NOTES ON ALGEBRAIC GEOMETRY MATH 202A. Contents Introduction Affine varieties 22


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1 NOTES ON ALGEBRAIC GEOMETRY MATH 202A KIYOSHI IGUSA BRANDEIS UNIVERSITY Contents Introduction 1 1. Affine varieties Weak Nullstellensatz Noether s normalization theorem Nullstellensatz Noetherian induction Generic points Morphisms Algebraically closed case Review First look at schemes 8 2. Projective varieties Segre embedding Graded rings and graded ideals Irreducible algebraic sets in P n Projective schemes Sheaves Presheaf Sheaf Affine varieties and affine schemes Affine varieties Prevarieties Morphisms Affine schemes 24 References 27 Introduction The aim of these notes is to develop basic algebraic geometry assuming commutative algebra. But I want to start with the set of Ω points in a variety over k which is the geometric version of a scheme over k. As I write these notes I see that this approach may go completely over the heads of beginning students. But, we will see. Date: October 29,
2 2 KIYOSHI IGUSA BRANDEIS UNIVERSITY 1. Affine varieties The basic object of study will be subsets of C n defined by polynomials equations with coefficients in a number field F (finite extension of Q). These are affine varieties over C but they also give affine schemes over F. More generally we take k to be any field and Ω to be an algebraically closed field extension of k of infinite transcendence degree. In other words Ω = k(x α ) is the algebraic closure of the field k(x α ) of rational functions on k with infinitely many variables X α. The first theorem that we want to prove is the weak Nullstellensatz which identifies the maximal ideals in the ring k[x 1,, X n ] Weak Nullstellensatz. Lemma Let a Ω n. Then the set of all f(x) k[x] so that f(a) = 0 is a prime ideal which we denote π(a) k[x]. (I.e., π is a mapping of sets π : Ω n Spec(k[X]).) Proof. This is a prime ideal since it is the kernel of the kalgebra homomorphism which sends f(x) to f(a). k[x] Ω Lemma π : Ω n Spec(k[X]) is surjective. Proof. Take any prime p k[x]. Then there is an embedding ϕ : k[x]/p Ω. Let a i = ϕ(x i ). Then f(x) = ϕ(f) = 0 iff f p. So, π(a) = p. Theorem π(a) is a maximal ideal if and only if all coordinates a i Ω of a are algebraic over k. Proof. Let ϕ a : k[x] Ω be given by evaluation at a so that π(a) = ker ϕ a. Then ϕ a (k[x]) = k[a 1, a 2,, a n ]. Since this is a subring of Ω, it is an integral domain finitely generated over k. So, the normalization theorem applies and this is a field iff all a i are algebraic over k. The usual weak Nullstellensatz follows. Corollary If k is algebraically closed then the only maximal ideal in k[x] are π(a) where a k n. (π(a) = (X 1 a 1,, X n a n )) Noether s normalization theorem. We need the following critically important theorem from commutative algebra. Theorem (Normalization Theorem). Let R be a finitely generated domain over a field k with transcendence degree d. Then R contains transcendental elements x 1,, x d over k so that R is an integral extension of the polynomial ring S = k[x 1,, x d ]. Exercise The proof in Mumford s book gives an algorithm for finding x 1,, x d. Use this algorithm to find S = k[x 1 ] in the case when R = k[x 2, X 3 ] k[x]. We also need, for now, the following lemma used in the proof of the GoingUp theorem. Lemma Let R be an integral extension of S. If R is a field then S is a field. Proof. Take any a 0 S. Then R contains b = 1/a. Since R is integral over S, there is a monic polynomial f(x) S[X] so that f(b) = 0: b n + c 1 b n c n = 0, c i S
3 Multiply by a n 1 to get: NOTES ON ALGEBRAIC GEOMETRY MATH 202A 3 So, b = 1/a S and S is a field. b + c 1 + c 2 a + c 3 a c n a n 1 = 0 Corollary Suppose R = k[a 1,, a n ] is a domain. Then R is a field if and only if every a i is algebraic over k. Proof. If the a i are algebraic over k then k[a 1,, a n ] = k(a 1,, a n ) is a field. Conversely, if some are transcendental, then, by the normalization theorem, R is an integral extension of a polynomial ring S = k[x 1,, X d ] with d > 0. But S is not a field. By the lemma, R is not a field Nullstellensatz. Let s review one construction from Commutative Algebra. Suppose that R is a ring and f R is not nilpotent. I.e., f n 0 for all n 0. Then M = {1, f, f 2, } is a multiplicative set and R f = R M is given by inverting f. { } a R f = f n where a f n = 0 iff af m = 0 for some m. In particular 1 f 0. Another description of R f is: R f = R[X]/(fX 1). For any subset S of k[x], let V (S) be the set of all x Ω n so that f(x) = 0 for all f S. Two extreme cases are: V ( ) = Ω n and V (1) =. Definition A subset Z Ω n is called an algebraic set defined over k if Z = V (S) for some S k[x]. Example SL(2, Ω) is the algebraic subset of Ω 4 defined over any k by SL(2, Ω) = {(a, b, c, d) Ω 4 : ad bc = 1} GL(2, Ω) is defined by the inequality ad bc 0. This does NOT define an algebraic subset of Ω 4. However, GL(2, Ω) is equivalent to the algebraic subset of Ω 5 (defined over any k) given by GL(2, Ω) = {(a, b, c, d, e) Ω 5 : (ad bc)e = 1}. The basic property of algebraic sets is: Theorem The sets V (S) form the closed subsets of a topology on Ω n. (This is called the Zariski topology or the ktopology.) Proof. The definition of the set V (S) can be parsed as follows: V (S) = V (f) f S So, V (S α ) = V ( S α ). The collection of sets {V (S)} is closed under arbitrary intersections. If S, T are two subsets of k[x] and ST = {fg : f S, g T } then V (ST ) = V (fg) = (V (f) V (g)) = V (g) = V (S) V (T ). f S,g T f S,g T f S V (f) g T So, the collection of sets {V (S)} is closed under finite unions. Since V ( ) = Ω n and V (1) =, the collection of sets {V (S)} satisfies the axioms for the closed sets in a topology on Ω n.
