Algebraic Geometry. Instructor: Stephen Diaz & Typist: Caleb McWhorter. Spring 2015


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1 Algebraic Geometry Instructor: Stephen Diaz & Typist: Caleb McWhorter Spring 2015 Contents 1 Varieties Affine Varieties Projective Varieties Homogenization and Dehomogenization Morphisms 18 3 Rational Maps Blowing Up
2 1 Varieties Algebraic Geometry Affine Varieties These notes were taken in Spring 2015 in a course taught by Professor Steven Diaz. In some places, notation/material has been changed or added. Any errors in this text should be attributed to the typist Caleb McWhorter and not the instructor or any referenced text. Any credit as to the organization of the topics, choice of examples, et cetera should be attributed to the referenced texts as well as the instructor. For the purposes of this course, all rings will be assume to be commutative rings with identity. All ring homomorphisms ϕ : R S are assumed to have the property that ϕ(1 R ) = 1 S. Furthermore, R is considered an ideal of R but is not considered a prime ideal (despite the fact that it actually is). Finally, the zero ideal is considered to be a prime ideal when it is; that is, when the ring is an integral domain, the zero ideal is prime, and when the ring has zero divisors the zero ideal is not a prime ideal. To define algebraic geometry, we could say that it is the study of the solutions of systems of polynomial equations... First, we shall discuss varieties. This was the initial methodology of algebraic geometry during the 1960s and before. However, the modern approach is via schemes. We shall come to this later in the course. Let k be a fixed algebraically closed field (as will be the general assumption in the course). Note that much of what shall immediately follow does not require the stringent restriction to algebraically closed fields (or in some cases even fields). Definition 1.2 (Affine nspace). Denote k n as the ordered ntuples of elements of k by A n (k) or A n k. However, this is often written An when the field k is clear. This is called the affine nspace over k. The polynomial ring for this space, k[x 1, x 2,, x n ], is denoted A. Note that p A n and f A so that f(p) k. This means that the elements of A can be thought of as functions from A n to k. Definition 1.3 (Zero Set). For f A, the zero set of f is the set {p A n f(p) = 0} and is denoted Z(f). For T A, we define Z(T ) = Z(f) = {p A n f(p) = 0 for all f T } f T Throughout these notes I, Caleb McWhorter, will take the short hand convention of writing Z(f 1, f 2,, f n ) when what should be written as Z({f 1, f 2,, f n }). Definition 1.4 (Algebraic Set). A subset Y of A n is called an algebraic set if and only if Y = Z(T ) for some T A. Notice that the above definitions hold for any ring R. Proposition 1.5. (i) If T A and J is the ideal generated by T, then Z(T ) = Z(J). So every algebraic set has the form Z(J) for some ideal J A. (ii) Every algebraic set is of the form Z(T ) is of the form Z(T ) for some finite set T.
3 3 C. McWhorter Proof: (i) First, we show that Z(T ) Z(J). Let p Z(T ). Then for all f T, we know that f(p) = 0. If g J, then g = n i=1 a if i, where a i A and f i T. As a i (p)f i (p) = 0, it is clear that g(p) = 0. Now we show that Z(J) Z(T ). If every g J is such that g(p) = 0, then f(p) = 0 for all f T as T J. (ii) This follows trivially from the previous part. Note that for part (a) of the previous proposition, k could be any ring. Since (b) follows immediately from (a), the same could be said of it. Proposition 1.6. Z(T 1 ) Z(T 2 ) = Z(T 1 T 2 ). Therefore, the finite union of algebraic sets is algebraic. Z ( i I T ) i = i I Z(T i). Therefore, the arbitrary intersection of algebraic sets is algebraic. = Z(1) and A n = Z(0). That is,, A n are algebraic sets. Proof: (i) First, we show that Z(T 1 ) Z(T 2 ) Z(T 1 T 2 ). Let p Z(T 1 ) Z(T 2 ). We know that f(p) = 0 for all f T 1 and g(p) = 0 for all g T 2. Therefore, if h T 1 T 2, we know that h = fg for some f T 1, g T 2. But then h(p) = f(p)h(p) = 0. We now show that Z(T 1 T 2 ) Z(T 1 ) Z(T 2 ). Suppose that p Z(T 1 T 2 ). Suppose that p / Z(T 1 ) Z(T 2 ). Without loss of generality, suppose that p / Z(T 1 ). Then there must be a f T 1 such that f(p) 0. For any g T 2, we know that f(p)h(p) = 0 0 = 0. As k is a field (hence an integral domain), we know that h(p) = 0. Therefore, p Z(T 2 ). But then p Z(T 1 ) Z(T 2 ), a contradiction. (ii) To see that Z ( i I T ) i = i I Z(T i), observe that this occurs if and only if every polynomial in every T i is such that f i (p) = 0, where p Z ( i I T i). That is, fi vanishes at p for all i. But this occurs if and only if p i I Z(T i). (iii) This is clear as f = 1 has no zeros and f = 0 is always zero. Notice that we can define a topology using Proposition 1.6 by defining the closed sets to be the algebraic sets in A n. Definition 1.7 (Zariski Topology). Define a topology on A n by letting the closed sets be the algebraic sets in A n. Equivalently, define a topology on A n by defining the open sets to be the complement of algebraic sets. This topology on A n is called the Zariski topology. Observe that in the above proposition, we only need k to be an integral domain. Recall that the Hilbert Basis Theorem says that k[x 1, x 2,, x n ] is a noetherian ring so that every ideal is finitely generated.
4 Algebraic Geometry 4 Example 1.8. What are the algebraic subsets of A 1? It is clear that and A 1 are algebraic subsets. The only other algebraic subsets are the finite point sets. Why? As k is a field, it a PID. The algebraic subsets are generated by an ideal, which in a PID are generated by single elements. As k is algebraically closed, this generating polynomial, say f, can be written as f(x) = c(x a 1 )(x a 2 ) (x a n ) for some c, a i A 1. But then Z(f) = {a 1, a 2,, a n }. Moreover, observe that this topology (the Zariski topology) is not Hausdorff. Example 1.9. In A n, any finite set is algebraic. Observe that the finite union of algebraic sets is algebraic. Therefore, it suffices to show that a single point is algebraic. Let (a 1, a 2,, a n ) A n. Observe that Z(x 1 a 1, x 2 a 2,, x n a n ) = {(a 1, a 2,, a n )}. If k is finite, then the Zariski topology is the discrete topology. If k is infinite, then the Zariski topology on A 1 is not T 2 because it is the complement of a closed set. Hence, it is necessarily infinite. Definition 1.10 (Irreducible). A nonempty subset Y of a topological space X is irreducible if it cannot be written as a union of two closed proper subsets Y = Y 1 Y 2 of Y in the induced topology. Otherwise, the subset Y is said to be reducible. Note that we do not consider to be irreducible. Example When k is infinite, A 1 k is irreducible as the only proper closed subsets are and the finite subsets. Example In A 2, Z(xy) = Z(x) Z(y) is reducible and can be pictured as Definition 1.13 (Affine Algebraic Variety). An affine algebraic variety (or more commonly written algebraic variety) is an irreducible closed subset of A n with the induced topology. An open subset of an affine variety is a quasiaffine variety. Definition Let Y A n be any subset. We define the ideal of Y to be I(Y ) = {f A f(p) = 0 for all p Y }. Observe we have two maps: { subsets of A n } I { subsets of k[x 1,, x n ]} Z However, these maps are not necessarily inverses of each other. Shortly, we will find subsets of these sets so that these maps are indeed inverses and we obtain a onetoone correspondence between these subsets. First, we confirm that the ideal generated by Y is indeed an ideal of A. Proposition Let Y A n be any subset. Then I(Y ) is an ideal in A. Proof: We know that 0 I(Y ) as 0 vanishes everywhere. Let f, g I(Y ). Then if p Y, we know that (f +g)(p) = f(p)+g(p) = 0+0 = 0. If p Y and h A, then (hf)(p) = h(p)f(p) = h(p) 0 = 0. Therefore, I(Y ) is an ideal of A. For the following proposition, we will need the following deep theorem.
