Math 418 Algebraic Geometry Notes

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1 Math 418 Algebraic Geometry Notes 1 Affine Schemes Let R be a commutative ring with 1. Definition 1.1. The prime spectrum of R, denoted Spec(R), is the set of prime ideals of the ring R. Spec(R) = {P R : P is prime}. Example 1.2. Let R = Z. Then Spec(Z) = {(p) : p Z is prime} {(0)}. Notice that the ideals (p) are maximal as well as prime. The ideal (0) is not maximal, and is contained in every other element (p) of Spec(Z). Example 1.3. Let R = F a field. Then Spec(F ) = {(0)}. Example 1.4. Let R = F [x], where F is a field. This ring is a principal ideal domain, and so its prime ideals are of the form (f(x)) where f(x) F [x] is irreducible. That is, Spec(F [x]) = {(f(x)) : f(x) F [x] is irreducible} {(0)}. The structure of Spec(F [x]) will depend on F. Example 1.5. Let R = Q[x]. Then we saw that Spec(Q[x]) = {(f(x)) : f(x) Q[x] is irreducible} {(0)}. For any q Q, the linear polynomial x q is irreducible, and so (x q) Spec(Q[x]). The polynomial x 2 +bx+c is irreducible whenever b 2 4c Q. In this case (x 2 +bx+c) Spec(Q[x]). There are polynomials of arbitrary degree in Spec(Q[x]). For example, (x n 2) Spec(Q[x]) for all n N. Example 1.6. Let R = C[x] (or F [x] where F is algebraically closed). In this case, the irreducible polynomials are exactly the linear polynomials. So Spec(C[x]) = {(x z) : z C} {(0)}. Note that the maximal ideals of C[x] are in one-to-one correspondence with the elements of C. Example 1.7. Let R = C[x, y]. This ring is not a principal ideal domain, however it is Noetherian, and so its ideals are finitely generated. The maximal ideals of R are of the form (x z, y w) where z, w C, and are in bijection with the points of C 2. Other prime ideals are generated by sets of irreducible polynomials, for example (x y) and (x y 2 ) are prime. Of course, (0) is always a prime ideal in an integral domain. We would like to view X = Spec(R) as a geometric space. The points in our space will be the elements of Spec(R). We will distinguish the points corresponding to maximal ideals, and call them the closed points of X. In order to view X as a geometric object, we will need what is called a topology on X. In other words, we will need to have some concept of an open neighborhood of a point in X. 1

2 Definition 1.8. For any subset I R, the vanishing set of I is the set of prime ideals of R containing I, that is V (I) := {P Spec(R) : I P }. We consider these sets to be closed subsets of X. For any element f R, the complement X f := V ({f}) c Spec(R) is called the basic open set of f. A subset U X is said to be open if it is a union of basic open sets. The open and closed sets described above are called Zariski open and Zariski closed. structure they give on X is called the Zariski topology on X. The map V takes subsets of R to subsets of X. It is not, however, injective. Lemma 1.9. If S is a subset of R, and I(S) is the ideal generated by S in R, then V (S) = V (I(S)). Proof. This is homework problem AM 15i. Based on this result, if we want to understand open and closed subsets of X, we can restrict our attention to ideals of R. The following lists some handy properties of V. Lemma Let I, J be ideals of R. Then 1. V (I J) = V (I) V (J). 2. V (I J) = V (I) V (J). 3. If I J, then V (J) V (I). Proof. Suppose P is prime and contains I and J. If P contains neither I nor J, then there exist i I and j J, neither of which are in P. This contradicts that ij I J P. Therefore V (I J) V (I) V (J). If P contains I or J then it certainly contains I J, so V (I) V (J) V (I J). Suppose now P contains I J. Then P contains both I and J, so P lies in V (I) V (J) On the other hand, if P lies in V (I) V (J), it contains both I and J, so it lies in V (I J). The third property is left as an exercise. Note that although the second property above holds for general subsets of R, the first does not. For example, if we take R to be the integers, I to be the natural numbers and 0, and J to be the negative numbers and 0, then all prime ideals contain I J = {0}. But no proper ideals contain I or J. In fact, even if we restrict to ideals of R, the map V will not be injective. For example, if we consider R = C[x], then V ((x)) = V ((x 2 )). This is because any prime ideal containing x 2 contains the product x x, and hence contains x itself. In fact, we could show by induction that V ((x)) = V ((x n )) for any positive integer n. In order to remedy this problem, we will need the following notion. Definition Let I be an ideal of R. The radical of I is r(i) := {r R : r m I for some k N}. As a special case, the nilradical of R is r((0)), in other words, it is the set of nilpotent elements of R. The 2