4 4 KIYOSHI IGUSA BRANDEIS UNIVERSITY For any closed subset Y of Ω n we take the induced topology. Exercise Show that the open sets Y f = {x Y : f(x) 0} form a basis for the Zariski topology on Y. Proof. (Using things that come later.) Let U be an open subset of Y with complement Z. Since k[x] is Noetherian, the ideal I(Z) is finitely generated by, say, f 1,, f m. If y U then f i (y) 0 for some i. So y Y fi U. So, the Y f are basic open sets in Y. For any S k[x], let (S) denote the ideal generated by S. (We allow ideals to be the whole ring. An ideal which does not contain 1 is called a proper ideal.) For any ideal a in k[x], the radical r(a) of a is defined to be the set of all f k[x] so that some power of f lies in a. The following statement is clear: Proposition For any subset S k[x]: V (S) = V ((S)) = V (r(s)). For any subset Z Ω n let I(Z) be the set of all f(x) k[x] so that f(z) = 0 for all z Z. Since f(z) = 0 iff f π(z), this is equivalent to: I(Z) = π(z) Therefore, I(Z) is an ideal in k[x]. Also, being an intersection of prime ideals, it is a radical ideal. Hilbert s Nullstellensatz states: z Z Theorem For any ideal a in k[x], I(V (a)) = r(a). Proof. (from the red book [4]) It is clear that I(V (a)) contains r(a). So, suppose g is not in r(a). Then we will find a point in x V (a) so that g(x) 0. 1) Since g is not in r(a), we can invert g in k[x]/a without making the ring trivial (1=0). The reason is that M = {1, g, g 2, } is a multiplicative set in k[x]/a which does not contain 0. [This is also proved in the last paragraph of the proof of Theorem in [4].] 2) Inverting g and modding out a is the same as adding one variable X n+1 and modding out the ideal J in R = k[x 1,, X n, X n+1 ] generated by a and 1 X n+1 g. Let a = (a 1,, a n+1 ) Ω n+1 be a point corresponding to a maximal ideal which contains the ideal J. (By the weak Nullstellensatz, all a i are algebraic over k.) 3) The point x = (a 1,, a n ) Ω n lies in V (a) but g(x) = 1/a n+1 0. Since a i are algebraic over k, π(x) = ker ϕ x is a maximal ideal in k[x] which contains a but does not contain g. Since g was an arbitrary element of k[x] not contained in r(a), this proves the following. Corollary For any ideal a in k[x], r(a) is the intersection of all maximal ideals containing a. Consequently, the Jacobson radical of any finitely generated algebra over any field is equal to it nilpotent radical. Corollary There is an order reversing bijection between the set of radical ideals in k[x] and the set of Zariski closed subsets of Ω n defined over k. Exercise Show that the closure (in the Qtopology) of the point (e, e 2 ) C 2 is the parabola {(z, z 2 ) z C}.