5 5 C. McWhorter Theorem 1.16 (Hilbert s Nullstellensatz). Let k be an algebraically closed field, let a be an ideal in A = k[x 1,, x n ], and let f A be a polynomial such that f vanishes at all points of Z(a). Then f n a for some integer n > 0. Proposition (i) If T 1 T 2 are subsets of A, then Z(T 1 ) Z(T 2 ). (ii) If Y 1 Y 2 are subsets of A n, then I(Y 1 ) I(Y 2 ). (iii) For any Y 1, Y 2 A n, we have I(Y 1 Y 2 ) = I(Y 1 ) I(Y 2 ). (iv) For any ideal J A, I(Z(J)) = J. (v) For any subset Y A n, we have Z(I(Y )) = Y, the closure of Y in the Zariski topology. Proof: (i) If p Z(T 2 ), then f(p) = 0 for all f T 2. But then f(p) = 0 for all f T 1. Therefore, p Z(T 1 ). (ii) If f I(Y 2 ), then f(p) = 0 for all p Y 2. But then f(p) = 0 for all p Y 1 so that f I(Y 1 ). (iii) We know that f I(Y 1 Y 2 ) if and only if f(p) = 0 for all p Y 1 Y 2. That is, if f(p) = 0 for all p Y 1 and f(p) = 0 for all p Y 2. But this occurs if and only if f I(Y 1 ) and f I(Y 2 ). However, this occurs if and only if f I(Y 1 ) I(Y 2 ). (iv) Recall that J = {f A f n J for some n > 0} and the result is immediate. (v) We know that Z(I(Y )) is closed as it is an algebraic set. We want to show that Z(I(Y )) Y. Suppose that p Y, then f(p) = 0 for all f I(Y ). But then p Z(I(Y )). We want to show that this is the smallest closed set contained in Y. Let W be a closed set. Say W = Z(J). Then I(W ) J. Then for all f J, f(p) = 0 for all p Z(J) = W. But then Z(J) = W so that f I(W ). Using (i), we know that Z(I(W )) Z(J). From (ii) with Y = W, we know Z(I(W )) W. Therefore, W = Z(I(W )). Now let W be any closed set containing Y. Then W Y and by (ii), we know I(W ) I(Y ). By (i), we know that Z(I(W )) Z(I(Y )). But from above, as W is closed, we know that W Z(I(Y )). Therefore, Z(I(Y )) is the smallest closed set containing Y. That is, Z(I(Y )) = Y. Remark Note that (iv) uses the Nullstellensatz, which truly requires k to be algebraically closed. See the examples below to see this. Example Let k = R and let J = (x 2 + x + 1) R[x]. Observe that Z(J) =. The function f(x, y) = 1 vanishes at all points of Z(J)  there are none. But no power of f(x, y) is in J.
6 Algebraic Geometry 6 Example Let k = R and let J = (x 2 + y 2 ) R[x, y]. Then Z(J) = {(0, 0)}. Then x vanishes at all points of Z(J) but no power of x is in J. Corollary There is a onetoone inclusion reversing correspondence between the algebraic sets in A n and radical ideals of A given by Y I(Y ) and J Z(J). Furthermore, the irreducible algebraic sets correspond to prime ideals. Proof: We have the following diagram: I { subsets of A n } { subsets of k[x 1,, x n ]} Z I { Algebraic sets } { radical ideals } Z So for any subset T A, we know that Z(T ) is algebraic. Therefore, we can restrict Z as claimed. The set I will map to radical ideals. Now let J be a radical ideal. We know that I(Z(J)) = J = J. Let Y be an algebraic set. Then Z(I(Y )) = Y = Y, as Y is an algebraic set  hence closed. Now given an algebraic set Y, we want to show that Y is irreducible if and only if I(Y ) is prime. We do this by showing the contrapositive: Y is reducible if and only if I(Y ) is not prime. Suppose that Y is reducible: Y = Y 1 Y 2. Now as Y is closed, we know that Y 1, Y 2 are closed in A n. Now Y 1 Y 2 nor does one contain the other as Y 1, Y 2 are proper subsets of Y. Therefore, we can choose p Y 1 such that p / Y 2. The fact that Y 2 is closed implies that Z(I(Y 2 )) = Y 2 so that there is a f I(Y 2 ) with the property f(p) 0. Let q / Y 1 and q Y 2.Then for all g I(Y ), we know that g(q) 0. But then f, g / I(Y ) as p, q Y. Finally, fg I(Y ) as f vanishes on Y 2 and g vanishes on Y 1 so that the product vanishes on the union Y 1 Y 2 = Y, so that I(Y ) is not prime. Now suppose that I(Y ) is not prime. Then there are f, g / I(Y ) but fg I(Y ). That is, we know that Z(f) Y, Z(g) Y but Z(fg) Y. We know that Z(f) Y, Z(g) Y are both proper closed subsets of Y or are empty. We also know that Y Z(fg) is Y as Z(fg) Y. But then ( ) Y = Y Z(fg) = Y Z(f) Z(g) = (Y Z(f)) (Y Z(g)) Neither of these sets on the right can be empty as then the original intersection would not be all of Y. Therefore, this shows that Y is reducible. Example A n is irreducible as it is the zero ideal in A, which is prime as A is an integral domain. That is, I(A n ) = {0}. Proposition Let k be a field and 0 f k[x 1, x 2,, x n ]. Then there exists p A n with f(p) 0. Proof: We use induction on n. The case where n = 1 follow trivially as any nonzero polynomial has finitely many roots and the field k is infinite. Now assume the statement is true for n. We prove
7 7 C. McWhorter the statement for n+1. Let f k[x 1, x 2,, x n+1 ]. Assume that f utilizes each of the n+1 variables, otherwise by the hypothesis we would be done. Write f as a m x m n+1 + a m 1x m a 1 x n+1 + a 0, where a i k[x 1,, x n ]. We know that f 0 for some a i 0. There are b 1, b 2,, b n A n such that a j (a 1, a 2,, a n ) 0. Then f(b 1, b 2,, b n, x n+1 ) k[x n+1 ] is nonzero. That is, there is a b n+1 A such that f(b 1, b 2,, b n, b n+1 ) 0. Note that the proposition is clearly false for a finite field. Let k = {x 1, x 2,, x n }. Then the polynomial f(x) = (x x 1 )(x x 2 ) (x x n ) k[x] is nonzero but vanishes at every point of k. Now given T 1, T 2 A, when is it that Z(T 1 ) = Z(T 2 )? If k is algebraically closed, then Z(T 1 ) = Z( T 1 ) = Z( T 1 ) Z(T 2 ) = Z( T 2 ) = Z( T 2 ) so that Z(T 1 ) = Z(T 2 ) if and only if T 1 = T 2. Notice that this takes a rather difficult geometric question and transforms it into a (hopefully simpler) algebraic question. Example Let f A be irreducible. Then f is prime as A is a UFD. But then Z(f) is irreducible. When we consider A n, if n = 2 then these irreducible functions are curves, for n = 3 they are surfaces, and for n > 3 they are called hyper surfaces. Example A maximal ideal M k[x 1,, x n ] must correspond to a minimal closed set. Therefore, M must be a point p = (a 1, a 2,, a n ). Then I(p) (x 1 a 1, x 2 a 2,, x n a n ). But I(p) can contain nothing more as k[x 1, x 2,, x n ]/(x 1 a 1, x 2 a 2,, x n a n ) = k and k is a field so that (x 1 a 1, x 2 a 2,, x n a n ) is maximal. Note that this needs k to be algebraically closed. For example, x R[x] is maximal because R[x]/x = C. But x has no zeros. Definition 1.26 (Affine