defined purely formally over any field without using the concept of limits and Taylor expansion works for polynomials over any field.) n f Thus f(p +tv) has a stationary point at t = 0 if and only if By using the gradient f = 0 can be re-written as ( f) P.v = 0. j=1 ( f, f,, f ), the condition x 1 x 2 x n P v j = 0. x j n j=1 f x j P v j = Definition. Let V A n (K) be an affine algebraic variety and let P V. A line l given in parametric form as {P + tv t K}, (v K n \ {0}), is said to be tangent to V at P if and only if ( f) P.v = 0 for every f I(V ). Definition. The tangent space to V at P, denoted by T P V is the union of the tangent lines to V at P and the point P. Theorem 3.1 Let V A n be an affine algebraic variety. Let f 1, f 2,..., f r be a set of generators for I(V ). Let P V and let v = (v 1, v 2,..., v n ) K n \ {0}. The line l = {P + tv t K} is tangent to V at P if and only if 33

34 J P v = 0, where J is the Jacobian matrix, f 1 f 1 x 1 x 2 f 2 f 2 J = x 1 x f r f r x 1 x 2 f 1 x n f 2 x n. f r x n and J P is J evaluated at P. In particular, T P V is an affine subspace of dimension n rankj P. Proof. Assume that l is tangent to V at P. Then ( f) P.v = 0 for every f I(V ), in particular, ( f i ) P.v = 0 for every i, 1 i r. ( f i ) P.v is exactly the ith component of J P v, therefore J p v = 0. Assume now that J P v = 0. Let f I(V ). Then f = r g i f i for some g i K[x 1, x 2,..., x n ], 1 i r. Now i=1 f = r (g i f i ) = i=1 r (( g i )f i + g i f i ) i=1 and so ( f) P = r (( g i ) P f i (P ) + g i (P )( f i ) P ) = i=1 r g i (P )( f i ) P. i=1 i=1 In the last step we used that f i (P ) = 0 for every i, 1 i r. Hence ( f) P.v = r g i (P )( f i ) P.v, but ( f i ) P.v = 0, because it is the ith component of J P v = 0, therefore ( f) P.v = 0. The set W = {v K n J P v = 0} is a linear subspace, therefore T P V = P +W is an affine subspace. Its dimension is dim T P V = dim W = n rankj P by the Rank-Nullity Formula. Example: Let V A 2 (C) be the cuspidal cubic curve defined by the equation y 2 x 3 = 0. It can be checked that y 2 x 3 is irreducible, so y 2 x 3 is a prime ideal, therefore it is radical and then the Nullstellensatz implies that that I(V ) = y 2 x 3. Therefore J = ( 3x 3 2y). 34

35 At P = (1, 1), J P = ( 3 2), therefore the line {P + tv t C} is tangent to V at (1, 1) if and only if 3v 1 + 2v 2 = 0. Hence T (1,1) V is the line of slope 3/2 through P with equation y = 3x/2 1/2. It agrees with the tangent line obtained by, for example, implicit differentiation. At Q = (0, 0), however, something strange happens. J Q = (0, 0), therefore every line through (0, 0) is a tangent line and T (0,0) V = A 2! Let v = (α, β) (0, 0) be the direction vector of a line through Q = (0, 0). Then Q + tv = (tα, tβ) and if we substitute x = tα, y = tβ into y 2 x 3 we obtain t 2 β 2 t 3 α 3 = t 2 (β 2 tα 3 ), which has a zero of multiplicity at least 2 at t = 0, so any line through (0, 0) is a tangent line by our definition. If β = 0, then t 2 (β 2 tα 3 ) has a zero of multiplicity 3 at t = 0, so the line y = 0 has a higher order tangency to the curve than a general line through (0, 0). Proposition Definition 3.2 Let V A n be an irreducible affine algebraic variety. If V, there exists a proper subvariety Z of V (Z V, but Z may be ) and an integer d 0 such that dim T P V = d for every P V \ Z and dim T P V > d for P Z. (In other words, dim T P V is constant on a nonempty Zariski open subset of V.) The dimension of V, denoted by dim V, is defined to be d. The points of V \Z, where dim T P V = dim V, are called nonsingular, while the points of Z, where dim T P V > dim V, are called singular and the set of singular points of V is denoted by Sing V. (Sometimes it is convenient to define dim = 1 and Sing =.) Proof. Let V m = {P V dim T P V m}, where m 0 is an integer. By Theorem 3.1, V m = {P V rank J P n m}. It is a theorem in linear algebra that a matrix has rank k if and only if all of its (k + 1) (k + 1) 35

36 minors vanish. Therefore V m = {P V all (n m + 1) (n m + 1) minors of J P vanish}. This shows that V m is the affine algebraic variety defined by the ideal generated by I(V ) and by the (n m + 1) (n m + 1) minors of J. We have V = V 0 V 1 V 2... V n V n+1 =. Choose d maximal such that V d = V. Then dim T P V = d for P V \V d+1 and dim T P V > d for P V d+1, so dim V = d and Sing V = V d+1. Example: If V is an affine subspace of A n, T P V = V at every point P V, so the dimension of V as a variety agrees with its dimension as an affine space and it has no singular points. In particular, A n is non-singular and has dimension n. Worked examples can be found in the separate handout at maths.manchester.ac.uk/~gm/teaching/math32062/singularities.pdf. Herwig Hauser s gallery at bildergalerie/gallery.html has many nice examples of singular surfaces. The combination of the preceding results gives the following procedure for finding the dimension and singular points of an irreducible variety V A n (K). 1. Choose a set of generators f 1, f 2,..., f r for I(V ). (In all the problems you will have to solve, the field will be algebraically closed and the defining ideal of V will be a radical ideal, so it will be equal to I(V ).) 2. Calculate the Jacobian matrix J. 3. For k = 0, 1, 2,..., find the points where all the (k + 1) (k + 1) minors of J vanish and also f 1 = f 2 =... = f r = 0. These are the points where rank J P k and therefore dim T P V n k. Do this until you find the smallest k such that dim T P V n k for every point P V. 4. Then dim V = n k and Sing V is the set of points where dim T P V > n k, i. e., the points of V where all the k k minors of J vanish. Definition. Let V be an arbitratry (not necessarily irreducible) affine algebraic variety. The dimension of V, dim V, is the maximum of the dimensions of the irreducible components of V. The local dimension of V at P V is defined to be the maximum of the dimension of the irreducible components of V containing P. P is called non-singular if and only if dim T P V is equal to the local dimension of V at P. P is called singular if and only if dim T P V is greater than the local dimension of V at P. The set of singular points of V is denoted by Sing V, just as in the irreducible case. 36

37 Fact: Sing V consists of the singular points of the individual irreducible components and of intersection points of different components. This means that in general, one needs to decompose the variety into its irreducible components in order to find the its dimension and its singular points, however, by the proposition below and by problem sheet 5, question 3, hypersurfaces over an algebraically closed field are an exception. Proposition 3.3 Let K be an algebraically closed field. Let H A n (K) be a hypersurface, that is, H = V( f ) for some non-constant polynomial f K[x 1, x 2,..., x n ], then dim H = n 1. Proof. f can be factorised as f = f α 1 1 f α 2 2 fr αr, where the f i, 1 i r, are irreducible polynomials such that f i is not a scalar multiple of f j if i j, and α i, 1 i r, are positive integers. The irreducible components are V( f i ), 1 i r, so if we prove the proposition for f irreducible, then it will imply the general case, because each component of H has dimension n 1. Let s assume that f is irreducible, then H is also irreducible and I(H) = f. ( f We have J =, f,..., f ). rank J P can only be 0 or 1 and if x 1 x 2 x n rank J P = 1 for some P H, then dim H = n 1. Assume to the contrary that rank J P = 0 for every P H. Then f f,, x 1 x 2 f..., I(H) = f, that is, all the partial derivatives are multiples of f. x n As their degree is smaller than the degree of f, the only way this can happen is if they are all 0. This is not possible in characteristic 0 since then f would be a constant, which has been excluded. In characteristic p it is possible for all the partial derivatives to be 0, it happens if and only if the exponent of every variable in every term is a multiple of p. ( d xn d x = nxn 1 = 0 if and only if p divides n.) Then f = i 1,i 2,...,i n α i1,i 2,...,i n x pi1 1 x pi 2 2 x pin n = ( i 1,i 2,...,i n ) p p αi1,i 2,...,i n x i 1 1 x i 2 2 x in n, so f is reducible, contradicting our assumption that it was irreducible. Theorem 3.4 Let V A n be an affine algebraic variety and let P V. Let m P = {F K[V ] F (P ) = 0} be the maximal ideal of K[V ] corresponding to P. Then dim T P V = dim(m P /m 2 P ). (The quotient ring K[V ]/m2 P is a finite dimensional vector space, m P /m 2 P is a subspace in it.) 37