5 NOTES ON ALGEBRAIC GEOMETRY MATH 202A Noetherian induction. Recall that k[x] is Noetherian, i.e., the ideals in k[x] satisfy the ACC: any ascending chain of ideals is eventually constant. Therefore, the closed subsets of Ω n satisfy the DCC: any descending chain of closed subsets is eventually constant. This allows for the following argument called Noetherian induction: For every statement about closed subsets of Ω n which is not true, there is a minimal counterexample. Proposition Every closed subset of Ω n is a union of finitely many irreducible closed subsets. (Irreducible means not the union of two proper closed subsets.) Proof. Suppose not. Then, by Noetherian induction, there is a minimal counterexample, say Y. Then Y is not irreducible. So, Y is the union of two proper closed subsets. Each of these is a union of finitely many irreducible closed subsets. So Y is also. Proposition The decomposition of a closed set Y as a union of irreducible closed subsets Y = Y j is unique. The unique pieces Y j of a closed set Y are the components of Y. Proof. If an irreducible X Y 1 Y 2 then, since X = (X Y 1 ) (X Y 2 ), X is contained in one of the Y i. So, if X i = Y j then X i Y j and Y j X k, etc., until we come back to the same set. So, one of the X i = Y j. Do it over and over until the proposition is proved. (Any X k is not contained in Y j since the X i do not contain each other.) Theorem The prime ideal correspond to the irreducible subsets under this correspondence. Proof. Suppose Y = Y 1 Y 2 then I(Y ) = I(Y 1 ) I(Y 2 ). But a prime ideal cannot be written as an intersection of two larger ideals. So, I(Y ) is not prime. Conversely, suppose Y is irreducible. Then, by the correspondence I(Y ) cannot be written as the intersection of two (or more) larger radical ideals. So, I(Y ) is prime by the following lemma. Lemma In any Noetherian ring A, any radical ideal is the intersection of finitely many prime ideals. Proof. By the Primary Decomposition Theorem, any ideal in a Noetherian ring is a finite intersection of primary ideals: a = q i. We claim that r(a) = r(q i ). The inclusion ( ) is clear. So, suppose that a r(q i ) for each i. Then, for each i there is an integer n i so that a n i q i. Let n be the maximum of these n i. Then a n q i = a. This proves the claim. The lemma follows since each r(q i ) is prime. This uses only the existence part of the Primary Decomposition Theorem which is easy: Lemma In any Noetherian ring A, every ideal is a finite intersection of primary ideals. Proof. Suppose not. Then there is a maximal counterexample, a. This ideal cannot be written as an intersection of two larger ideals. Otherwise, each of these is an intersection of primary ideals making a an intersection of these primary ideals. We claim that a is primary. To show this, take two elements a, b A so that ab a but no power of a or b lies in a. For each n 1 let I n = (a : a n ) be the set of all c A so that a n c a. Then I n is an ascending chain of ideals so I n = I n+1 for some n. Let J = (a, a n ), K = (a, b). Since a n, b / a these ideals properly contain a. We will show that J K = a giving a contradiction (showing that a is primary). Let c J K. Since c J, c = a n x + y where y a. Since ak a, ac a. So, a n+1 x a. So, x I n+1 = I n. So, a n x a. So, c a.
6 6 KIYOSHI IGUSA BRANDEIS UNIVERSITY 1.5. Generic points. Theorem The closure of any point x Ω n is an irreducible algebraic set defined over k. Conversely, for any irreducible algebraic subset Y Ω n, there is an element y Y (called a generic point of Y ) so that Y = y. Proof. The first statement is an exercise. The second statement follows from Lemma Namely, Y, being irreducible corresponds to a prime ideal p = I(Y ). By Lemma 1.1.2, there is a point y Ω n so that π(y) = p. Then y = V (p) = Y. Corollary Two point x, y Ω n map to the same prime ideal π(x) = π(y) if and only if x = y. Exercise (1) Show that x y iff x = y is an equivalence relation on Ω n. (2) Show that x y iff π(x) π(y). (3) Show that x is a finite set iff π(x) is a maximal ideal. (4) When is x Ω n a closed point? (What if k is not perfect?) 1.6. Morphisms. Let Y Ω n be an irreducible closed subset defined over k. Then a regular function f : Y Ω is any mapping which is the restriction to Y of a mapping Ω n Ω given by a polynomial in k[x 1,, X n ]. The coordinate ring Γ(Y ) of Y is the ring of all regular functions Y Ω. If two polynomials give the same regular function on Y then their difference is zero on Y and conversely. So, Γ(Y ) = k[x]/i(y ). Since Y is irreducible, I(Y ) is a prime ideal and Γ(Y ) is an integral domain. If y Y is a general point then f : Y Ω is uniquely determined by f(y). So, Γ(Y ) is isomorphic to the image of y under k[x]. Given Y, Z irreducible closed subsets of Ω n, Ω m resp, a morphism Y Z is a mapping f : Y Z so that there exists m polynomials f i k[x 1,, X n ] so that f(y) = (f i (y)) for all y Y. There is an induced map: f : Γ(Z) Γ(Y ) defined on any g Γ(Z) k[t 1,, T m ] by f (g) = g f given on each x Ω n by f (g)(x) = g(f 1 (x),, f m (x)). Theorem Hom(Y, Z) = Hom k alg (Γ(Z), Γ(Y )). Proof. Given any ϕ : Γ(Z) Γ(Y ), let f ϕ : Y Z be given by the m polynomials ϕ(t j ) k[x] where t 1,, t m : Z Ω m Ω are the m projection maps restricted to Z: Then f ϕ : Γ(Z) Γ(Y ) is given by f ϕ (x) = (ϕ(t j )(x)) f ϕ(t j ) = t j f ϕ = t j (f ϕ ) = ϕ(t j ). In other words, f ϕ(t j ) = ϕ(t j ). Since the t j generate Γ(Z), f ϕ = ϕ. Conversely, if we start with f : Y Z and ϕ = f then So, f ϕ = f. f ϕ = (ϕ(t j )) = (f (t j )) = (t j f) = (f j ). Proposition The ideal I(f(Y ))/I(Z) is equal to ker f Γ(Z). Proof. g ker f iff g f = 0 iff g(im f) = 0 iff g I(f(Y )).