Coordinate Ring). If Y A n is an affine algebraic set, we define the affine coordinate ring A(Y ) of Y to be A/I(Y ). We think of the elements of A(Y ) as functions f : Y k given by polynomials. Now we ask ourselves the question, when do f, g A give the same function f, g : Y k? If f g vanishes on Y, then f g I(Y ). Therefore, f, g induce the element of A/I(Y ). Remark We know that A(Y ) is an integral domain if and only if Y is irreducible. This is because A/I(Y ) is an integral domain if and only if I(Y ) is prime. Furthermore, A(Y ) has no nilpotent elements as I(Y ) is a radical ideal. We know that any finitely generated kalgebra that is an integral domain is A(Y ) for some variety Y. To see this, suppose that B is a finitely generated kalgebra that is also an integral domain. We have a surjection f : k[x 1, x 2,, x n ] B. But then k[x 1, x 2,, x n ]/ ker f = B, which is an integral domain so that ker f is prime. Set Y A n so that Y = Z(ker f). Then we know that I(Y ) = I(Z(ker f)) = ker f = ker f is prime. Example We know that A(p) = k, where p is a point p A n. Note that this require k be algebraically closed.
8 Algebraic Geometry 8 Definition 1.29 (Noetherian Topological Space). A space X is noetherian if every sequence Y 1 Y 2 of closed subsets, there is an integer r such that Y r = Y r+1 =. That is, X satisfies the descending chain condition for closed subsets. Example We know that A n with the Zariski topology is noetherian. Given Y 1 Y 2, we have a chain I(Y 1 ) I(Y 2 ). But we must have I(Y r ) = I(Y r+1 ) = for some r as k[x 1,, x n ] is noetherian. But then Y r = Y r+1 =. Any algebraic set is necessarily noetherian as any closed subsets of it are closed in A n. Is R n with its usual topology noetherian? The answer is no. Let Y i = [a i, b i ], where a i = i ( 1 j j=1 2) and b i = 2 a i. This creates a family {Y i } of nested closed subsets which does not stabilize. Proposition In a noetherian topological space, every nonempty set of closed sets has a minimal element. Proof: Let S be a nonempty set of closed sets. Choose Y 1 S. If Y 1 is minimal, we are done. If not, choose Y 2 S such that Y 1 Y 2. If Y 2 is minimal, we are done. If not, choose Y 3 S such that Y 2 Y 3. This process creates a chain Y 1 Y 2 Y 3 of closed sets. As the space is noetherian, this chain must stabilize at some integer, sa r. Then Y r = Y r+1 = so that Y r is a minimal element. The following proof should greatly remind the reader of the proof that Z is a UFD. Proposition In a noetherian topological space X, every nonempty closed subset Y can be expressed as a finite union Y = Y 1 Y r of irreducible closed subsets of Y. If we require that Y i Y j for i j, then this expression can be written uniquely. The Y i are called the irreducible components of Y. Proof: Let S be the set of all nonempty closed subsets of X that do not have such an expression. Assume that S. As X is noetherian, S must have a minimal element. Let Y S be a minimal element. We know that Y is not irreducible or Y = Y would be such an expression. Suppose that Y = A B, where A, B are both proper closed subsets of Y. By the minimality of Y, A, B / S. Then we can write A = A 1 A 2 A s, where each of the A i is irreducible closed subset of A. Similarly, we have a decomposition B = B 1 B 2 B t, where the B i are irreducible closed subsets of B. But then we know that Y = A 1 A 2 A s B 1 B 2 B t so that Y is a union of irreducible closed subsets of Y. This contradicts the fact that Y S. Therefore, S = so that all nonempty closed subsets of X have such an expression. Now given Y X a nonempty closed subset of X, decompose Y as in the statement of the theorem. If Y i Y j for i j, drop Y j so that one obtains a finite union. Now suppose that we require Y i Y j for i j, we want to show that the expression for Y is unique. Assume to the contrary that there are two such expressions for Y : Y = Y 1 Y 2 Y s = Y 1 Y 2 Y t We know that Y i, Y i are irreducible and that for i j, Y i Y j, Y i Y j. We know that Y i = Y Y i = Y i (Y 1 Y t ) = (Y i Y 1) (Y i Y t )
9 9 C. McWhorter But Y i is irreducible, so for some p 1, 2, 3,, t, we know that Y i = Y i Y p. But then Y p Y i. Mutatis mutandis, we obtain Y p Y j for some j = 1, 2, 3,, s. Therefore, Y i Y p Y j so that Y i Y j, a contradiction. So it must be that j = i. This shows that Y i = Y p. Without loss of generality, renumber so that i = p = 1. We claim that Y Y 1 = Y 2 Y j and Y Y 1 = Y 2 Y t. Demonstrating one proves the other mutatis mutandis, so we shall prove the first. We know that Y 2 Y s is a closed set containing Y Y 1. We need show that it is the smallest such closed set. Let W be any closed set containing Y Y 1. Fix some i 2 so that W Y i Y 1. We know Y i = (Y i W ) (Y i Y 1 ) but Y i is irreducible so either Y i = Y i W or Y i = Y i Y 1. If Y i = Y i Y 1, then Y i Y 1, a contradiction. Therefore, Y i = Y i W. But then W Y i for all i so that W Y 2 Y s. Then by induction, we obtain the necessary uniqueness. Corollary Every algebraic set in A n can be uniquely expressed as a union of varieties, no one containing another. Definition 1.34 (Dimension). If X is a topological space, we define the dim of X, denoted dim X, to be the supremum of all integers such that there is a chain Z 0 Z 1 Z 2 Z n of distinct irreducible closed subsets of X. We define the dimension of an affine or quasiaffine variety to be its dimension as a topological space. Example We know that dim A 1 = 1 as the only closed subsets of A 1 are the empty set, finite sets, and A 1. So the irreducible closed subsets of A 1 are points and the longest possible chain is {point} A 1. Remark It is usually very hard to prove directly that a maximal chain is necessarily the maximal chain. However, lower bounds are fairly easy to come up with. If Y X, then we know that dim Y dim X since Y Z 0 Z n X then dim X n dim Y. For any algebraic sets, algebra helps compute the dimension of a space by converting chains of irreducible closed subsets to a chain of prime ideals! Definition 1.37 (Height). In a ring A, the height of a prime ideal P is the supremum of all integers n such that there is a chain P 0 P 1 P n = P of distinct prime ideals. Define dim A, called the Krull dimension of A, to be the supremum of the heights of all chains of prime ideals of A. Remark The above definition is just the algebraic version of the dimension defined for varieties preceding it. Proposition If Y is an affine algebraic set, then dim Y = dim A(Y ). Proof: We know that A(Y ) = k[x 1,, x n ]/I(Y ). So if Y is an affine algebraic set in A n, then the closed irreducible subsets of Y correspond to the prime ideals of A = k[x 1,, x n ] containing I(Y ). These correspond to the prime ideals in A(Y ). So dim Y is the length of the longest chain of prime ideals in A(Y ), which is its dimension.