3 Note that r(r(i)) = r(i) for every ideal I of R, so we may call an ideal whose radical is itself a radical ideal. Lemma For any ideal I of R, V (r(i)) = V (I). Proof. This proof is homework problem AM 15i. There is another characterization of radical ideals which will be useful to us. Proposition If I is an ideal in R, then r(i) is the intersection of all prime ideals containing I. In particular, the nilradical of R is the intersection of all prime ideals of R. Proof. First, if x r(i) then x m I for some positive integer m. Hence x lies in every prime ideal containing I. This shows one containment. Now we must show that the intersection of all prime ideals containing I lies in r(i). Suppose first the I = (0), and r(i) is the nilradical of R. We wish to show that the intersection of all prime ideals in R contains only nilpotent elements. Suppose then that f is not nilpotent. Consider the set of ideals Σ which contain no power of f. Since f is not nilpotent, the 0 ideal is in Σ. Since Σ is not empty, by Zorn s Lemma it contains a maximal element under inclusion, call it M. If x, y M then the ideal (x) + M contains M, so f m (x) + M for some positive integer m. Similarly, f n (y) + M for some positive integer n. Therefore f n+m (xy) + M, and xy M. This shows M is prime, so f is not in the nilradical of R. Now for a general ideal I, we can apply the previous argument to the ring R/I to get the result. Corollary If I and J are ideals of R such that V (I) = V (J), then r(i) = r(j). In particular, the map V is an injective map from the set of radical ideals of R to subsets of X. Definition The space X = Spec(R) together with the Zariski topology is called an affine scheme. We may now wish to consider maps between different affine schemes. There is a close relationship between such morphisms and ring homomorphisms. Definition Let X = Spec(R) and Y = Spec(S). A morphism of affine schemes f : X Y is a map of sets given by a ring homomorphism f : S R as follows: For any prime ideal P X, f(p ) := (f ) 1 (P ). The map f is called the pull-back map of f. Note that this map is well defined, since if xy (f ) 1 (P ), then f (xy) P, hence at least one of (f ) 1 (x), (f ) 1 (y) lies in (f ) 1 (P ). It is important to note, however, that if Q Y is prime, f (Q) need not be prime. Example Suppose we start with the quotient map f : C[x, y] C[x, y]/(xy). We can write any ideal of C[x, y]/(xy) as I + (xy) for an ideal I of C[x, y] containing (xy). Therefore, the map f is simply the inclusion map of prime ideals of C[x, y] containing (xy) back into C[x, y]. Note that the maximal ideals containing (xy) are those of the form (x, y a) and (x b, y) for some a, b C. This example shows that in a sense, we can think of the scheme Spec(C[x, y]/(xy)) as a subscheme of Spec(C)[x, y]. We will now make this precise. 3

4 Definition A subscheme of an affine scheme X = Spec(R) is a scheme Y = Spec(S) such that there is a ring homomorphism i : R S which induces the inclusion i: Y X. Lemma Let I be an ideal of R. i: Spec(R/I) Spec(R). The quotient map i : R R/I induces an injection This shows that there is a correspondence between ideals of R and subschemes of Spec(R). However, this correspondence is not one-to-one. As we ve seen, the prime ideals containing one ideal may be the same as the prime ideals containing another. The following proposition gives the desired one-to-one correspondence. Proposition The subschemes of X = Spec(R) are in one to one correspondence with the radical ideals of R. 2 Polynomial Rings Let k be an algebraically closed field, such as C. We will now consider the special case that R = k[x 1,..., x n ]. In this case, Spec(R) is sometimes referred to as affine n-space, and is often denoted A n k. This case has many advantages. One is that R is a finitely generated