38 Proof. Let s introduce the notation f = f + I(V ) for f K[V ]. Let P = (a 1, a 2,..., a n ) and let F m P. F = f for a suitable f K[V ] and F (P ) = f(p ) = 0. We noted in the proof of the Nullstellensatz (Theorem 1.7) that f(p ) = 0 if and only if f x 1 a 1, x 2 a 2,..., x n a n, therefore there exist polynomials g i K[x 1, x 2,..., x n ] such that f = n (x i a i )g i in K[x 1, x 2,..., x n ]. Hence f = n (x i a i )g i in K[V ]. x i a i m p for every i, i=1 1 i n, obviously, therefore m p = x 1 a 1, x 2 a 2,..., x n a n K[V ]. In the rest of the proof we shall assume P = (0, 0,..., 0) for simplicity, then we have m p = x 1, x 2,..., x n K[V ]. We can also assume that T P V is the subspace x 1 = x 2 =... = x k = 0 of K n, where k = n dim T P V, this can be achieved by a linear change of co-ordinates. This means that the reduced row echelon form of J P is , with 1s on the diagonal in the first k rows and all the other entries are 0. The rows of the reduced row echelon form are linear combinations of the rows of J P. By applying the same linear combinations to the generators of I(V ) we obtain a set of generators f 1, f 2,..., f r for I(V ) such that the Jacobian of this set of generators is the reduced row echelon form above, therefore f i = x i + (terms of degree 2) for 1 i k, and f i only contains terms of degree 2 for k + 1 i r. Let s define a function D : m p K n k in the following way. Let F m p. Then F = f for some f x 1, x 2,..., x n K[x 1, x 2,..., x n ]. f can be written as f = (α k+1, α k+2,..., α n ). n i=1 i=1 α i x i + (terms of degree 2) and we set D(F ) = We shall prove that D is a surjective linear map with kernel m 2 p, then it will follow that m P /m 2 P = K n k, so dim(m P /m 2 P ) = dim Kn k = n k. Step 1. First we need to show that the vector (α k+1, α k+2,..., α n ) does not depend on the choice of f, so that D is indeed a function. 38

39 Let g K[x 1, x 2,..., x n ] be another polynomial such that F = g. Then f g I(V ), so there exist h i K[x 1, x 2,..., x n ], 1 i r, such that f g = r i=1 h i f i. As f i = x i + (terms of degree 2) for 1 i k, and f i only contains terms of degree 2 for k + 1 i r, the sum r h i f i has no linear terms in x k+1, x k+2,..., x n, so the coefficients α k+1, α k+2,..., α n are uniquely determined and therefore D(F ) is indeed a function. Step 2. D is clearly linear, D(F + G) = D(F ) + D(G) for any F, G m P and D(λF ) = λd(f ) for any F m P and any λ K. n Step 3. D is surjective, since D( α i x i ) = (α k+1, α k+2,..., α n ) for any α k+1, α k+2,..., α n K. i=k+1 Step 4. ker D = {F m p D(F ) = 0} by definition. We shall prove that ker D = m 2 p. Let F = f, f = n i=1 α i x i + (terms of degree 2) as in the definition of D. Then F ker D if and only if α k+1 = α k+2 =... = α n. This implies that f k α i f i only contains terms of degree 2, i. e., it is an element of the i=1 ideal {x i x j 1 i j n} K[x 1, x 2,..., x n ]. This is equivalent to the existence of polynomials g ij K[x 1, x 2,..., x n ], 1 i j n, such that i=1 f k α i f i = g ij x i x j i=1 1 i j n in K[x 1, x 2,..., x n ]. Then in K[V ], we have f = g ij x i x j, 1 i j n i. e., F = f {x i x j 1 i j n} = m 2 P ker D m 2 p. Conversely, let s now assume that F m 2 p. Then F = G ij x i x j 1 i j n K[V ]. This shows that for suitable G ij K[V ], 1 i j n. Now let g ij K[x 1, x 2,..., x n ], 1 i j n, be such that g ij = G ij, then we can take f = g ij x i x j 39 1 i j n

40 and it satisfiesf = F. As f has no linear terms, we have D(F ) = (0, 0,..., 0), therefore F ker D. This shows m 2 p ker D, by combining it with the previous result we obtain ker D = m 2 p. We have proved that D is a surjective linear map m p K n k with kernel m 2 p, therefore m p /m 2 p = K n k, in particular dim(m P /m 2 P ) = dim Kn k = n k = dim T P V. Corollary 3.5 Let ϕ : V W be an isomorphism between affine algebraic varieties. Then dim T P V = dim T ϕ(p ) W for every P V, therefore dim V = dim W and ϕ(sing V ) = Sing W. In particular, a singular variety cannot be isomorphic to a non-singular variety. Proof. By Theorem 2.1, ϕ induces a morphism ϕ : K[W ] K[V ] and by Corollary 2.2, if ϕ is an isomorphism of varieties then ϕ is an isomorphism of rings. It easy to check ϕ also gives a bijection between m P K[V ] and m ϕ(p ) K[W ] and also between m 2 P K[V ] and m2 ϕ(p ) K[W ]. Hence m p/m 2 p and m ϕ(p ) /m 2 ϕ(p ) are isomorphic vector spaces, therefore by Theorem 3.4, dim T P V = dim m p /m 2 p = dim m ϕ(p ) /m 2 ϕ(p ) = dim T ϕ(p )W. The other assertions follow from this, since dimension and singularity are defined in terms of the dimensions of the tangent spaces at the points of the variety. Examples: 1. The affine line A 1 (C) and the cuspidal cubic curve defined by the equation y 2 x 3 = 0 in A 2 (C) are not isomorphic, since the former has no singular point, while the latter has a singular point at (0, 0). 2. Let W 1 = V( xy, xz, yz ) A 3 (C) and let W 2 = V( xy(x + y) ) A 2 (C) y W 1 1 W x The irreducible components of W 1 are the co-ordinate axes, V( x, y ), V( x, z ) 40

41 and V( y, z ). The irreducible components of W 2 are the lines V( x ), V( y ) and V( x + y ). Both varieties consist of three lines meeting at a point, the origin, but we shall prove that W 1 and W 2 are not isomorphic. Let s calculate T (0,0,0) W 1 and T (0,0) W 2. xy, xz, yz ) and xy(x + y) are radical ideals, so they are equal to I(W 1 ) and I(W 2 ), respectively. y x 0 The Jacobian of W 1 is J(W 1 ) = z 0 x. At (0, 0, 0), J (0,0,0) (W 1 ) = 0 z y , so it has rank 0 and therefore dim T (0,0,0) W 1 = 3. (T (0,0,0) W 1 = A 3 can be proved even without calculating the Jacobian. Any f I(W 1 ) is identically 0 on the three co-ordinate axes, therefore these axes are tangent lines. If we know that T (0,0,0) W 1 is an affine subspace, then it follows that T (0,0,0) W 1 = A 3 as no other affine subspace contains all three co-ordinate axes.) The Jacobian of W 2 is J(W 2 ) = ( 2xy + y 2 x 2 + 2xy ). At (0, 0), J (0,0) (W 2 ) = ( 0 0 ), so it has rank 0 and therefore dim T(0,0) W 2 = 2. If there existed an isomorphism ϕ : W 1 W 2, then we would have ϕ(0, 0, 0) = (0, 0), as (0, 0, 0) and (0, 0) are uniquely characterised as the intersection points of the components in W 1 and W 2, respectively. However, isomorphisms preserve the dimension of the tangent space and dim T (0,0,0) W 1 = 3 dim T (0,0) W 2 = 2, therefore there does not exist an isomorphism between W 1 and W 2. The rest of the material in this section is not examinable. For any vector space X over K, the dual space X is defined to be the vector space of all linear maps X K with the obvious operations. For any linear map α : X Y, there is a dual map α : Y X. What Theorem 3.4 really proves, without mentioning dual spaces explicitly, is that T P V = (m P /m 2 P ). While any two finite dimensional vector spaces of the same dimension are isomorphic, the isomorphism between T P V and (m P /m 2 P ) is natural, it can be defined without reference to bases. Given an arbitrary morphism ϕ : V W, by Theorem 2.1 we have a morphism of rings ϕ : K[W ] K[V ], which also induces a linear map m ϕ(p ) /m 2 ϕ(p ) m P /m 2 P. Taking duals, we obtain the differential map dϕ P : T P V = (m P /m 2 P ) T ϕ(p ) W = (m ϕ(p ) /m 2 ϕ(p ) ). dϕ P is similar to the differential map betwen tangent spaces in differential geometry. 41