10 Algebraic Geometry 10 Theorem Let k be a field and let B be an integral domain which is a finitely generated kalgebra. The dimension of B is the transcendence degree of the quotient field K(B) of B over k. Furthermore, for any prime ideal P in B, we have height P + dim B/P = dim B Proof: Matsumura, Commutative Algebra, Ch 5, 14. Remark The condition that B is finitely generated is absolutely necessary! The fact that B need be an integral domain is so that the quotient field be well defined. Proposition dim A n = n Proof: We know that the dimension of A n is the dimension of its affine coordinate ring: k[x 1,, x n ], which is n following immediately from the proposition above. Lemma Any nonempty open subset of an irreducible space is irreducible and dense. Proof: Suppose that Y is irreducible and let U Y be an open subset of Y. Assume that U Y is not dense in Y. Then the set Y \ U Y is closed. But then Y = U Y \ U, contradicting the fact that Y is irreducible. Therefore, U is dense. We need only show that it is irreducible. Suppose that U is reducible. We can then find closed sets Y 1, Y 2 such that U = (U Y 1 ) (U Y 2 ) and U Y i Y. Then each Y i is a proper closed subset of Y. Therefore, we know that U Y 1 Y 2. But this is a closed set. So we know that U Y 1 Y 2 = Y 1 Y 2. But U = Y so that we have contradicted the fact that Y is irreducible. Lemma If Y is an irreducible and closed subset of X, then Y is an irreducible space in X. Proof: Suppose that Y is not irreducible. Then we can find closed subsets Y 1, Y 2 of X with Y Y 1, Y Y 2 but Y = (Y Y 1 ) (Y Y 2 ) Since Y Y, Y = (Y Y 1 ) (Y Y 2 ) = Y (Y 1 Y 2 ) Without loss of generality, assume that Y = Y Y 1. Then Y Y 1 so that Y Y 1 = Y 1, a contradiction. The first lemma says that a quasiaffine is irreducible and dense in the variety it is open in. The second lemma states that the closure of a quasiaffine variety is the whole affine variety it is sitting in. We put these two statements together to obtain the following: Proposition If Y is a quasiaffine variety, then dim Y = dim Y.
11 11 C. McWhorter Proof: Suppose that Z 0 Z 1 Z n is a sequence of distinct closed irreducible subsets of Y. We look Z 0 Z 1 Z n, which from the second lemma above we know must be a sequence of closed irreducible subsets of Y. We need to show that these are still distinct. Suppose that Z i Z i+1. We want to show that Z i Z i+1. As Z i, Z i+1 are closed in Y, there are Y i, Y i+1 closed subsets of Y with Z i = Y i Y and Z i+1 = Y i+1 Y. Then Z i Z i+1 so that Y i+1 \ Y i and Z i+1 \ Y i. But then Z i Z i+1. From above, we know that dim Y dim Y = dim A(Y ). This shows that dim Y is finite. Therefore, we can choose a maximal chain Z 0 Z 1 Z n of distinct closed irreducible subsets of Y with n = dim Y. We know that Z 0 must be a point, otherwise we could lengthen the chain on the left by setting Z 0 = {p}, where p Z 0. Furthermore, we know that Z n = Y, otherwise we could extend the chain on the right by choosing Z n = Z n {p}, where p Y \ Z n. We claim that Z 0 Z 1 Z n is a maximal chain in Y. Again, we know that Z 0 is a point and that Z n = Y. We need only show that nothing new can be inserted at the nonends of the chain. Suppose that Z i W Z i+1, where W is closed and irreducible. We know that the Z i are maximal so that Z i Y W Y Z i+1 Y. Therefore, we must have W Y = Z i or W Y. But we know that W Y = W. So that W = Z i or W = Z i+1, a contradiction. Therefore, Z 0 Z 1 Z n is a maximal chain of distinct closed irreducible subsets of Y. We know Z 0 corresponds to a point, which corresponds to a maximal ideal M of the coordinate ring A(Y ) of Y. Furthermore, the Z i correspond to prime ideals contained in M so that height M = n. But we also know that A(Y )/M = k. Therefore, using Theorem 1.40, we know that n = dim A(Y ) = dim Y. But then dim Y = dim Y. Corollary (a) Let k be a field and let B be an integral domain which is a finitely generated kalgebra. Let p be prime ideal of B. Let p 0 p 1 p n = p be a maximal chain of prime ideals in p. Then for every i = 0, 1, 2,, n, the height of p i = i. Therefore, maximal chains compute height. (b) Let Y be an affine variety and Y 0 Y 1 Y n = Y a maximal chain of irreducible closed subsets of Y. Then dim Y = n. Proof: (a) We proceed by induction on i. If i = 0, then as the chain is maximal, there are no primes contained in p 0 so that the height of p 0 = 0. Now assume that the result is true up to i, we prove the result for i + 1. By assumption, height P i = i. Using Theorem 1.40, we have p i + dim B/p i = dim B dim B/p i = dim B i In B/p i, p i+1 /p i has height 1. In B, there are no primes between p i and p i+1 (otherwise, the original chain would not be maximal). Using Theorem 1.40, we obtain 1 + B/p i /p i+1 /p i = dim B/p i 1 + B/p i+1 = dim B i B/p i+1 = dim B (i + 1)