k-algebra. In particular, it is Noetherian. Therefore the ideals we wish to study will always be finitely generated. Further, techniques exist to compute the radical of an ideal in this setting. In particular, this is where the study of Gröbner bases intersects algebraic geometry. In order to study this specific case, we first want to understand the closed points of A n k. This result is foundational to the field. Theorem 2.1 (Hilbert s Nullstellensatz). The maximal ideals of R are the ideals (x 1 a 1,..., x n a n ) where a 1,..., a n k. Proof. This proof will assume that k = C, but the result holds for any algebraically closed field k. Note that it is clear that ideals of the form I = (x 1 a 1,..., x n a n ) are maximal, since the quotient R/I = C. Suppose then that I is a maximal ideal. The ring R is Noetherian, so we may write I = (f 1,..., f m ) as being generated by a finite number of polynomials f 1,..., f m R. We can define a new field K to be the extension of Q obtained by adjoining all coefficiants of the polynomials f 1,..., f m to Q. This is a subfield of C. Define I 0 to be the maximal ideal I K[x 1,..., x n ] in K[x 1,..., x n ]. The quotient K[x 1,..., x n ]/I 0 is a field. In fact, it is a field which embeds in C. If we define a i to be the image of the class x i + I 0 in this embedding, then f i (a 1,..., a n ) = 0 for each i = 1,..., m. Hence I 0 (x 1 a 1,..., x n a n ). Corollary 2.2. If I is a proper ideal of k[x 1,..., x n ], then V (I) is nonempty. Proof. This follows from Hilbert s Nullstellensatz, and the fact that every proper ideal is contained in a maximal ideal. Now we have a one-to-one correspondence between the points in A n k and the maximal ideals of k[x 1,..., x n ]. We know further that there is a bijection between radical ideals of k[x 1,..., x n ] and subschemes of A n k, given in one direction by the map V. We can describe the inverse of V nicely in this case. 4

5 Definition 2.3. If X is a subset of points in A n k, we define the ideal of X to be I(X) := {f k[x 1,..., x n ] : f(p ) = 0 for all P X}. If X is a subset of points in A n k, then V (I(X)) = X. However, it is not always true that for an ideal J of R, I(V (J)) = J. For example, in A 2 k, I(V ((x2 ))) = (x). However, if we restrict our attention to radical ideals, I and V are inverse. Corollary 2.4. If J is a radical ideal of R. then I(V (J)) = J. Proof. Clearly J I(V (J)). Our ring is Noetherian, so we can write J = (f 1,..., f m ) for some polynomials f i k[x 1,..., x n ]. Suppose now that f I(V (J)). We can then include f into the ring k[x 1,..., x n, t], then consider the ideal A = J + (ft 1) in this ring. A point in A n+1 k is in V (A) if and only if it is both in V (J) in A n k and it is in V (ft 1). But f I(V (J)) so both conditions cannot be met at the same time. Therefore V (A) =. By the previous corollary of Hilbert s Nullstellensatz, this implies that A = k[x 1,..., x n, t]. Therefore there are g 0,..., g m k[x 1,..., x n, t] such that g 0 (ft 1) + g 1 f g m f m = 1. Let N be the highest power of t appearing on the left hand side. We can multiply both sides of the equation by f N to get f N g 0 (ft 1) + f N g 1 f f N g m f m = f N. Because f k[x 1,..., x n ], we can replace each g i (x 1,..., x n, t) with G i (x 1,..., x n, ft). Then G 0 (x 1,..., x n, ft)(ft 1) + G 1 (x 1,..., x n, ft)f G m (x 1,..., x n, ft)f m = f N. If we now consider this polynomial in the quotient k[x 1,..., x n, t]/(ft 1), we have G 1 (x 1,..., x n, 1)f G m (x 1,..., x n, 1)f m = f N. Since k[x 1,..., x n ] can be included into this quotient ring, hence the equation holds there as well. It shows that f N J. Since r(j) = J, f J. The ideals in polynomial rings are generated by a finite number of polynomials. The previous results show that we can identify closed sets in A n k as the solution sets to a finite list of polynomials in k[x 1,..., x n ]. Definition 2.5. A set X A n k is an algebraic set if it is a solution set to a finite number of polynomial equations in k[x 1,..., x n ]. Note that given an algebraic set X A n k, we can view X as an affine subscheme of An k, as it is X = Spec(k[x 1,..., x n ]/I(X)), and with the quotient map by the ideal I(X) inducing the inclusion map of X into A n k. Example 2.6. In A 2 C, the solutions set to one polynomial is called a plane curve. This name can be a bit misleading, since a plane curves is in fact two dimensional over R. For example, an elliptic curve, an important object of study not only in algebraic geometry but also in number theory and cryptography, is the solution set to a degree 3 equation. In fact, such a curve is a torus (doughnut)! 5