42 Isomorphisms can be characterised by using the differential map, ϕ is an isomorphism if and only if it is bijective and dϕ P is an isomorphism for every P V. The dimension of a variety can be defined in other ways as well, which are more suited for certain purposes, but are less useful for practical calculations. 1. dim V is equal to the maximum d for which there exists a chain of irreducible subvarieties V 0 V 1... V d V. Moreover, if V is irreducible, any such chain that cannot be extended by inserting further subvarieties has the same length. From this definition it is clear that the dimension of a proper subvariety of an irreducible variety is smaller than the dimension of the whole variety, which is hard to prove by using tangent spaces. 2. If V is irreducible, then dim V is transcendence degree of K(V ) over K, i. e., the maximal number of elements x 1, x 2,..., x d K(V ) which do not satisfy any non-zero polynomial with coefficients in K. As a consequence of an algebraic result called the Noether Normalisation Lemma, any variety is birationally equivalent to a hypersurface, therefore dim V = d if and only if V is birationally equivalent to a hypersurface in A d+1. This interpretation of the dimension shows that the dimension of irreducible varieties is invariant under birational equivalence, not just under isomorphism. Fact: For any irreducible variety V in characteristic 0, there exist a nonsingular variety Ṽ and a birational morphism ϕ : Ṽ V such that ϕ is an isomorphism outside Sing V. This is called the desingularisation of V. The existence of a desingularisation is a very hard theorem, for which the Japanese mathematician Heisuke Hironaka received the Fields Medal in The problem is still open in prime characteristic in dimension 3. 42

43 4 Projective space and projective varieties One of the motivations for projective geometry is to understand perspective. It is well-known phenomenon that parallel lines such as railway tracks appear to meet at a point, called the vanishing point. It was not until the Renaissance that painters were able to give a geometrically correct representation of objects. The painting below, The delivery of keys by Pietro Perugino in the Sistine Chapel (source Entrega_de_las_llaves_a_San_Pedro_(Perugino).jpg) is a good example of the use of perspective, the parallel lines on the ground converge towards a point near the centre of the painting. Projective geometry gives a nicer theory than affine geometry in many ways, for example, two lines in the projective plane always meet, or more generally two curves defined by degree m and degree n equations, resp., have mn intersection points counted with multiplicity. Projective geometry also has important practical applications, images are combined by applying projective transformations in photo stitching, which is used in panoramic photos, Google Street View and virtual reality. 43

44 4.1 Basic properties of projective space and projective varieties The definition of projective space is also motivated by the idea that if a camera is at the origin of the co-ordinate system, all points of a ray through the origin will be mapped to the same point of the image, so points of the image correspond to lines through the origin. Definition. The n-dimensional projective space over a field K, denoted by P n (K) or by P n if the field is understood, is the set of equivalence classes of K n+1 \ {(0, 0,..., 0)} under the equivalence relation (x 0, x 1,..., x n ) (λx 0, λx 1,..., λx n ) for any λ K \ {0}. (The points of P n correspond to lines through the origin in K n+1.) The equivalence class of a point (X 0, X 1,..., X n ) K n+1 \ {(0, 0,..., 0)} is denoted by (X 0 : X 1 :... : X n ). X 0, X 1,... X n are called homogeneous co-ordinates on P n. Two ways of looking at projective space 1. Let U 0 = {(X 0 : X 1 :... : X n ) P n X 0 0}. (X 0 : X 1 :... : X n ) = (1 : X 1 /X 0 :... : X n /X 0 ) in U 0 and X 1 /X 0, X 2 /X 0,..., X n /X 0 can take arbitrary values in K, so the points of U 0 are in bijection with the points of A n. The set P n \ U 0 = {(X 0 : X 1 :... : X n ) P n X 0 = 0} is clearly a copy of P n 1. Therefore P n = A n P n 1 as a set. If n = 1, then U 0 = {(X 0 : X 1 ) P 1 X 0 0} = { ( 1 : X 1 X 0 ) X1 K, X 0 0} = {(1 : x) x K}, so we can identify U 0 with A 1 via (X 0 : X 1 ) X 1 /X 0 A 1. P 1 \ U 0 = {(X 0 : X 1 ) P 1 X 0 = 0} = {(0 : X 1 ) X 1 K \ {0}} = {(0 : 1)}, because (0 : X 1 ) = (0 : 1) for any K \ {0}. This is the decomposition, P 1 = A 1 P 0, as P 0 consists of just a single point. We can identify A 1 with K and denote the single point of P 0 by, this way we can identify P 1 with K { }. If n = 2, then U 0 = {(X 0 : X 1 : X 2 ) P 2 X 0 0} = { ( 1 : X 1 : X 2 ) X1, X 2 K, X 0 0} X 0 X 0 = {(1 : x : y) x, y K} 44

45 so we can identify U 0 with A 2 via (X 0 : X 1 : X 2 ) (X 1 /X 0, X 2 /X 0 ) A 2. The complement of U 0 is P 2 \U 0 = {(X 0 : X 1 : X 2 ) P 2 X 0 = 0} = {(0 : X 1 : X 2 ) (X 1, X 2 ) K 2 \{(0, 0)}}, this is a copy of P 1. To understand these points, consider the parallel lines x+y = 0 and x+y 2 = 0 in A 2 = U 0. As x = X 1 /X 0 and y = X 2 /X 0, we can rewrite the equations in terms of homogeneous co-ordinates as X 1 + X 2 = 0 and X 1 + X 2 2 = 0. X 0 X 0 X 0 X 0 After multiplying by X 0 we get X 1 + X 2 = 0 and X 1 + X 2 2X 0 = 0. The solutions of this system of linear equations are X 0 = 0, X 1 = X 2, which correspond to the point (0 : 1 : 1) P 1. Therefore these parallel lines in A 2 = U 0 intersect in P 2 \ U 0. Any other line parallel to them also contains (0 : 1 : 1). 0:1: x y x 2 x y 0 Similarly, all lines with direction vector (X 1, X 2 ) intersect at (0 : X 1 : X 2 ) P 2 \ U 0. This is true in general for arbitrary n, the points of P n \U 0 = P n 1 correspond to directions of lines in A n or equivalence classes of parallel lines. (0 : X 1 :... : X n ) is the point where all the lines with direction vector (X 1, X 2,..., X n ) in A n meet. (The single point of P 0 corresponds to the only line in A 1.) For 45

46 this reason, these points are often called points at infinity, but this does not mean that they are intrinsically different, the disctinction depends on the choice of co-ordinates. The youtube video (also linked to from the Animations and videos section of the course website) contains a good explanation of the projective plane. 2. The other approach is is to consider the sets U i = {(X 0 : X 1 :... : X n ) P n X i 0}, 0 i n. Each one is a copy of A n and their union is P n, since there is no point (0 : 0 :... : 0). These n + 1 copies are glued together by identifying their points via rational maps ϕ ij : U i U j so that P U i is the same point of P n as ϕ ij (P ) U j. For example, if n = 1, let t = X 1 /X 0 be the co-ordinate on U 0, and u = X 0 /X 1 the co-ordinate on U 1. We have the rational maps ϕ 01 : U 0 U 1, t 1/t, and ϕ 10 : U 1 U 0, u 1/u. A point t U 0 is the same point in P 1 as ϕ 01 (t) U 1, whenever ϕ 01 (t) is defined and similarly, a point u U 1 is the same point in P 1 as ϕ 10 (u) U 0, whenever ϕ 10 (u) is defined. If n = 2, then let x = X 1 /X 0, y = X 2 /X 0 be the co-ordinates on U 0, u = X 0 /X 1, v = X 2 /X 1 the co-ordinates on U 1 and s = X 0 /X 2, t = X 1 /X 2 the co-ordinates on U 2. Now we have u = 1/x and v = y/x, which give the rational map ϕ 01 : U 0 U 1, (x, y) (1/x, y/x), so that (x, y) U 0 and ϕ 01 (x, y) U 1 are the same point in P 2 whenever the latter is defined. The point (2 : 3 : 4) P 2 is the same as (1 : 3/2 : 2), (2/3 : 1 : 4/3) or (1/2 : 3/4 : 1), therefore (3/2, 2) U 0, (2/3, 4/3) U 1 and (1/2, 3/4) U 2 correspond to the same point of P 2 and they get identified in the gluing process. In general, if F K[X 0, X 1,..., X n ] and P = (X 0 : X 1 :... : X n ) P n, F (P ) cannot be defined, since F (λx 0, λx 1,..., λx n ) will give different values for different values of λ K \{0}. However, if F is homogeneous of degree d, then F (λx 0, λx 1,..., λx n ) = λ d F (X 0, X 1,..., X n ), so we can tell whether F (P ) = 0 or not, and this is enough to define projective algebraic varieties. Definition. An ideal I K[X 0, X 1,..., X n ] is called homogeneous if and only if it can be generated by homogeneous elements. (Warning: This does not mean that all elements of the ideal are homogeneous polynomials.) 46