12 But then we have Algebraic Geometry 12 height p i+1 + dim B/p i+1 = dim B height p i+1 + dim B (i + 1) = dim B height p i+1 = i + 1 Therefore, the result follows by induction. (b) Given a maximal chain of irreducible closed subsets Y 0 Y 1 Y n = Y of Y, an affine space, we look at the functions vanishing on the Y i ; that is, we look at the I(Y i ) s. Since this is a chain of irreducible closed subsets, this corresponds to a sequence of prime ideals in A(Y ). We know that I(Y ) = 0. Furthermore, I(Y 0 ) must be maximal. Furthermore, we know that A(Y )/I(Y 0 ) = k. Then height I(Y 0 ) + dim A(Y )/I(Y 0 ) = dim A(Y ) But we know that dim A(Y )/I(Y 0 ) = 0 as k is a field and the only prime ideal is 0. But then height I(Y 0 ) = dim A(Y ) = n. Theorem 1.47 (Krull s Hauptidealsatz). Let A be a noetherian ring and f A be an element which is neither a zero divisor nor a unit. Then every minimal prime ideal p containing f has height 1. Proposition A noetherian integral domain is a UFD if and only if every prime ideal of height 1 is principal. Proposition A variety Y in A n has dimension n 1 if and only if it is the zero set Z(f) of a single irreducible f A. Proof: If f is an irreducible polynomial, we know that Z(f) is a variety as (f) is prime. But then I(Z(f)) = f. But as (f) is prime, it must be minimal prime ideal containing f. But then (f) has height 1. Using the fact that height(f) + dim A/(f) = dim A. We know that dim A = n so dim A/(f) = A(Z(f)) = 1. Suppose that dim Y = n 1. We know that I(Y ) = p. But then height p + dim A/p = dim A height p + (n 1) = n so that height p = 1. Since k[x 1, x 2,, x n ] is a UFD, we know that p is principal. But then p = (f) and f must be irreducible for some f A Projective Varieties Definition 1.51 (Graded Ring). Let R be a ring. A grading of R is an expression of the additive group (R, +) as an internal direct sum R = i=0 R i (R i = 0 is permissible) with the property that if a R i and b R j, then ab R i+j, written R i R j R i+j. A graded ring is a ring together with a fixed grading. An element of R d is said to be homogeneous of degree d.
13 13 C. McWhorter Example Let R = k[x 1, x 2,, x n ]. Let R i = {f R all monomials in f having degree i} {0}. Example In C[x, y], x 2 + xy 17y 2 is homogeneous of degree 2 while x 3 xy is not. Let R be a graded ring. Each 0 f has a unique expression f = f 0 + f f d, where f i is homogeneous of degree i, f d 0. Proposition Let R be a graded ring and I R an ideal. Then the following are equivalent: (i) I is generated by homogeneous elements. (ii) I = I R i as a group. (iii) Given f R, f = f 0 + f f d, where f i homogeneous of degree i, then f I if and only if f i I for all i. An ideal satisfying any of the above conditions is said to be a homogeneous or graded ideal. Proof: The equivalency of (ii) and (iii) follows immediately from the definition of the direct sum. To see that (iii) implies (i), write f I as f = f 0 + f f d, where f i are homogeneous of degree i. Then I is generated by the f i s for all f s. To see (i) implies (iii), say that I is generated by homogeneous elements {f α } α A, each f α homogeneous of degree d α. Choose F I, then F = n i=1 a if αi, where a i R. So F = F F d, where F i is homogeneous of degree i. (This proof needs to be finished!) Proposition Let R be a graded ring. (i) If I, J are homogeneous ideals of R, then I J, IJ, I J, and I are all homogeneous ideals. (ii) Let I be a homogeneous ideal. Then I is prime if and only if given two elements f, g R, if fg I then f I or g I. (iii) If f, g R are nonzero. Then R is an integral domain then fg is homogeneous if and only if f, g is homogeneous. Proof:? For a homogeneous ideal I R i, we denote this by I i. We know that R/I = R i / I i = Ri /I i Furthermore, R/I is naturally a graded ring with (R/I) i = R i /I i. Assume that R is noetherian graded ring and I R a homogeneous ideal (i) I can be generated by homogeneous elements. (ii) I can be generated by finitely many elements. Can I be generated by finitely many homogeneous elements.
14 Algebraic Geometry 14 In fact, the answer to both these questions is yes. We now can discuss projective spaces. We can think of projective spaces as an affine plane where parallel lines meet at infinity. Definition 1.56 (Projective Space). Let k be a (algebraically closed) field. Projective nspace over k, denoted P n k or Pn, is defined to be the set of equivalence classes of (n+1) tupes of (a 0, a 1,, a n ) of elements of k, not all zero, under the equivalence relation given by (a 0,, a n ) (λa 0,, λa n ) for all λ k. That is, P n is the quotient of the set A n+1 \ {(0, 0,, 0)} under the equivalence relation which identifies points lying on the same line through the origin. An element of P n is called a point. If p P n is a point, then any (n + 1)tuple (a 0,, a n ) in the equivalence class of p is called the set of homogeneous coordinates for p. The set of homogeneous coordinates of (a 0,, a n ) is denoted [a 0,, a n ]. So P n is the set of lines through the origin in A n+1 or the set of onedimensional subspaces of A n+1. We can form the polynomial ring S = k[x 0,, x n ] into a graded ring by taking S d to be the set of linear combinations of monomials of total weight d in x 0, x 1,, x n. If f S is a polynomial, we cannot necessarily define a function on P n because of the nonuniqueness of homogeneous coordinates. However, this can be fixed if f is homogeneous. Suppose that f is a homogeneous polynomial of degree d, then f(λa 0,, λa n ) = λ d f(a 0,, a n ). Then the property of f being zero or not depends only on the equivalence class [a 0,, a n ]. Considering this, f gives a function from P n to {0, 1} by f(p) = 0 if f(a 0,, a n ) = 0 and f(p) = 1 if f(a 0,, a n ) 0. Then if T is any set of homogeneous elements of S, we can define the zero set of T to be Z(T ) = {p P n f(p) = 0 for all f T }. If a is a homogeneous ideal of S, we define Z(a) = Z(T ), where T is the set of all homogeneous elements in a. Definition Given a homogeneous polynomial f S = k[x 0,, x n ], we define the zeros of f to be Z(f) = {p P n f(p) = 0}. If T is a set of homogeneous elements of S, we define the zero set of T to be Z(T ) = {p P n f(p) = 0 for all f T }. If a is a homogeneous ideal of S, we define Z(a) = Z(T ), where T is the set of all homogeneous elements in a. A subset Y P n is algebraic if and only if Y = Z(T ) for some T S of homogeneous elements. Proposition Let T S consist of homogeneous elements. Let I be the ideal generated by T (as S is noetherian, T has a finite generating set), then Z(I) = Z(T ). Proof: This proof is the same as the affine case mutatis mutandis. One need only be careful using the set of homogeneous points. Thus, every algebraic subset of P n is of the form Z(I), where I is a homogeneous ideal and Z(f 1,, f n ), where f i is a homogeneous polynomial. Proposition The union of two algebraic sets is algebraic. The intersection of any family of algebraic sets is an algebraic set. Finally, the empty set and the whole space is an algebraic set. Proof: This is again the same as in the affine case, mutatis mutandis. Definition The Zariski topology on P n is the topology formed by taking the open sets to the complement of algebraic sets.