6 We have discussed morphisms of affine schemes. We can say exactly what these look like for algebraic sets. Proposition 2.7. If X A m k and Y An k are algebraic sets, then f : X Y is a morphism of algebraic sets if and only if we can write f(p ) = (f 1 (P ),..., f n (P )) for polynomials f 1,..., f n k[x 1,..., x m ] and P X. Proof. Suppose f : k[x 1,..., x n ]/I(Y ) k[x 1,..., x m ]/I(X) is a ring homomorphism. Then f must restrict to the identity on k, hence it is also a k-algebra homomorphism. It is completely determined by the n polynomials g i = f(x i ) k[x 1,..., x m ]. For any point P X, f(p ) = (g 1 (P ),..., g n (P )) Y. 2.1 Affine varieties We now wish to study the geometric properties of algebraic sets which can be determined using algebra. The first property we will study is irreducibility. Definition 2.8. An algebraic set X is reducible if it can be written as a union X = X 1 X 2 of nonempty algebraic sets. Otherwise, X is called irreducible. If X is reducible and X 1 and X 2 can be chosed to be disjoint, then X is disconnected. Otherwise X is connected. There is a special name for irreducible algebraic sets in A n k. Definition 2.9. An irreducible algebraic set in A n k is called an (affine) algebraic variety. We will now show that the points of Spec(k[x 1,..., x n ]) are in bijection with the irreducible algebraic varieties in A n k. In other words, the subvarieties of An k are the vanishing sets of prime ideals of k[x 1,..., x n ]. Proposition If I is an ideal of k[x 1,..., x n ], then V (I) is irreducible if and only if I is prime. Proof. First suppose V (I) is irreducible. If the product fg I, then (fg) I, so V (fg) V (fg). Then since V (fg) = V (f) V (g), X = (V (f) X) (V (g) X). Both V (f) X and V (g) X are algebraic sets, so X = V (f) X or X = V (g) X. Hence f or g is in I. This shows I is prime. Now suppose I is a prime ideal. If we can write X = X 1 X 2 for algebraic sets X 1 = V (I 1 ) and X 2 = V (I 2 ), for radical ideals I 1 and I 2, then I = I 1 I 2. Since I is prime, I = I 1 or I = I 2. Hence X is irreducible. Every algebraic set can be written as a finite union of algebraic varieties. In this way, Spec(k[x 1,..., x n ]) contains the building blocks of all the algebraic sets in A n k. 2.2 Dimension When thinking of a geometric object we can visualize, we have an intrinsic sense of dimension. We know that a line is one-dimensional, a square and a triangle are two-dimensional, and we are three-dimensional. But in this abstract setting, we will often describe our varieties as solutions sets of several polynomials in some large dimensional space. We need a definition of dimension that fits in this setting. 6

7 Definition The dimension of an algebraic set X is the largest number n so that there is a chain of closed irreducible subsets of X X 0 X 1 X n 1 X n = X. Example The chain shows that the dimension of A n k exactly n. 0 A 1 k An 1 k A n k is at least n. It is hard to show, however, that the dimension is We can change the question of dimension into an algebraic question. By using the map I we can instead look for the longest chain of ideals I(X) I(X n 1 ) I(X 1 ) I(X 0 ) k[x 1,..., x n ]. In particular, we could require that the subsets X i are irreducible, and look for the longest chains of prime ideals of this form. Or rather, we could look at the longest chain of prime ideals in the ring k[x 1,..., x n ]/I(X). I(X n 1 )/I(X) I(X 1 )/I(X) I(X 0 ) k[x 1,..., x n ]/I(X) Definition The Krull dimension of a ring R is the largest n such that there is a chain of prime ideals P 0 P 1 P n. It is immediate that the dimension of the algebraic set X in A n k dimension of the ring k[x 1,..., x n ]/I(X). is the same as the Krull Example The ring k[x] is a principal ideal domain. Hence the longest possible chain of prime ideals is 0 (f) where f is irreducible. This shows A 1 k is one-dimensional. Example Suppose