47 Definition. Let I K[X 0, X 1,..., X n ] be a homogeneous ideal. The projective algebraic variety defined by I is the set V(I) = {(X 0 : X 1 :... : X n ) P n F (X 0, X 1,..., X n ) = 0, F I, F homogeneous}. Two ways of looking at projective varieties 1. Let s consider the variety V = V( X 1 X 2 X 2 0 ) P 2. If X 0 0, we can divide X 1 X 2 X 2 0 = 0 by X 2 0 to get (X 1 /X 0 )(X 2 /X 0 ) 1 =, which can be written as xy 1 = 0 in terms of the affine co-ordinates x = X 1 /X 0 and y = X 2 /X 0 on U 0 = A 2, so V 0 = V U 0 is a hyperbola. The points V \ V 0 are the points of V with X 0 = 0. By substituting X 0 = 0 into X 1 X 2 X 2 0 = 0 we obtain X 1 X 2 = 0, therefore the points at infinity are (0 : 1 : 0), which corresponds to lines parallel to the x-axis, and (0 : 0 : 1), which corresponds to lines parallel to the y-axis. The asymptotes of the hyperbola pass through these points at infinity. V consists of the hyperbola with two points at infinity corresponding to the asymptotes. 4 2 xy x 2 Starting with the equation xy 1 = 0 we can recover X 1 X 2 X0 2 = 0 by substituting x = X 1 /X 0 and y = X 2 /X 0 into xy 1 = 0 and multiplying it by X0. 2 Let s now consider the variety V = V( X1 2 X 0 X 2 ) P 2. If X 0 0, we can divide X1 2 X 0 X 2 = 0 by X0 2 to get (X 1 /X 0 ) 2 (X 2 /X 0 ) = 0, which can be written as x 2 y = 0 in terms of the affine co-ordinates x = X 1 /X 0 and y = X 2 /X 0 on U 0 = A 2, so V 0 = V U 0 is a parabola. The points V \ V 0 are the points of V with X 0 = 0. By substituting X 0 = 0 into X1 2 X 0 X 2 = 0 47

48 we obtain X 2 1 = 0, therefore the only points at infinity is (0 : 0 : 1), which corresponds to lines parallel to the y-axis. Unlike the hyperbola, the parabola has no asympotes, but for large x and y, its tangent direction gets approaches the direction of the y-axis. 0:0: y x Starting with the equation x 2 y = 0 we can recover X 2 1 X 0 X 2 = 0 by substituting x = X 1 /X 0 and y = X 2 /X 0 into x 2 y = 0 and multiplying it by X 2 0. Given a homogeneous polynomial F K[x 1, x 2,..., x n ], F [X 0, X 1,..., X n ] X deg F 0 = F [1, x 1, x 2,..., x n ] where x i = X i /X 0, 1 i n, is called the dehomogenisation of F with respect to X 0. If we dehomogenise all homogeneous elements of an homogeneous ideal J K[x 1, x 2,..., x n ], we obtain an ideal defining the affine algebraic variety V 0 = V U 0 A n, called the affine piece X 0 0 of V = V(J) P n. If we start with a polynomial f(x 1, x 2,..., x n ), X deg f 0 f(x 1 /X 0, X 2 /X 0,..., X n /X 0 ) K[X 0, X 1,..., X n ] is a homogeneous polynomial, the homogenisation of f. Homogenisation can also be done at the level of ideals, the projective algebraic variety in P n defined by the ideal generated by the homogenisation of the elements of an ideal J K[x 1, x 2,..., x n ] is called the projective closure of V(J) A n. The 48

49 affine piece X 0 0 of the projective closure is exactly V(J), while the points at infinity correspond to asymptotic directions of V(J) as we have seen in the example. Homogenisation and dehomogenisation are one-sided inverses of each other. Homogenisation followed by dehomogenisation always yields the same polynomial, and similarly for varieties, taking the projective closure of an affine variety and then the affine piece X 0 0 gives the original variety. Dehomogenising and then homogenising a homogeneous polynomial will give the polynomial divided by the highest power of X 0 dividing it. (For example, the dehomogenisation of X 0 X 1 is x 1, whose homogenisation is just X 1.) For varieties this means that taking projective closure of the affine piece X 0 0 of a projective algebraic variety will give the union of irreducible components of the original variety not contained in the hyperplane X 0 = 0. (Irreducibility for projective varieties will be defined later, but it is completely analogous to the affine case.) If a projective variety V has no irreducible components contained in the hyperplane X 0 = 0, then V is the projective closure the affine piece V 0 = V U 0 and the points of V correspond to the points of V 0 and the asympotic directions of V The other approach is to consider all the affine pieces V i = V U i, 0 i n, of a projective variety V. These are affine varieties and they are glued together by identifying the points via the rational maps ϕ ij : U i U j defined previously. For example, by homogenising the equation y 2 x 3 x 2 = 0 of the nodal cubic in A 2, we obtain X 0 X 2 2 X 3 1 X 0 X 2 1 = 0. The projective curve defined by this equation is the projective closure V, whose affine piece V 0 = V U 0 is original affine nodal cubic curve. Substituting X 0 = 0 into the equation gives X 3 1 = 0, therefore the only point with X 0 = 0 is (0 : 0 : 1), the asymptotic direction of the y-axis. x and y can be expressed as x = X 1 /X 0, y = X 2 /X 0 in terms of the homogeneous co-ordinates. The affine co-ordinates on the other affine pieces are u = X 0 /X 1, v = X 2 /X 1 on U 1 and s = X 0 /X 2, t = X 1 /X 2 on U 2. Dehomogenising X 0 X 2 2 X 3 1 X 0 X 2 1 = 0 with respect to X 1 gives uv 2 1 u = 0, dehomogenising it with respect to X 2 gives s t 3 st 2 = 0. Therefore the projective curve has the affine pieces shown below. 49

50 y 2 x 3 x 2 = 0 uv 2 1 u = 0 s t 3 st 2 = x u s They are glued together by the maps ϕ 01 : U 0 U 1, (x, y) (1/x, y/x) and ϕ 02 : U 0 U 2, (x, y) (1/y, x/y). Definition. Let F K[X 0, X 1,..., X n ] be a polynomial. The degree i homogeneous part of F, denoted by F [i], is the sum of all the terms of degree i in F. If i < 0 or i > deg F, F [i] is defined to be 0. Example: Let F = X 0 +X 0 X 2 X 2 1 +X 0 X 1 X 2 K[X 0, X 1, X 2 ], then F [0] = 0, F [1] = X 0, F [2] = X 0 X 2 X 2 1 and F [3] = X 0 X 1 X 2. Lemma 4.1 The ideal I K[X 0, X 1,..., X n ] is homogeneous if and only if for any F I, all the homogeneous parts of F are also elements of I. Proof. Assume that I is homogeneous, let F 1, F 2,..., F r be a set of homogeneous generators for I. Let F I, then F = r F i G i for some G i K[X 0, X 1,..., X n ], 1 i r. The F [d] = r F i G [d deg Fi ] I for every d, 0 d deg F, so all the homogeneous parts of F are also elements of I. Assume now that F I implies that F [0], F [1],..., F [deg F ] are also in I. Let F 1, F 2,..., F r be an arbitrary set of generators for I. We claim that the set {F i[j] 1 i r, 0 j deg F i } generates I. On one hand F i = deg F i F i[j] is contained in the ideal generated by this set for each i, 1 i r, so this ideal contains I. On the other hand, each of the generators is in I, so the ideal generates by them is a subset of I. Therefore the ideal generated by this set is exactly I and since these generators are homogeneous, I is a homogeneous ideal. Proposition 4.2 (Cf. Proposition 1.3) (i) Let V 1 = V(I 1 ), V 2 = V(I 2 ),..., V k = V(I k ) be projective algebraic i=0 i=1 j=0 50