7 NOTES ON ALGEBRAIC GEOMETRY MATH 202A Algebraically closed case. Let s compare the definitions and the last theorem to the standard setup. Assume in this subsection that k is algebraically closed and let Y 0 = Y k n and Z 0 = Z k m. These are the sets of closed points in Y, Z. Lemma If Y Ω n is an irreducible algebraic set then Y = Y 0. Proof. It suffices to show that I(Y 0 ) = p = I(Y ). Since k is algebraically closed, Y 0 contains all points x Y so that π(x) k[x] is a maximal ideal. The condition x Y is equivalent to the condition π(x) p. Since the Jacobson radical of k[x]/p is equal to its nilradical, we have: I(Y 0 ) = π(x) = π(x) = r(p) = p So, x Y π(x) p Y 0 = V (I(Y 0 )) = V (p) = Y. A morphism f : Y 0 Z 0 is defined to be any set mapping which is the restriction to Y 0 of a polynomial function k n k m given by m polynomials in n variables. Theorem Hom(Y 0, Z 0 ) = Hom(Y, Z) = Hom k alg (ΓZ, ΓY ). Proof. Since Y = Y 0, any continuous map f : Y 0 Z 0, composed with inclusion Z 0 Z extends in at most one way 1 to a continuous mapping Y Z. However, f is given by polynomial equations which extend to a map Y Ω m which necessarily have image in Z. Conversely, any mapping Y Z given by polynomials with coefficients in k will send Y 0 into Z 0. So, Hom(Y 0, Z 0 ) = Hom(Y, Z). The other isomorphism was proved earlier. Proposition For any nonempty open subset U of Y, Y is the closure of U Y 0. Proof. The lemma implies that every nonempty open subset of Y meets Y 0. If the closure of U Y 0 is not equal to Y then its complement V is a nonempty open set in Y. Since Y is irreducible U V is nonempty. So, U V Y 0 is nonempty. Contradiction. Corollary For any open subset U of Y, any mapping U Y 0 k given locally as a rational function with coefficients in k extends uniquely to a regular function U Ω and all regular functions U Ω take U Y 0 into k. Proof. Let f : U Y 0 k and, for each x U Y 0, suppose that there is a basic open neighborhood V = Y g U of x and h ΓY so that f(y) = h(y)/g(y) m for all y V Y 0. Then h/g m is one extension of f to V. Since V Y 0 is dense in V, this is the only extension. The second statement is clear. These theorems show that we do not lose any information by restricting functions to the k points Y 0 when k is algebraically closed. However, we are throwing away important and useful parts of the structure of the affine variety and of morphisms of affine varieties by restricting to just the closed points. 1 This is the Hausdorff property of Z that we need to discuss later.
8 8 KIYOSHI IGUSA BRANDEIS UNIVERSITY 1.8. Review. Before discussing projective space, let us briefly review the concepts. (1) k is any field and Ω is a very large field containing k. (2) We consider functions f : Ω n Ω given by polynomials f k[x]. (3) There is an order reversing 11 correspondence between radical ideals (a = r(a) in k[x] and closed subsets Z of Ω n given by V (a) := {y Ω n : f(y) = 0 f a} a = I(Z) := {f k[x] : f(z) = 0} (4) Under this bijection the prime ideals correspond to the irreducible subsets (those not equal to a union of two proper closed subsets). (5) Every irreducible set Y contains an element y called a generic point so that y = Y. I(Y ) = p = I(y) = π(y) = ker(ev y : k[x] Ω) (If y Y is not generic then y = Z Y is an irreducible proper subset.) (6) If Y Ω n, Z Ω m are irreducible, a morphism f : Y Z is a function given by polynomial equations. (7) Hom(Y, Z) = Hom k alg (Γ(Z), Γ(Y )) where Γ(Y ) = k[x] I(Y ) is the ring of all polynomial functions f : Y Ω. (8) For any open subset U Y, (Y irreducible), a regular function on U is defined to be a mapping f : U Ω so that f is given locally by rational functions f(x) = g(x) h(x). These functions form a ring Γ(U) First look at schemes. The idea is that all generic points of an irreducible set Y are the same. Each of them represents the general point of Y. We do this all the time when teaching calculus: Take the curve Y R 2 given by the equation y = x 2. The general point of this curve is the point (x, x 2 ) where x is a variable. But x is a dummy variable which means that (z, z 2 ) also represents a general point on the parabola Y. In algebraic geometry, we replace the expression Let x be a variable with the more precise expression: Let T 1 be a transcendental element over R. Then y = (T 1, T1 2 ) is a general point in Y. Recall that Ω is the algebraic closure of the field k(t 1, T 2, ). These extra transcendental elements are useful to write down other general elements, such as: [ T1 T 2 T 3 T 4 ] This is a 2 2 matrix whose entries are general elements of R. The point is that y = (T 2, T2 2 ) is also a general point in Y. So, we should think of all these points as being the same point expressed in terms of different dummy variables. Definition For any (commutative) ring R let Spec(R) denote the topological space whose elements are the prime ideals of R. We use the notation [p] for the point in the space Spec(R) corresponding to p. The basic open subsets of X = Spec(R) are the sets X f = {[p] : f / p} for any f R. If we think of p as representing the ring homomorphism ϕ p : R R/p then f / p ϕ p (f) 0.