15 15 C. McWhorter Definition 1.61 (Projective Algebraic Variety). A projective algebraic variety (or projective variety) is an irreducible algebraic set in P n with the induced topology. An open subset of a projective variety is a quasi projective variety. The dimension of a projective or quasi projective variety is the dimension as a topological space. If Y P n, the homogeneous ideal of Y in S, denoted I(Y ), is the ideal generated by {f S f homogeneous and f(p) = 0 for all p Y }. If Y is an algebraic set, we define the coordinate ring of Y to be S(Y ) = S/I(Y ). Let H i = Z(x i ) and U i = P n \ Z(x i ). Observe that the U i form an open cover of P n. ( n ) C U i = i=0 n Ui C = i=0 n H i = i=0 n Z(x i ) = {p P n p = [a 0,, a n ] with a i = 0} i=0 So that H i = Z(x i ) = {[a 0,, a n ] P n a i = 0} and U i = {[a 0,, a n ] P n a i 0}. Each point in U i has a unique set of homogeneous generating coordinates with a 1 in the ith spot. Any such point in U i looks like A n. So we can define ϕ i : A n U i by ϕ(a 1,, a n ) = [a 1,, 1,, a n ]. Then the map ϕ i is a homeomorphism of U i with the induced topology to A n with the Zariski topology. If f S is a linear homogeneous polynomial, then the zero set of f, Z(f), is called a hyperplane. We denote the zero set of x i by H i. We let U i be the open set P n \ H i. Then P n is covered by the open sets U i as p = (a 0,, a n ) is a point, then at least ( one of the a i 0. Therefore, p U i. We can define a mapping ϕ i : U i A n by (a 0,, a n ) a0 a i,, an a i ), where (a 0,, a n ) U i and the a i /a i is omitted in the image point. It is clear that ϕ i is well defined since the ratios a i /a i are independent of the choice of homogeneous coordinates. Proposition The map ϕ i is a homeomorphism of U i with its induced topology to A n with its Zariski topology. Corollary If Y is a projective (or quasi projective) variety, then Y is covered by the open sets Y U i which are homeomorphic to affine (or quasi affine) varieties via the mapping ϕ i defined above Homogenization and Dehomogenization Let S = k[x 0,, x n ] and A = k[x 1,, x n ]. Let S i be the set of homogeneous elements of S. To make notation simpler, we will do this for i = 0. However, one can do it for any i = 0, 1, 2,, n. Let α : S A be a ring homomorphism given by evaluation at x 0 = 1. That is, α(f(x 0,, a n )) = f(1, x 1,, x n ). Observe that we have α(fg) = α(f)α(g) and α(f + g) = α(f) + α(g) when α is restricted to S i ; that is, when it is the function α : S i A. The ring map α is called dehomogenization. ( Now let β : A S i be given by f(x 1,, x n ) x deg f 0 f x1 x 0,, xn x 0 ). We look at a monomial of f: x i 1 1 x i 2 2 x in n, where i 1 + i i n deg f. Then we have x deg f 0 i x x i 2 x i 1 1 x i 1 xin n 0 x in 0 = x deg f (i 1+ +i n) 0 x i 1 1 x i 2 2 x in n
16 Algebraic Geometry 16 having degree the same as that of f. However, this is not a ring homomorphism. This map does have the property β(fg) = β(f)β(g): deg f+deg g β(fg) = x0 f ( x1,, x ) ( n x1 g,, x ) n x 0 x 0 x 0 x 0 However, β(f + g) need not be β(f) + β(g) as β(f) + β(g) need not be homogeneous. Now write F = a i0 i n x i 0 0 x in n. Assume that F is not divisible by x 0. Then α(f ) = a i0 i n x i 1 1 x in n. If F is homogeneous, once we know i 1,, i n, we know i 0. ( ) i1 ( ) in β(α(f )) = x d x1 xn 0 ai0 i n x 0 = a i0 i n x i 0 0 x in n = F Furthermore, we know that α(x m 0 F ) = α(f ) so that β(α(xm 0 F )) = F. Consider a point (a 1, a 2,, a n ) A n and ϕ 0 (a 1,, a n ) = [1, a 1,, a n ] U 0 P n. Let f k[x 1,, x n ] and F k[x 0,, x n ], where F is homogeneous. Then f(a 1,, a n ) = 1 deg F f ( ) a 1 1,, an 1 = β(f)[1, a1,, a n ]. We also know F (1, a 1,, a n ) = α(f )(a 1,, a n ). It is our goal to show that ϕ 0 : A n U 0 P n is a homeomorphism onto its image. We have already seen that this map is injective and surjective. Suppose that Y A n is closed. Then Y = Z(T ), where Z k[x 1,, x n ]. We know that β(t ) = {β(f) f T }. We claim that β 0 (Y ) = Z(β(T )) U 0. Now suppose that W W 0 is closed in U 0. We know that W = U 0 W, where W is closed in P n. But then W = Z(T ), where T S n. Furthermore, α(t ) = {α(f ) F T } and ϕ 1 0 (W ) = Z(α(T )). We can perform this for any i = 0, 1,, n. We restate the result from above. Corollary If Y is a projective variety (or quasi projective variety), then Y is covered by open sets. We make a few more notes. Observe that A n = U i P n, U i = P n \ H i, where H i = Z(x i ), and H i = {[a 0,, a i 1, 0, a i+1,, a n ]} where not all the a i = 0. But this is precisely P n 1. Then in some sense we have P n = A n P n 1. Observe that P 1 is a space where we can think of parallel lines meeting at infinity. Consider the space A 2. Let Ax + By + C = 0, with at least one of A, B is nonzero and Ax + By + D = 0, where C D is a parallel line. Consider homogeneous lines. Then Z ( A X Z + B Y Z + C) = AX +BY +CZ = AX + BY + DZ. Then (C D)Z = 0 so Z = 0. So we have AX + BY = 0 and [λb, λa, 0] is a point in P 2 and it is. Often, one wants to look at a curve in different patches and see how these have to fit together. Example Consider the Veronese embedding with n = 1 and d = 2: ν : P 1 P 2 given by [a 0, a 1 ] [ a 2 0, a 0 a 1, a }{{}}{{} 2 1 ]. Now we have XZ = Y 2. Therefore, either X 0 or Z 0. Suppose }{{} X Y Z that X 0, without loss of generality assume that X = 1. Then we have Z = Y 2. Then [1, Y, Y 2 ] is the image of [1, Y ]. If Z 0, without loss of generality assume that Z = 1. Then X = Y 2 so that [Y 2, Y, 1] is the image of [Y, 1]. But then the image of ν is Z(X, Z Y 2 ), which is closed. This mapping is injective. We have inverses if one of a 0, a 1 are nonzero. If a 0 0, say a 0 = 1. Then we have [1, a 1 ] [1, a 1, a 2 1 ]. If a 1 = 1, then we have [a 0, 1] [a 2 0, a 0, 1]. x 0