51 varieties in P n. Then V 1 V 2... V k = V(I 1 I 2... I k ) = V(I 1 I 2... I k ) is also a projective algebraic variety. (ii) Let V α = V(I α ), α A be projective algebraic varieties in P n. Then V α = V ( ) I α α A is also a projective algebraic variety. α A Proof. Just imitate the proof of Proposition 1.3 (i) and (ii). Warning: There is no direct analogue of Proposition 1.3 (iii) for projective varieties because P m P n is very different from P m+n. Definition. The homogeneous ideal of a set Z P n is the ideal I(Z) K[X 0, X 1,..., X n ] generated by the set {F K[X 0, X 1,..., X n ] F homogeneous, F (X 0,..., X n ) = 0 (X 0 :... : X n ) Z}. Theorem 4.3 (Projective Nullstellensatz, cf. Theorem 1.7) Let K be an algebraically closed field and let J K[X 0, X 1,..., X n ] be a homogeneous ideal. (i) V(J) = if and only if J = K[X 0, X 1,..., X n ] or J = X 0, X 1,..., X n. (ii) I(V(J)) = J unless J = X 0, X 1,..., X n. Idea of proof. For any homogeneous ideal J K[X 0, X 1,..., X n ] we can consider the projective algebraic variety V(J) defined by in P n and also the affine algebraic variety defined by J in A n+1, called affine cone on V(J). This gives an almost bijective correspondence between projective algebraic varieties in P n and certain affine algebraic varieties in A n+1, the only failure of bijectivity is that K[X 0, X 1,..., X n ] and X 0, X 1,..., X n both define the empty set as a projective variety. Definition. A projective algebraic variety V is reducible if and only if it can be written as V = V 1 V 2, where V 1, V 2 are also projective algebraic varieties, V 1 V V 2. If V is not reducible, it is called irreducible. Proposition 4.4 (Cf. Theorem 1.8) Every projective algebraic variety V can be decomposed into a union V = V 1 V 2... V k such that every V i, 1 i k, is an irreducible projective algebraic variety and V i V j for i j. The decomposition is unique up to the ordering of the components. The V i, 1 i k, are called the irreducible components of V. 51

52 Proof. Completely analogous to the proof of Theorem 1.8. Lemma 4.5 A homogeneous ideal I K[X 0, X 1,..., X n ] is prime if and only if for any homogeneous polynomials F, G K[X 0, X 1,..., X n ], F G I implies F I or G I. Proof. If I is prime then F G I implies F I or G I for all F, G K[X 0, X 1,..., X n ] by the definition of a prime ideal. Let s assume I is a homogeneous ideal which is not prime. Then there exist polynomials P / I, Q / I such that P Q I. Let d 0 be the minimal integer such that P [d] / I. There exists such a d since if P [j] I for every j 0, then P = deg P P [j] I, too. Similarly, let e 0 be the minimal integer j=0 such that Q [e] / I. Then (P Q) [d+e] = d+e P [j] Q [d+e j]. If j < d, then P [j] I, j=0 so P [j] Q [d+e j] I, too. If j > d, then then Q [d+e j] I as d + e j < e, so P [j] Q [d+e j] I, too. Therefore all the terms in the sum except for P [d] Q [e] are in I, (P Q) [d+e] I by Lemma 4.1 since P Q I and I is homogeneous. Hence P [d] Q [e] I, too, but P [d] / I and Q [e] / I. Therefore F = P [d] and G = Q [e] are homogeneous polynomials with the property that F, G / I but F G I. Proposition 4.6 (Cf. Proposition 1.9) A projective algebraic variety V is irreducible if and only if I(V ) is prime. Proof. Imitate the proof of Proposition 1.9 and use Lemma

53 4.2 Tangent spaces, dimension and singularities There are two possible approaches. 1. One can define tangent lines and imitate the development of the affine theory. The line through the points P Q consists of the points λp + µq, where (λ : µ) P 1, and this line is tangent to V at P if and only if for any homogeneous polynomial F I(V ), F (λp + µq) contains no degree 1 term in µ. 2. Let V P n be an irreducible projective algebraic variety, and let P V. Let (X 0 : X 1 : X 2 :..., X n ) be homogeneous co-ordinates on P n. Choose i such that P is not contained in the hyperplane X i = 0. Let V i = {(X 0 : X 1 :... : X n ) V X i 0} be the affine piece X i 0 of V. The tangent space T P V is defined as the projective closure of T P V i, the local dimension of V at P is the local dimension of V i at P, and the P is a singular point of V if and only if it is a singular point of V i. (It needs to be proved that these definitions are independent of the choice of i, which can be done by an improved version of Theorem 3.5.) Tangent spaces, dimension and singular points of projective varieties can be calculated by using the Jacobian matrix in the same way as in the affine case. Example: Find the singular points, if any, of the curve C defined by the equation Y 2 Z X 3 X 2 Z = 0 in P 2 (C). Let F = Y 2 Z X 3 X 2 Z, then F X = 3X 2 2XZ, F Y = 2Y Z and F Z = Y 2 X 2. The rank of the Jacobian J is 0 where all 3 partial derivatives vanish, and 1 elsewhere. The points P C where rank J p = 0 are the solutions of F = F X = F Y = F Z = 0. F Y = 0 implies Y = 0 or Z = 0. If Y = 0, then F Z = 0 implies X = 0, and we get the point (0 : 0 : 1), which is indeed on C. (In projective space, (0 : 0 : 1) = (0 : 0 : Z) for any Z 0.) If Z = 0, then F X = 0 implies X = 0, and then F Z = 0 implies Y = 0, but (0 : 0 : 0) is not a point in P 2. Therefore rank J = 0 and dim T P C = 2 at (0 : 0 : 1), and rank J = 1 and dim T P C = 1 at all other points of C. Hence dim C = 1 and (0 : 0 : 1) is the only singular point of C. If we take the affine piece Z 0 of C, we get the familiar nodal cubic curve y 2 x 3 x 2 = 0, which has a singular point at (0, 0), corresponding the singular point of the projective curve. 53

54 4.3 Functions, rational maps and morphisms Definition. The homogeneous co-ordinate ring of a projective variety V P n is K[V ] = K[X 0, X 1,..., X n ]/I(V ). If V is irreducible, the function field of V, K(V ), is the set of equivalence classes of fractions F/G, where F, G K[V ], G 0, F, G are homogeneous of the same degree, and two fractions F 1 /G 1, F 2 /G 2 are equivalent if and only if F 1 G 2 = F 2 G 1 in K[V ]. Warning: Unlike in the affine case, the elements of K[V ] are not functions on V. As I is a homogeneous ideal, we can also define homogeneous elements in K[V ] degree j as the images of the homogeneous polynomials of degree j in K[X 0, X 1,..., X n ] and then every element of K[V ] can be written uniquely as the sum of homogeneous elements just like polynomials. The elements of K(V ) are partially defined functions on V. If F, G are both homogeneous of degree d, then and F (λx 0, λx 1,..., λx n ) = λ d F (X 0, X 1,..., X n ) G(λX 0, λx 1,..., λx n ) = λ d G(X 0, X 1,..., X n ), so the value of F/G does not depend on the representantive of the point (X 0 : X 1 :... : X n ) P n chosen. Similarly, this value does not change if another fraction representing the same element of K(V ) is chosen as long as the denominator does not vanish. Definition. Let V P m (K) and W P n (K) be projective algebraic varieties and let V be irreducible. A rational map Φ : V W is a partial function defined by an equivalence class of (n + 1)-tuples of homogeneous elements of K[V ] of the same degree, (Φ 0 : Φ 1 :... : Φ n ) (Ψ 0 : Ψ 1 :... : Ψ n ) if and only if Φ i Ψ j = Φ j Ψ i for every i, j, 0 i, j n. Φ : V W is defined at P V if and only if Φ can be represented by (Φ 0 : Φ 1 :... : Φ n ) such that Φ i (P ) 0 for some i, 0 i n, in this case we require that Φ(P ) = (Φ 0 (P ) : Φ 1 (P ) :... : Φ n (P )) W. A rational map Φ : V W is a morphism if and only if it is defined at every point of V. (This definition of a morphism is not really the right one if K is not algebraically closed, but it will do for our purposes. If K is algebraically closed, our definition is analogous to the characterisation of morphisms of affine algebraic varieties as rational maps which are defined everywhere, see problem sheet 4, question 5, but for morphisms of affine algebraic varieties there is a simpler definition which works over any field.) 54