9 NOTES ON ALGEBRAIC GEOMETRY MATH 202A 9 When R = k[x 1,, X n ] and p = π(y) = ker ev y then R/p Ω and (1.1) [p] X f ϕ p (f) = f(y) 0 y / V (f). In the example that we have been discussing so far, R is a finitely generated domain over the field k. This has the form R = k[x] I(Y ) where Y is irreducible. (Finitely generated over k means there is an epimorphism k[x 1,, X n ] R. R being a domain means the kernel is a prime ideal p. We know that such ideals are given uniquely by p = I(Y ) where Y Ω n is irreducible.) Lemma Let R be any finitely generated domain over an field. Let f R be represented by a polynomial f(x). Let X = Spec(R). Then, for any y Y, π(y) = p X f y Y f. (Recall that Y f = {y Y : f(y) 0}.) In other words, π 1 (X f ) = Y f where Proof. This follows from (1.1): π : Y Spec(R) y Y f y / V (f) [p] X f. Theorem There is a continuous epimorphism π : Y X = Spec(R) = Spec(k[X]/I(Y )) sending each point y Y to the prime ideal π(y) = ker ev y : k[x] Ω. Furthermore, π is open (the image of any open set in Y is open in Spec(R)). So, Spec(R) has the quotient topology with respect to π (V Spec(R) is open iff π 1 (V ) is open in Y ). Proof. Prime ideals in R = k[x] I(Y ) are p I(Y ) where p is a prime ideal in k[x] containing I(Y ). But p = π(x) for some x Ω n and I(Y ) p iff x y = Y. Therefore, π : Y X = Spec(R) is surjective. The lemma shows that π is continuous. The lemma also shows that π is open since π sends every basic open set Y f in Y to an open set X f in X = Spec(R). But any open surjection is a quotient map. So, Spec(R) has the quotient topology. What is the difference between different generic points of the same Y? (These are all elements of π 1 (π(y)).) Theorem For any y Ω n there is a 11 correspondence between π 1 (π(y)) (the set of generic points of y = Y ) and the set of all kalgebra monomorphisms ϕ : R = k[x]/π(y) Ω. Proof. Let p = π(y). Then x π 1 (π(y)) iff π(x) = p. For each such x we have ϕ = ev x : k[x]/p Ω Conversely, given any ϕ : k[x]/p Ω, we get back the point x since its coordinates are given by x i = ϕ(x i ). Since ev x (X i ) = x i, this gives a bijection x ϕ.