P is a prime ideal in the ring k[x, y]. If P is maximal, we know P = (x a, y b) for some a, b k. Otherwise, P must be finitely generated, and k[x, y]/p must be an integral domain. In fact, we can show P = (f) for some irreducible polynomial f, and so the longest possible chain of prime ideals is 0 (f) (x a, y b), showing A 2 k is two-dimensional. Example Suppose f k[x, y] is not a unit. Then (f) is contained in some maximal ideal (x a, y b) for a, b k. The existence of a chain of prime ideals (f) P 1 P 2 (x a, y b) would contradict that A 2 k is two-dimensional. By the lattice isomorphism theorem, k[x, y]/(f) is one-dimensional. So V (f) is a one-dimensional subvariety (a curve). 2.3 Plane Curves As seen in the previous subsection, any nonconstant polynomial f(x, y) k[x, y] defines a onedimensional subvariety. Definition A plane curve is the vanishing set of a single polynomial in A 2 k. 7

8 The curve V (f) will be irreducible if and only if the polynomial f(x, y) is irreducible in k[x, y]. For example, the polynomial f(x, y) = x 2 + y defines an irreducible curve which we might view as an upside-down parabola. However, f(x, y) = xy defines the union of the two coordinate axes. However, both curves share a property, called their degree. Definition The degree of a plane curve is the degree of the defining equation of the curve. It can also be defined as the number of intersection points between the curve and a general line, but we would need more machinery to make this alternate definition precise. Notice that the reducible plane curve V (xy) is a union of irreducible degree 1 plane curves, V (x) and V (y). It is always true that degree is additive in this way. The point (0, 0) is special on the curve V (xy). It is where the two irreducible components of the curve meet. We would like a way to detect special points on a curve. First, we look at some examples. Example Let f(x, y) = y 2 x 3 x 2. This is a degree 3 irreducible curve. Despite being irreducible, there is a special point on this curve. (This curve is called a nodal cubic.) Example Let f(x, y) = y 2 x 3. Again, this is an irreducible degree 3 curve. The special point on this curve is different in some sense from the the special point on the curve in the previous example. (This curve is called a cuspidal cubic). How can we measure this. Definition A point P = (a, b) on a curve V (f(x, y)) is a multiple point if f x (P ) = f y (P ) = 0. If we write f(x, y) = f m + + f n where f i is a sum of monomials of degree exactly m, then the multiplicity of f(x, y) at P = (0, 0) is defined to be m. If we write f m = L r 1 1 Lf j j for linear monomials L i, then the lines defined by the L i are called the tangent lines to f(x, y) at P = (0, 0). If f m is the product of m distinct lines, P is called a node of f(x, y). Example When f(x, y) = y 2 x 3 x 2, f x (x, y) = 3x 2 2x and f y (x, y) = 2y. These both vanish only at the point (0, 0). The multiplicity of f at (0, 0) is 2. We can write y 2 x 2 = (y x)(y + x) as the product of two distinct tangent lines, so (0, 0) is a node. Example When f(x, y) = y 2 x 2, again f x and f y both vanish at the point (0, 0). The multiplicity is again 2 at this point. However, f 2 = y 2 is a product of the same line twice, so (0, 0) is not a node. If P = (a, b) is not (0, 0), then we can apply the methods of the previous definition to f(x + a, y + b) to determine is P is a multiple point, and what its tangent lines are. There is another more algebraic way to test for multiple points, which can be applied more generally to points on affine schemes. Definition Let M be the maximal ideal in k[x, y]/i(x) defining the point P in X. Then M/M 2 is the cotangent space to P, and the dimension of M/M 2 as a k-vector space is the dimension of the cotangent space at P. Definition A point P on X is a singular point if the dimension of the cotangent space at P is larger than the dimension of X. Example Let f(x, y) = y 2 x 3. The maximal ideal (x, y) + (f) in k[x, y]/(f) defines the point (0, 0). As a k=vector space, (x, y)/(x, y) 2 + (f) is generated by the two elements x and y. On the other hand, if g(x, y) = y x 2 is smooth at (0, 0), the k-vector space (x, y)/(x, y) 2 + (g) is already generated by x, since y = x 2 = 0 in this space. 8