55 A rational map Φ : V W is dominant if and only if its image is not contained in any proper subvariety of W. A rational map Φ : V W is a birational equivalence if and only if it has an inverse rational map Ψ : W V such that Ψ Φ = id V, Φ Ψ = id W. In this case V and W are called birationally equivalent. A projective variety is called rational if and only if it is birationally equivalent to P n for some n. A morphism Φ : V W is an isomorphism if and only if it has an inverse morphism Ψ : W V such that Ψ Φ = id V, Φ Ψ = id W. In this case V and W are called isomorphic. Worked examples of rational maps and morphisms between projective varieties can be found on a separate handout at ac.uk/~gm/teaching/math32062/projexamples.pdf. Remarks. 1. Rational maps between projective varieties can be defined in terms of (n + 1)-tuples of elements of K(V ), just like in the affine case, but it is more common to write them in terms of elements of K[V ]. 2. The same degree condition in the definition of the rational map means that Φ 0, Φ 1,..., Φ n in one representation of Φ must be homogeneous of the same degree, but this common degree can be different for different representations of Φ. 3. If K[V ] has unique factorisation into irreducibles (e. g., V = P m so K[V ] = K[X 0, X 1,..., X n ]), then every rational map can be represented in a form (Φ 0 : Φ 1 :... : Φ n ) such that Φ 0, Φ 1,..., Φ n have no common factors and this representation is unique up to a scalar factor. It is sufficient to consider this only representation of Φ because all others are of the form (F Φ 0 : F Φ 1 :... : F Φ n ) for some homogeneous element F K[V ]. 4. Morphisms are functions, whereas rational maps are only partially defined functions, defined outside a proper subvariety. 5. The image of a projective algebraic variety under a morphism is also a projective algebraic variety. Theorem 2.4 and Corollary 2.5 can be adapted to the context of projective varieties. Warning: There is no equivalent of Theorem 2.2 and Corollary 2.3 for morphisms between projective varieties. Theorem 4.7 (Cf. Theorem 2.4) Let V P m (K) and W P n (K) be irreducible projective algebraic varieties. Let (Y 0 : Y 1 : Y 2 :... : Y n ) homogeneous co-ordinates on P n and assume that W does not lie in the hyperplane Y 0 = 0. There exists a bijection between dominant rational maps 55

56 φ : V W and field homomorphisms α : K(W ) K(V ) preserving K given by the following constructions: for a rational map Φ : V W, define Φ : K(W ) K(V ), Φ (f) = f Φ, and for a homomorphism α : K(W ) K(V ) preserving K, define the corresponding rational map by (1 : α(y 1 /Y 0 ), α(y 2 /Y 0 ),..., α(y n /Y 0 )). Sketch of the proof. Let (X 0 : X 1 : X 2 :... : X m ) be homogeneous coordinates on P m. We can assume that V does not lie in the hyperplane X 0 = 0. Let V 0 be the affine piece X 0 0 of V and W 0 the affine piece Y 0 0 of W. Homogenisation with respect to X 0 gives an isomorphism K(V ) K(V 0 ). Similarly, homogenisation with respect to Y 0 gives an isomorphism K(W ) K(W 0 ). If Φ = (Φ 0 : Φ 1 :... : Φ n ), then ϕ = (Φ 1 /Φ 0, Φ 2 /Φ 0,..., Φ n /Φ 0 ) is a rational map ϕ : V 0 W 0, ϕ : K(W 0 ) K(V 0 ) gives Φ : K(W ) K(V ) via the isomorphisms K(V ) = K(V 0 ) and K(W ) = K(W 0 ). Conversely, given α : K(W ) K(V ), there is a corresponding homomorphism K(W 0 ) K(V 0 ), which gives a rational map V 0 W 0 by Theorem 2.4. This rational map can be extended to the rational map V W described in the statement of the theorem. Corollary 4.8 (Cf. Corollary 2.5) Φ : V W is a birational equvalence if and only if Φ : K(W ) K(V ) is an isomorphism of fields. V and W are birationally equivalent if and only if K(V ) and K(W ) are isomorphic as extensions of K. A variety V is rational if and only if K(V ) = K(t 1, t 2,..., t k ) for some k. 56

57 4.4 Projective transformations a 00 a 01 a 0n a 10 a 11 a 1n Definition. Let A =..... be an (n + 1) (n + 1) invertible. a n0 a n1 a nn matrix with entries from the field K. The projective transformation defined by A is the rational map Φ : P n P n, Φ(X 0 : X 1 :... : X n ) = (Y 0 : Y 0 Y 1 Y n X 0 X 1 Y 1 :... : Y n ), where. = A.. Y i can be written as Y i = a ij X j, X n so each Φ i is given by homogeneous degree 1 polynomials, therefore Φ is indeed a rational map. (We shall prove that the invertibility of A implies that projective transformations are, in fact, morphisms.) For any λ 0 the matrix λa defines the the same projective transformation as A. The projective transformations of P n form a group, called the projective linear group, denoted by P GL(n + 1, K). It is the quotient of GL(n + 1, K), the group of all invertible (n + 1) (n + 1) matrices by the normal subgroup consisting of scalar multiples of the identity matrix. Definition. Two sets V, W P n are projectively equivalent if and only if there exists a projective transformation Φ : P n P n such that Φ(V ) = W. Projective equivalence is an equivalence relation. Examples: 1. The projective closure of the hyperbola xy 1 = 0 has equation X 1 X 2 X0 2 = 0, while the projective closure of the parabola y x 2 = 0 has equation X1 2 X 0 X 2 = 0. The projective transformation (X 0 : X 1 : X 2 ) (X 1 : X 0 : X 2 ), corresponding to the matrix 1 0 0, maps one to the other. In fact, all ellipses, parabolas and hyperbolas in P(R) or in P(C) are projectively equivalent. 2. Any two sets of 4 distinct points in P 2 with no 3 of them collinear are projectively equivalent, this follows from question 2 on problem sheet 6. However, the interior of the quadrilateral formed by one set of 4 points may not be mapped to the interior of the quadrilateral formed by the other set of 4 points. 57

58 Proposition 4.9 Let n be a positive integer, let A be an invertible (n + 1) (n+1) matrix and let Φ : P n P n be the projective transformation defined by A. Then Φ is an isomorphism P n P n (an automorphism of P n ). X 0 0 X 1 Proof. As A is invertible, A. = 0. implies X 0 = X 1 =... = X n = 0, X n 0 therefore Φ is defined at every point of P n, so it is a morphism. Let Ψ be the projective transformation defined by A 1. A 1 is also invertible, so by the previous argument, Ψ is also morphism P n P n. Let Φ(X 0 : X 1 :... : X n ) = (Y 0 : Y 1 :... : Y n ) and let Ψ(Y 0 : Y 1 :... : Y n ) = Y 0 Y 1 (Z 0 : Z 1 :... : Z n ), then. = A. and. = A 1. = X 0 X 1 X n Y n., therefore Ψ Φ = id P n. Similarly, Φ Ψ = id Pn, so Φ and Ψ are X 0 X 1 X n inverses of each other, therefore they are isomorphisms. Remark. By this theorem, projective transformations are automorphisms of P n. The converse is also true over algebraically closed fields, the automorphisms of P n are exactly the projective transformations, see Theorem 4.12 for the case n = 1. Projective transformations of P 1 are of particular interest. If we identify P 1 with K { } by (X 0 : X 1 ) X 0 /X 1 K (X 1 0) and ( (X) 0 : 0), then a b the projective transformation Φ given by the matrix can be written c d as Φ(z) = az + b with ad bc 0. (Arithmetic involving or division by 0 cz + d is handled in the obvious sensible way.) These maps are also called fractional linear maps or Möbius transformations. Definition. Let z 1,z 2, z 3, z 4 K { }, with at least 3 of them distinct. The cross ratio of z 1,z 2, z 3, z 4 is defined as (z 1, z 2 ; z 3, z 4 ) = (z 1 z 3 )(z 2 z 4 ) (z 1 z 4 )(z 2 z 3 ) 58 Z 0 Z 1 Z n Y 0 Y 1 Y n