10 10 KIYOSHI IGUSA BRANDEIS UNIVERSITY Example Suppose that k = R. Then any R = R[X]/p has two kinds of maximal ideals: (1) m so that R/m = R. Since there is only one Ralgebra homomorphism R Ω, π 1 (m) = {x} is a closed point. These are the elements of Y R n. (Y = V (p)) (2) m so that R/m = C. There are exactly two Ralgebra monomorphisms C Ω (with the same image). So, π 1 (m) has two points which lie in Y C n and must be complex conjugates of each other. Everyone knows that the parabola Y = X 2 is missing its point at infinity and this is one of the reasons for introducing projective space. But most people don t realize that the circle is also missing a point at infinity: Example What is Spec(R) where R = R[X 1, X 2 ] (X X2 2 1)? Here k = R and Y = V (X X2 2 1) Ω2. The real points of Y form a circle of radius 1. The complex points of Y form a hyperbola since the solution set of X X 2 2 = (X 2 + ix 1 )(X 2 ix 1 ) = 1 in C 2 is given by B = 1 A where A = X 2 + ix 1, B = X 2 ix 1. X 2 R 2 B X 2 A C 2 Y R 2 X 1 Y C 2 X 1 Note that, in Y C 2, A is an arbitrary nonzero complex number, B = 1 A and X 1, X 2 are given by X 2 = 1 2 (A + B) and X 1 = 1 2i (A B). In (A, B)cordinates, (A, B) is real if A C lies on the unit circle. The point (A, B) = (z, 1/z) is, in (X 1, X 2 ) coordinates, given by ( ( 1 (X 1, X 2 ) = z 1 ), 1 ( z + 1 )) 2i z 2 z with complex conjugate: ( ( ) 1 1 2i z z, 1 ( z + 1 )) 2 z giving A = 1/z. The mapping z 1/z interchanges the outside of the circle and the inside minus the origin. There is also one more point in Spec(R) corresponding to the unique minimal prime ideal 0. Thus Spec(R) is the punctured disk union an open point: Spec(R) = (generic point)
11 NOTES ON ALGEBRAIC GEOMETRY MATH 202A Projective varieties For any field F, the standard definition of projective space P n (F ) is that it is the set of one dimensional F vector subspaces of F n+1. For example, P 1 (R) is the set of lines through the origin in R 2. These are specified by slope 0 θ < π. So, P 1 (R) = R/πZ is a circle. However, there are many points missing. As an affine scheme, the circle implicitly also contains all points in the punctured disk (the interior of the circle) and a generic point. As a projective variety, the puncture will be filled in! In order to construct the scheme theoretic version of ndimensional projective space P n (k) we need to take P n (Ω). This is defined to be the space of all one dimensional Ω vector spaces in Ω n+1. This is also the quotient space of Ω n+1 \0 modulo the identification x ax for all a 0 Ω. This is a union of the open subsets U i = {[x 0,, x n ] P n (Ω) : x i 0} where [x 0,, x n ] is the equivalence class of the point (x 0,, x n ) Ω n+1 \0. A closed subset of P n (Ω) (in the ktopology) is the set of zeros of a finite collection of homogeneous polynomials f i k[x 0,, X n ]. The following comments prove the statements which Mumford says are obvious. Definition A homogeneous ideal in k[x 0,, X n ] is an ideal generated by homogeneous polynomials. Proposition An ideal a is homogeneous iff it contains the homogeneous components f [k] of each of its elements f. Proof. If a satisfies this conditions. Then it is generated by homogeneous polynomials. Conversely, suppose that a is generated by homogeneous polynomials f i. Then each element of a has the form gi f i = g [k] i f i Since each summand g [k] i f i is in a the condition is satisfied. So, the condition is equivalent to the definition. Proposition The radical of any homogeneous ideal is homogeneous. Proof. Suppose that f r(a). Then f n a. Since a is homogeneous, this means the highest degree term f [k] of f is in the radical of a. Subtracting this we see by induction that each homogeneous component of f lies in r(a). So, r(a) is homogeneous. Now we can follow all steps in the proof of Proposition 4 in [4]. Theorem Y I(Y ) and a V (a) gives an order reversing bijections between closed algebraic subsets of P n (Ω) and homogeneous radical ideals in k[x] not equal to (X 0,, X n ). Example Let f = XW Y Z. This is a homogeneous polynomial of degree 2 in k[x, Y, Z, W ]. Let Σ = V (XW Y Z) P 3 (Ω). This may be more familiar when we write it this way: Σ = {[ ] x, y z, w } : det = 0 Since f cannot be factored as a product of polynomials of lower degree (necessarily linear), (XW Y Z) is prime and Σ is irreducible. Last week we discussed regular functions on open subsets of irreducible sets f : U Ω. By definition these are functions which are
12 12 KIYOSHI IGUSA BRANDEIS UNIVERSITY given locally by rational functions f(x) = g(x) h(x). Mumford emphasized that, in some cases, the same fraction cannot always be used throughout the set U. Here is an example. {[ ] } {[ ] } x, y x, y Σ X = : x 0, Σ z, w Y = : y 0 z, w Let U = Σ X Σ Y. Let f : U Ω be given by { z f = x on Σ X on Σ Y on the intersection Σ X Σ Y w y we have that x, y are both nonzero and z x = zw xw = zw yz = w y so f is well defined. (We forgot to check the case when w = 0. But the equation xw = yz implies that z = 0 in that case and we get z x = 0 = w y.) This is an example of a regular function U Ω for which there is no single rational functional expression which works on all of U Segre embedding. Mumford uses this example (in another book [5]) to introduce the following construction. Definition The Segre embedding is given by σ : P n P m P n+m+nm ([X i ], [Y j ]) [Z ij = X i Y j ]. In other words, each of the n + 1 homogeneous variables X i is multiplied by each of the m + 1 variables Y j to get (n + 1)(m + 1) variables Z ij = X i Y j : For example, when n = m = 1 we get: Ω n+1 Ω m+1 Ω (n+1)(m+1) P 1 (Ω) P 1 (Ω) P 3 (Ω) ([S, T ], [U, V ]) [}{{} SU, }{{} SV, }{{} T U, }{{} T V ] X Y Z W Theorem The Segre embedding is a monomorphism with image σ(p n P m ) = V (Z ij Z kl Z il Z kj ) Corollary The image of the Segre embedding is irreducible. Proof. The idea is to use the Ωtopology. Since k Ω, there are more Ωclosed set than kclosed sets. If a kclosed set is irreducible as an Ω closed set then it is irreducible. Since Ω = Ω we are in the realm of standard algebraic geometry. For each point [y] P m (Ω), the mapping P n (Ω) P n+m+nm (Ω) given by [x] σ([x], [y]) is continuous. (This is only true for [y] P m (k) in the ktopology). It is easy to see that P n (Ω) is irreducible. This implies that the image of σ is irreducible (even if σ were not an embedding). Another proof of the same thing can be given using graded rings and graded ideals.