17 17 C. McWhorter Example Consider the map ψ : P 1 P 1 P 3 given by Then we have W Z XY = 0. [a 0, a 1 ] [b 0, b 1 ] [a 0 b }{{} 0, a 0 b 1, a }{{} 1 b 0, a }{{} 1 b 1 ] }{{} W X Y Z There are three ways to define the Zariski topology on P r P s. 1. We know that P r is covered by the affine open sets U r = A r. But then P r P s is covered by the sets {U i U j } 1 i r,1 j s. Then we can give each U i U j the Zariski topology it gets from A r+s. 2. On P r, we take the homogeneous coordinates x 0,, x r and on P s we take the homogeneous coordinates y 1,, y s. A polynomial f k[x 0,, x r, y 1,, y s ] is said to be bihomogeneous of degree de if and only if it is homogeneous of degree d in x s and degree e in y s. We have λ d µ e f(a 0,, a r, b 0,, b s ) so it makes sense to ask if f(p, q) = 0 for (p, q) P r P s. Then we say Y P r P s is closed if and only if Y = Z(T ) = f T Z(f) for some set T k[x 1,, x r, y 1,, y s ] of bihomogeneous polynomials. 3. We can give P r P s the topology is must have for the Segre embedding to be a homeomorphism onto its image. All of three above methods yield the exact same Zariski topology on P r P s.
18 2 Morphisms Algebraic Geometry 18 Note that any affine variety is also a quasiaffine variety and any projective variety is also a quasiprojective variety. Definition 2.1 ((Affine) Regular). Let Y be a quasiaffine variety in A n. A function f : Y k is regular at a point p Y if and only if there is an open neighborhood Y with p U Y and polynomials g, h A = k[x 1,, x n ] such that h is nowhere 0 on U and f = g/h on U. We say that f is regular on Y if and only if it is regular at every point of Y. Remark 2.2. The previous definition can be summed up as follows: A function is regular if and only if it is locally a rational function. Lemma 2.3. A regular function is continuous when k is identified with A 1 k with the Zariski topology. Proof: We show that f 1 of closed sets is closed. We know the closed sets in A 1 k are the empty set, A 1 k, and the finite point sets. We know that f 1 (A 1 k ) = Y and f 1 ( ) =. Therefore, it is sufficient to show that f 1 ({a}) is closed for any a k. We check this locally. If Y has an open cover {U i } and f 1 ({a}) U i is closed in U i for all i, then f 1 ({a}) is closed. For p Y, find an open neighborhood U p of p such that on U p, f = g/h on U, where g, h k[x 1 ] and h is nonzero on U p. Then f 1 ({a}) U p = { q U p g(q) } h(q) = a = U p Z(g ah) where the equality holds as g(p)/h(p) = a if and only if (g ah)(p) = 0. But U p Z(g ah) is closed in the Zariski topology. Remark 2.4. If F, G k[x 1,, x n ] are homogeneous polynomials of the same degree d. Then for p P n, we know that F (p)/g(p) is a well defined element of k as F (λa 0,, λa n ) G(λa 0,, λa n ) = λd F (a 0,, a n ) λ d G(a 0,, a n ) = F (a 0,, a n ) G(a 0,, a n ) Definition 2.5 ((Projective) Regular). Let Y be a quasiprojective variety in P n. A function f : Y k is regular at a point p Y if and only if there is an open neighborhood U with p U Y such that there are homogeneous polynomials g, h k[x 1,, x n ] of the same degree and h nowhere 0 on U such that f = g/h on U. We say that f is regular on Y if it is regular at every point. Proposition 2.6. Let Y P n be a quasiprojective variety and f : Y k a function. Let p Y and assume p U i = P n \ Z(x i ). When we think of U i as A n, Y U i is a quasiaffine variety and by restriction, we get a function f Y Ui : Y U i k. Then f is regular at p under the projective definition if and only if f is regular at p under the affine definition. Proof: Without loss of generality, assume that i = 0. For the forward direction, assume that we have U an open set such that p U Y and homogeneous polynomials g, h k[x 1,, x n ] of the
19 19 C. McWhorter same degree such that h 0 on U and f = g/h on Y. We know that on U U 0, which is open in Y U 0, g(a 0,, a n ) h(a 0,, a n ) = g(1,, a n) h(1,, a n ) But g(1,, a n ), h(1,, a n ) k[x 1,, x n ], so that f is regular under the affine definition. For the reverse direction, assume that f Y U0 is regular at p under the affine definition. Let U be an open set such that p U Y U 0 and let g, h k[x 1,, x n ] such that h 0 on U and f = g/h on U. Note that U is open in Y also. Now let G, H be the homogenizations of g, h, respectively. If they are not of the same degree, multiply by the appropriate degree of x n 0 so that G, H have the same degree. Call these homogeneous polynomials of the same degree G, H. For any point q = [1, a 1,, a n ] U, we know that f(q) = G(1, a 1,, a n ) H(1, a 1,, a n ) = G(λ, λa 1,, λa n ) H(λ, λa 1,, λa n ) for λ 0. It is clear that H is still nonzero on U. But then f is regular under the projective definition. Remark 2.7. As in the case of quasiaffine varieties, a regular function is continuous. This follows from the equivalence of the affine and projective definitions and the fact that quasiprojective varieties can be covered by quasiaffine varieties. One checks continuity locally. An important consequence of the proposition above is that if f, g are regular functions on a variety X and f = g on the same nonempty open subset U X, then f = g everywhere. Remember that in a variety nonempty open sets are dense. The set Z(f g) is closed in X but contains a dense subset so that Z(f g) = X. But then f = g on X. Definition 2.8 (Morphism). Let k be a fixed algebraically closed field. A variety over k (or simply a variety) is any affine, quasiaffine, projective, or quasiprojective variety. If X, Y are two varieties, a morphism ϕ : X Y is a continuous map such that for every open set V Y and for every regular function f : V k, the function fϕ : ϕ 1 (V ) k is regular. One need check that the composition of morphisms is a morphism (it is clear that their composition is continuous being the composition of continuous functions). In particular, this means that we have a category whose objects are varieties and whose morphisms are exactly the morphisms defined above. Definition 2.9 (Isomorphism). A morphism ϕ : X Y is an isomorphism if and only if there exists a mprihqsm ψ : Y X such that ϕψ = 1 Y and ψϕ = 1 X. Note that an isomorphism is necessarily bijective and bicontinuous so that isomorphism are also homeomorphisms. However, a bicontinuous function need not be an isomorphism! Remark A morphism is a function given locally by tuples of rational functions. Proposition Let X P n and Y P m be quasiprojective varieties and f : X Y a function. Then f is a morphism if and only if given p X, there is an open set U with p U X such that f(u) is contained in some affine open set U i P m.