59 (Again, arithmetic involving or division by 0 is handled in the obvious sensible way.) ( contains a good description of the cross ratio and its properties.) Proposition 4.10 Let z 1, z 2, z 3, z 4 K { } with at least 3 of them distinct. Then (z 1, z 2 ; z 3, z 4 ) = (ϕ(z 1 ), ϕ(z 2 ); ϕ(z 3 ), ϕ(z 4 )) for any projective transformation ϕ : P 1 P 1. Proof. ϕ(z) = az + b cz + d for some a, b, c, d K, ad bc 0. If c = 0, 1 cz + d + a c. In either case, ϕ can be composed of maps of the form z z + α (α K), z λz (λ K \ {0}) and z 1/z. It is easy to see that the first two preserve the cross ratio and ( 1 ( ) 1 )( 1 1 ) ( )( ) z3 z 1 z4 z 2 1 z 1 z 3 z 2 z 4 z = ( )( ) 1 z 3 z 2 z = ( )( 4 ) 1 1 z4 z 1 z3 z 2 ϕ(z) = a d z + b bc ad, while if c 0, ϕ(z) = d c z 1, 1 z 2 ; 1 z 3, 1 z z 1 z 4 z 2 z 3 z 1 z 4 z 2 z 3 = (z 3 z 1 )(z 4 z 2 ) (z 4 z 1 )(z 3 z 2 ) = (z 1 z 3 )(z 2 z 4 ) (z 1 z 4 )(z 2 z 3 ) = (z 1, z 2 ; z 3, z 4 ). All three types of maps preserve the cross ratio, therefore so does ϕ. Proposition 4.11 Let z 1, z 2, z 3, w 1, w 2, w 3 be elements K { } such that z 1, z 2, z 3 are pairwise distinct and w 1, w 2, w 3 are also pairwise distinct. Then there exists a unique projective transformation ϕ : P 1 P 1 such that ϕ(z i ) = w i for i = 1, 2, 3. In particular, any two sets of 3 distinct points in P 1 are projectively equivalent. Proof. Let ϕ 1 (z) = (z 1, z 2 ; z 3, z), then ϕ 1 is a projective transformation and ϕ 1 (z 1 ) =, ϕ 1 (z 2 ) = 0, ϕ 1 (z 3 ) = 1. Similarly, let ϕ 2 (z) = (w 1, w 2 ; w 3, z), then ϕ 2 is a projective transformation and ϕ 2 (w 1 ) =, ϕ 2 (w 2 ) = 0, ϕ 2 (w 3 ) = 1. Therefore ϕ = ϕ 1 2 ϕ 1 is a projective transformation and it satisfies ϕ(z i ) = w i for i = 1, 2, 3. To prove uniqueness, we note that since ϕ(z i ) = w i for i = 1, 2, 3, we have (z 1, z 2 ; z 3, z) = (ϕ(z 1 ), varphi(z 2 ); ϕ(z 3 ), z) = (w 1, w 2 ; w 3, ϕ(z)) for any z K { }. The left-hand side is ϕ 1 (z), the right-hand side is ϕ 2 (ϕ(z)), therefore ϕ(z) = ϕ 1 2 (ϕ 1 (z)) for any z K { }, so ϕ = ϕ 1 2 ϕ 1 is the only projective transformation which satisfies ϕ(z i ) = w i for i = 1, 2, 3. Remarks: 1. If we write ϕ(z) = az + b, then the coefficients a, b, c, d are cz + d only determined up to multiplication by a non-0 scalar, but the function ϕ 59

60 is unique. 2. For a higher dimensional analogue, see problem sheet 6, question 4. Example: Find a projective transformation ϕ : P 1 (C) P 1 (C) such that ϕ( 3) = 4, ϕ( 1) = 0, ϕ(0) = 1. Solution 1. Let ϕ 1 (z) = ( 3, 1; 0, z) = ϕ 2 (z) = (4, 0; 1, z) = ( 3)( 1 z) ( 3 z)( 1) = 3z + 3 z + 3. Let (3)( z) (4 z)( 1) = 3z. The projective transformation z + 4 we want is ϕ = ϕ 1 2 ϕ 1. ϕ 1 2 (z) = 4z z + 3, so ϕ(z) = 4( ) 3z+3 z+3 3z = z+3 12z+12 z+3 6z+12 z+3 = 12z z + 12 = 2z + 2 z + 2. Solution 2. This is the recommended method as it is faster. As ϕ preserves the cross ratio, ϕ(z) is characterised by the property that ( 3, 1; 0, z) = (4, 0; 1, ϕ(z)), i. e., 3z + 3 z + 3 = 3ϕ(z), (3z + 3)( ϕ(z) + 4) = 3ϕ(z)(z + 3). ϕ(z) + 4 Solving this equation for ϕ(z) gives ϕ(z) = 2z + 2 z + 2. Theorem 4.12 Any isomorphism P 1 (C) P 1 (C) is a projective transformation, i. e., of the form ϕ(x 0 : X 1 ) = (ax 0 + bx 1 : cx 0 + dx 1 ), where a, b, c, d C with ad bc 0. Proof. Assume that Φ : P 1 P 1 is an isomorphism. By Proposition 4.11, there exists a projective transformation Ψ such that Ψ(Φ(0 : 1)) = (0 : 1), Ψ(Φ(1 : 0)) = (1 : 0) and Ψ(Φ(1 : 1)) = (1 : 1). Let Θ = Ψ Φ. Θ can be written as Θ(X 0 : X 1 ) = (Θ 0 (X 0, X 1 ) : Θ 1 (X 0, X 1 )) with Θ 0, Θ 1 coprime homogeneous polynomials of the same degree d. Θ is bijective, in particular it is injective, so there is a unique (X 0 : X 1 ) P 1 such that Θ(X 0 : X 1 ) = (0 : 1), only (0 : 1) has this property, therefore Θ 0 (X 0, X 1 ) = αx d 0 for some α C \ {0}. Similarly, Θ 1 (X 0, X 1 ) = βx d 1 for some β C \ {0}. Now Θ(1 : 1) = (1 : 1) implies α = β, so we may assume α = β = 1 and then Θ(X 0 : X 1 ) = (X d 0 : X d 1 ). If ε is any dth root of unity, then Θ(ε : 1) = (1 : 1) = Θ(1 : 1), so the injectitivity of Θ implies that d = 1, Θ = id P 1. Therefore Φ = Ψ 1 is a projective transformation. Remark. This proof also works over any algebraically closed field of characteristic 0 and the theorem is, in fact, true without any restriction on the field, but the general proof requires Galois theory. 60

61 5 Geometry in the plane One of the nice features of the projective plane is that any two lines meet at exactly one point. We shall see that over an algebraically closed field, this can be generalised to arbitrary curves, they have exactly the expected number of intersection points if we count the intersections with appropriate multiplicity. This is analogous to counting the zeros of a polynomial with multiplicity in the Fundamental Theorem of Algebra. Definition. Let V be an irreducible affine variety and let P V. The local ring of V at P is O V,P = {ϕ K(V ) ϕ is defined at P }. This means that O V,P consists of the rational functions which can be written as ϕ = f/g with g(p ) 0. The function ε P : O V,P K, ϕ ϕ(p ) is a ring homomorphism. It is surjective, therefore its kernel m V,P = {ϕ O V,P ϕ(p ) = 0} is a maximal ideal. Any element of O V,P \ m V,P is invertible, so any proper ideal of O V,P must be contained in m V,P, therefore m V,P is the unique maximal ideal of O V,P. (In algebra, a local ring means a ring with a unique maximal ideal.) Definition. Let f, g K[x, y] and let P A 2. The intersection multiplicity of f, g at P is I P (f, g) = dim O A 2,P / f, g. If C and D are curves in A 2, then let I(C) = f and I(D) = g, and the intersection multiplicity of C and D at P is I P (C, D) = I P (f, g). For homogeneous polynomials in K[X, Y, Z] or for curves C and D in P 2, their intersection multiplicity at P P 2 is defined to be the intersection multiplicity of their dehomogenisations or of the corresponding affine curves in an affine piece of P 2 containing P. If f K[x], then the multiplicity of f at a K is the dimension of O A 1,a/ f, so the intersection multiplicity defined above is indeed a generalisation of the multiplicity of a zero of a polynomial at a point. Key properties of I p (C, D): 1. I P (C, D) is finite if and only if C and D have no common component containing P. 2. I P (C, D) = 0 if and only if P / C or P / D. 3. I P (C, D) = 1 if and only if P is a non-singular point of both C and D and the tangent lines to C and D at P are different. 4. I P (C, D) > 1 if and only if P is a singular point of at least one of C and D, or P is a non-singular point of both C and D but the tangent lines to C and D at P are the same. 61