13 NOTES ON ALGEBRAIC GEOMETRY MATH 202A Graded rings and graded ideals. Definition A graded ring is a ring R which is decomposed as an infinite direct sum: R = R 0 R 1 R 2 so that R n R m R n+m. Elements in R d are said to be homogeneous of degree d. The condition R n R m R n+m means deg(fg) = deg f + deg g if f, g are homogeneous. The equation R = R d means that every element of R is equal to a sum of homogeneous components which all lie in R. The basic example is R = k[x 1,, X n ]. A graded ring can be constructed from its pieces as follows. Let R 0 be a ring and let R n be R 0 modules. Let µ n,m : R n R m R n+m be an Rbilinear mapping which is commutative, associative and gives the action of R 0 on R n when m = 0. Then R = R n is a graded ring. A graded ideal can be constructed using the same idea. Definition A graded ideal in a graded ring R = R d is a sequence of R 0 submodules I n R n so that I n R m + R n I m I n+m. In that case I = I n is a homogeneous ideal in R = R n : It is clearly an ideal. It is homogeneous since it is generated by elements of I n. (Keep in mind that there is one element 0 which is in all I n and all R n but, otherwise, the sets R n are disjoint.) Conversely, any homogeneous ideal in a graded ring is a graded ideal. The grading makes it easier to talk about projective algebraic sets: V (I) = {[x] P n (Ω) : f(x) = 0 f I d d} Definition Let R = R n, S = S m be graded kalgebras. Then a graded k homomorphism of degree d is a homomorphism of kalgebras ϕ : R S so that ϕ(r n ) S dn for all n. (Note that all rings are Zalgebras.) It is easy to see that the kernel of any graded homomorphism of graded rings is an ideal. Going back to our example, one way to show that an ideal is prime is to show that it is the kernel of a graded homomorphism. In our case, I = (XW XY ) is the kernel of the (degree 2) graded homomorphism: ϕ : k[x, Y, Z, W ] k[s, T, U, V ] ϕ(x) = SU ϕ(y ) = SV ϕ(z) = T U ϕ(w ) = T V It is clear that XW Y Z ker ϕ. To show that ker ϕ = I = (XW Y Z), we should compare: ker ϕ d : R d S 2d and I d = (XW Y Z)R d 2 Lemma If R = k[x 0,, X n ] then R 0 = k and R d a vector space over k of dimension ( ) n+d n with basis given by the monomials X α = X a 0 0 Xan n where α := a i = d and all a i 0.
14 14 KIYOSHI IGUSA BRANDEIS UNIVERSITY We say that X α has multidegree α = (a 0,, a n ). In the case at hand we have: ( ) d + 3 dim R d = = d + 1 (d + 3)(d + 2) 3 6 ( ) d + 1 dim I d = dim R d 2 = = d + 1 d(d + 1) 3 6 and the difference is dim R d /I d = d (d2 + 5d + 6 d 2 d) = (d + 1) 2. On the other hand, ϕ(r d ) is spanned by monomials S a T b U c V e where a + b = d = c + e. There are (d + 1) 2 such monomials. So, by linear algebra, I d = ker ϕ d. So, I = ker ϕ is a prime ideal. (Maybe not the most efficient way to prove this. But, when you get used to it, you can do this calculation very fast.) Example Here is an example in [4]. The twisted cubic is V (I) P 3 (Ω) where I is the kernel of the degree 3 graded homomorphism ϕ : k[x, Y, Z, W ] k[s, T ] ϕ(x) = S 3 ϕ(y ) = S 2 T ϕ(z) = ST 2 ϕ(w ) = T 3 Since ϕ is graded, the kernel I is a graded ideal in k[x, Y, Z, W ]. The claim is: I = (XZ Y 2, Y W Z 2, XW Y Z) Certainly, these lie in ker ϕ. It is also easy to see that ϕ d : R d S 3d is surjective. Furthermore, any two monomials in the inverse image of S a T 3d a differ by a element of I d. So, I = ker ϕ. (Proof: We can assume a 3d a. Then for any monomial X i Y j Z k W l in ϕ 1 (S a T 3d a ), we have k + l 0. So, we can exchange all X terms using the relations XZ Y 2, XW Y Z modulo I d. Then we have i = 0. Next we can eliminate either Y or W in the expression using Y W = Z 2. We are left with either Y Z or Z W which is unique.) The graded homomorphism ϕ : k[x, Y, Z, W ] k[s, T ] gives a morphism f ϕ : P 1 (Ω) P 3 (Ω) defined by f ϕ ([s, t]) = [s 3, s 2 t, st 2, t 3 ] This is welldefined since, if we replace [s, t] with the equivalent [s, t ] = [λs, λt] then f ϕ ([s, t ]) = [λ 3 s 3, λ 3 s 2 t, λ 3 st 2, λ 3 t 3 ] Claim: Let Σ denote the image of this mapping. Then Σ = V (I) where I = (XZ Y 2, Y W Z 2, XW Y Z). Proof: It is clear that Σ V (I) since each generator of I is a homogeneous polynomial which is zero on the image of f ϕ. Conversely, suppose that [x, y, z, w] V (I). Then