20 Algebraic Geometry 20 Proof: Assume that f is a morphism. Let p X with f(p) U i. We know that f 1 (U i ) is an open neighborhood of p X. Furthermore, we know that f(f 1 (U i )) U i. Without loss of generality, assume that i = 0. We think of U 0 as A m with coordinates Y 1,, Y m. Each of the Y i is a regular function on Y U 0. By the definition of morphism, Y i f is regular on f 1 (U i ) X. By possibly shrinking f 1 (U i ) to a smaller open neighborhood p X, we can assume Y i f = b i /c i, where b i, c i k[x 1,, x n ] and c i (p) 0. But then locally near p, f is given by [ b1 (q) Q c 1 (q),, b ] m(q) c m (q) for q near p. Example Take the Veronese embedding: ν : P 1 P 2 given by [x 0, x 1 ] [x 2 0, x 0, x 1, x 2 1 ]. On U 0 and U 1, we have U 0 : [1, x 1 /x 0 ] [1, x 1 /x 0, (x 1 /x 0 ) 2 ] U 1 : [x 0 /x 1, 1] [(x 0 /x 1 ) 2, x 0 /x 1, 1] Proposition Let X P n and Y P m be quasiprojective varieties and f : X Y be a function. Then f is a morphism if and only if given any p X, there is an open neighborhood U of p and m homogeneous polynomials F 0, F 1,, F m k[x 0,, x n ], all homogeneous of the same degree, such that for all q U f(q) = [F 0 (q), F 1 (q),, F m (q)] Proof: Suppose that f is a morphism. Then f is locally given by rational functions so that there exists a p U X such that q [1, g 1 (q),, g m (q)], where the g i are regular functions on U. Shrink U appropriately and assume g i = F i /G i, where F i, G i are homogeneous polynomials of some degree and are in k[x 0,, x n ] and G i 0 for all i on the shrunken U. We know that [ q 1, F 1(q) G 1 (q),, F ] m(q) G m (q) After multiplication by G = lcm(g 1,, G m ), we obtain q [G(q), F 1 (q)g(q),, F m (q)g(q)] where each of the entries is of degree deg G (the terms F i, G i are of the same degree so that F i /G i is of degree 0). This is the desired form. Now we show the reverse direction. First, note that all the F i being homogeneous of the same degree makes the map well defined. Let p X. Let f(p) U i, say i = 0. We know that F i (p) 0 for all i. Replace U by U \ Z(F 0 ). Then on this new open neighborhood, f is given by [ q 1, F 1(q) F 0 (q),, F ] m(q) F 0 (q)
21 21 C. McWhorter Example We show that the Veronese embedding ν : P 1 P 2 is an isomorphism onto its image. We know that [x 0, x 1 ] [x 2 0, x 0x 1, x 2 2 ]. We know further that { ν 1 [Y 0, Y 1 ], on U 0 ([Y 0, Y 1, Y 2 ]) = [Y 1, Y 2 ], on U 2 On U 0 P 1, this maps to U 2 in P 2. Using the fact that x 0 0, we obtain [x 0, x 1 ] ν [x 2 0, x 0 x 1, x 2 2] ν 1 [x 2 0, x 0 x 1 ] = [x 0, x 1 ] On U 1 P 1, we know that this goes to U 2 in P 2. Using the fact that x 1 0, we obtain However, on U 0 P 2, [x 0, x 1 ] ν [x 2 0, x 0 x 1, x 2 2] ν 1 [x 0 x 1, x 2 1] = [x 0, x 1 ] [Y 0, Y 1, Y 2 ] ν 1 ν [Y 0, Y 1 ] [Y0 2, Y 0 Y 1, Y2 2 ] But we only claimed that ν 1 works when restricted to the image of ν. The image of ν satisfies Y 0 Y 2 = Y 2 1. So using the fact that Y 0 0 on U 0, we obtain On U 2 P 2, we have [Y 2 0, Y 0 Y 1, Y 2 2 ] = [Y 0, Y 1, Y 2 ] [Y 0, Y 1, Y 2 ] nu 1 ν [Y 1, Y 2 ] [Y1 2, Y 1 Y 2, Y2 2 ] = [Y 0 Y 2, Y 1 Y 2, Y2 2 ] = [Y 0, Y 1, Y 2 ] Definition Let Y be a variety. We denote by O(Y ) the ring of all regular functions on Y. If p is a point of Y, we define the local ring of p on Y, O p,y or simply O p, the ring of all germs of regular functions on Y near p. That is, an element of O p is a pair U, f, where U is an open subset of Y containing p and f is a regular function on U. We identify two such pairs U, f and V, g if f = g on U V. For this definition, there are many things to check: O(Y ) is a ring under pointwise addition and multiplication. Generally, if Y is any set and R is any ring, all functions Y R is a ring under pointwise addition and multiplication. This also begs another question: given regular functions f, g, are f ± g, fg regular functions under pointwise operations? Indeed, this is so and is simple to check. We need also check that the above identification is an equivalence relation. 1. Reflexive: U, f U, f is trivial. 2. Symmetric: If U, f V, g, then it is trivial that V, g U, f 3. Transitivity: Suppose that U, f V, g and V, g W, h. Then we know that f = g on U V and g = h on V W. Then we know that f = h on U V W. As a nonempty open subset of an irreducible space is irreducible and U V W is open, it must be that f = h on U W so that U, f W, h.
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