62 Rules for calculating I P (f, g): 1. I P (f, g) only depends on the ideal f, g, so f, g can be replaced by another pair of polynomials that generate the same ideal. 2. I P (f 1 f 2, g) = I P (f 1, g) + I P (f 2, g). 3. If f(p ) 0 or g(p ) 0 then I P (f, g) = If neither f, nor g has a repeated factor that vanishes at P, the curves f = 0 and g = 0 are non-singular at P and have distinct tangent lines at P, then I P (f, g) = If g = ax + b y (a, b K) and P = (x 0, y 0 ), then I P (f, g) is equal to the multiplicity of x 0 as a root of f(x, ax + b) = 0. (Of course, it can also be applied with the roles of x and y reversed.) Examples: 1. Calculate the intersection multiplicity I (0,0) (x 3 + x 2 y 2, x + y). If we substitute y = x into x 3 +x 2 y 2 we obtain x 3, so by rule 5, I (0,0) (x 3 + x 2 y 2, x + y) = Calculate the intersection multiplicity By using rules 1 and 2 I (0,0) (x 3 + x 2 y 2, x 2 2xy + y 2 x 3 ). I (0,0) (x 3 + x 2 y 2, x 2 2xy + y 2 x 3 ) = I (0,0) (x 3 + x 2 y 2, (x 2 2xy + y 2 x 3 ) + (x 3 + x 2 y 2 )) = I (0,0) (x 3 + x 2 y 2, 2x(x y)) = I (0,0) (x 3 + x 2 y 2, x) + I (0,0) (x 3 + x 2 y 2, x y). I (0,0) (x 3 +x 2 y 2, x) = I (0,0) ( y 2, x) = 2I (0,0) (y, x) = 2 by rules 1, 2 and 4, and similarly I (0,0) (x 3 +x 2 y 2, x y) = I (0,0) (x 3 +x 2 y 2 (x y)(x+y), x y) = 62

63 I (0,0) (x 3, x y) = 3I (0,0) (x, x y) = 3, hence I (0,0) (x 3 + x 2 y 2, x 2 2xy + y 2 x 3 ) = Theorem 5.1 (Bézout) Let K be an algebraically closed field, and let F, G K[X, Y, Z] be homogeneous polynomials with no common factor of positive degree. Then P P 2 I P (F, G) = deg F deg G. Remark. It is relatively easy to prove by using the resultant that the number of intersection points without considering multiplicity is at most deg F deg G over any field, and quite often this is sufficient. Definition. Let f = a m x m + a m 1 x m a 1 x + a 0 and g = b n x n + b n 1 x n b 1 x + b 0 be polynomials in K[x]. The resultant of f and g is defined to be the (m + n) (m + n) determinant a m a m 1 a m 2... a 1 a a m a m 1... a 2 a 1 a a m... a 3 a 2 a 1 a a m a m 1 a m 2 a m 3... a 1 a a Res(f, g) = m a m 1 a m 2... a 2 a 1 a 0 b n b n 1 b n 2... b 2 b 1 b b n b n 1... b 3 b 2 b 1 b b n 2 b n 3 b n 4 b n 5... b b n 1 b n 2 b n 3 b n 4... b 1 b b n b n 1 b n 2 b n 3... b 2 b 1 b 0 The resultant has the property that Res(f, g) = 0 if and only if a m = b n = 0 or if f and g have a common factor of positive degree, which is equivalent 63

64 to having a common zero if K is algebraically closed. (For more information on resultants, see and http: //mathworld.wolfram.com/resultant.html.) We can dehomogenise F and G with respect to Z by introducing x = X/Z and y = Y/Z, let the f, g K[x, y] be the polynomials obtained this way. We can write f(x, y) = a m (y)x m + a m 1 (y)x m a 1 (y)x + a 0 and g(x, y) = b n (y)x n + b n 1 (y)x n b 1 (y)x + b 0, where the coefficients a i, 0 i m, and b j, 0 j n, are polynomials in y, which we can consider as elements of the field K(y). Using these a i, 0 i m, and b j, 0 j n, in the above determinant we obtain the resultant of f and g with respect to x, denoted by Res x (f, g) to indicate that we are considering f and g as polynomials in x. Res x (f, g) is a polynomial in y. As F and G have no common factors, neither do f and g, not even in K(y)[x], therefore Res x (f, g) is not the 0 polynomial. By considering the total degrees of f and g, we have deg a i deg f i for every i, 0 i m deg b j deg g j for every j, 0 j n. Therefore every term in the determinant defining Res x (f, g) has degree at most deg f deg g, so deg Res x (f, g) deg f deg g deg F deg G. If f(x 0, y 0 ) = g(x 0, y 0 ) = 0, then the 1-variable polynomials in x f(x, y 0 ) = a m (y 0 )x m + a m 1 (y 0 )x m a 1 (y 0 )x + a 0 and g(x, y 0 ) = b n (y 0 )x n + b n 1 (y 0 )x n b 1 (y 0 )x + b 0 have x 0 as a common zero, therefore their resultant is 0, but this resultant is just Res x (f, g)(y 0 ). In other words, the y co-ordinates of the intersection points of the curves f(x, y) = 0 and g(x, y) = 0 are among the zeros of Res x (f, g). As Res x (f, g) is a non-0 polynomial of degree at most deg F deg G in y, there are at most deg F deg G possible values for the y co-ordinate of the intersection points. By swapping the roles of x and y, we can similarly obtain that there are at most deg F deg G possible values for the x co-ordinate of the intersection points of the curves f(x, y) = 0 and g(x, y) = 0. Therefore the number of intersection points is finite. The intersection points of the affine curves f(x, y) = 0 and g(x, y) = 0 correspond exactly to the intersection points of the projective curves F (X, Y, Z) = 0 and G(X, Y, Z) = 0 with Z 0. Z is not a common factor of F and G, therefore at least one of the curves F (X, Y, Z) = 0 and G(X, Y, Z) = 0 only intersects the line Z = 0 in finitely many points. Therefore the total number of intersection points of F (X, Y, Z) = 0 and G(X, Y, Z) = 0 is finite. Now let s change co-ordinates in such a way that in the new homogeneous co-ordinates (X : Y : Z ) none of the intersection points lie on the line Z = 0 and after dehomogenisation, the x axis is not parallel to any of the lines connecting two intersection points. Now we do not lose any intersection 64

65 points by the dehomogenisation and disctinct intersection points have distinct y co-ordinates, so if we repeat the above argument, we obtain that the number of intersection points is at most deg F deg G. This argument works over any field K. In the last step we may need to extend the field to find an appropriate change of co-ordinates, but a bigger field can only increase the number of intersection points, so if their number is still at most deg F deg G after the field extension, the number of intersection points over the original field K is at most deg F deg G. Example: By Bézout s Theorem any two curves in P 2 meet in at least one point. This can be used to prove that P 2 and the quadric surface Q = V( XY ZW ) P 3 are not isomorphic. They were shown to be birationally equivalent in Example 4 of uk/~gm/teaching/math32062/projexamples.pdf. However, Q contains the lines X = Z = 0 and Y = W = 0, which do not intersect, so if Q and P 2 were isomorphic, the images of these non-intersecting lines would give curves in P 2 which do not intersect, which is impossible. In fact, the set of lines {αx + βw = αz + βy = 0 (α : β) P 1 } is contained in Q and no two distinct lines in this set intersect. Similarly, no two elements of the set {αx + βz = αw + βy = 0 (α : β) P 1 } intersect, but every element of the first set intersects every element of the second set in exactly one point. The first picture below shows the two families of lines on an affine piece of XY ZW = 0, the second one the families of lines on the projectively equivalent hyperboloid. No two red lines intersect, similarly no two blue lines intersect, but every red line meets every blue line. 65

66 The two pictures below show real life applications of the lines on the hyperboloid, the bridge across Corporation Street between the Arndale Centre and Marks&Spencer and statues in Manlleu, Catalonia. The youtube video (also linked to from the Animations and videos section of the course website) includes other examples of hyperboloids in architecture. Theorem 5.2 Let P 1, P 2, P 3, P 4 and P 5 be distinct points of P 2 and assume that no 4 of them are collinear. Then there exists a unique conic (curve defined by a degree 2 equation) in P 2 passing through P 1, P 2, P 3, P 4, P 5. Proof. First we prove the existence of such a curve. A degree 2 homogeneous polynomial F K[X, Y, Z] has 6 coefficients. The condition that the curve defined by F passes through a given point gives a homogeneous linear equation for the coefficients. Five homogeneous linear equations in 6 unknowns always have a non-trivial (not all 0) solution, therefore there exists a degree 2 homogeneous polynomial F such that F (P i ) = 0 for every i, 1 i 5. F cannot be the square of a polynomial of degree 1, as the points are not collinear, therefore the curve C = V( F ) defined by F is indeed a conic. Now for the proof of uniqueness let s assume that the curve D = V( G ) defined by the homogeneous degree 2 polynomial G also passes through P 1, P 2, P 3, P 4, P 5. By Bézout s Theorem, this can only happen if F and G have a common factor of positive degree, otherwise C and D could only have at most 4 points in common. If this common factor has degree 2, then F and G are scalar multiples of each other, so C = D. The only other possibility is that F and G have a common factor of degree 1, so F = LM and G = LN, where L, M and N are homogeneous polynomials of degree 1. The line L = 0 can 66

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