15 NOTES ON ALGEBRAIC GEOMETRY MATH 202A 15 xz = y 2, yw = z 2, xw = yz. Also, either x 0 or w 0 (otherwise x = y = z = w = 0 which is not allowed). Suppose w 0. Then we may assume w = 1. Then, y = z 2 and x = yz = z 3. So, is in the image of f ϕ. [x, y, z, 1] = [z 3, z 2, z, 1] = f ϕ ([z, 1]) 2.3. Irreducible algebraic sets in P n. We need to verify that the arguments used in the affine case carry over to the homogeneous case. Lemma A homogeneous ideal a in k[x] is prime iff it satisfies the following. For any two homogenous polynomials f, g, if fg a then either f a or g a. Remark This can be rephased (and generalized) in two equivalent ways: (1) A graded ring R = R n is a domain iff there are no nonzero elements f R n, g R m so that fg = 0. (2) A graded ideal I = I d in a graded ring R = R d is prime iff for all f R n \I n, g R m \I m, fg R n+m \I n+m. Proof. Prime ideals certainly have this property. Conversely, suppose that a is homogeneous and has this property. Let f, g be arbitrary polynomials so that fg a. Let f [n], g [m] be the highest degree terms in f, g. Then f [n] g [m] a since it is the highest degree term in fg. So, either f [n] or g [m] lies in a, say f [n] a. Then (f f [n] )g a. By induction, either f f [n] a (which implies f a) or g a. So, a is prime. By a similar argument we have the following. Lemma A homogeneous ideal a in k[x] is primary iff it satisfies the following. For any two homogenous polynomials f, g, if fg a then either f r(a) or g a. Write this in terms of graded ideals in any graded ring. Proof. We may assume that none of the homogeneous components of g lie in a. Then f [n] g [m] a implies that f [n] r(a). Let k 1 be minimal so that (f [n] ) k g a. Then (f f [n] )(f [n] ) k 1 g = (f [n] ) k 1 fg (f [n] ) k g a and (f [n] ) k 1 g / a. So, f f [n] r(a) (which implies f r(a) by induction on the number of homogeneous components of f). Using these lemmas, the proof in the affine case carries over to show the following. Proposition Every homogeneous radical ideal in any graded Noetherian ring is the intersection of finitely many homogeneous prime ideals. Theorem Irreducible closed subsets of P n (Ω) correspond to homogeneous prime ideal in k[x] not equal to (X 0,, X n ). Furthermore, every closed subset of P n (Ω) can be expressed uniquely as a finite union of irreducible subsets. Note: the uniqueness is a topological property. The proof of Proposition works in any topological space.
16 16 KIYOSHI IGUSA BRANDEIS UNIVERSITY 2.4. Projective schemes. The purpose of this section is to construct the projective analogue of Spec(R) and prove the analogue of Theorem We continue using the color code: red for schemes, blue for varieties. Definition Let R = R d be a graded ring. Then P roj(r) is defined to be the following topological space (together with a sheaf which will be constructed later). As a set P roj(r) is the set of homogeneous prime ideals in R which are not equal to the ideal d>0 R d. In the special case R = k[x 0,, X n ], this is the ideal (X 0, X 1,, X n ) which was excluded before. The topology on P roj(r) is given as follows. For any f R, let V (f) = {P P roj(r) : f P } These sets are defined to be closed. General closed sets are intersections of these sets: V (S) = V (f) = {P P roj(r) : S P } f S In the case R = k[x] we have a bijection: irreducible closed subsets of P n (Ω) P P roj(k[x]) Theorem There is a continuous epimorphism π : P n (Ω) P roj(k[x]) given by π([x]) = P where P P roj(k[x]) corresponds to the irreducible set [x]. Proof. xxx
17 NOTES ON ALGEBRAIC GEOMETRY MATH 202A Sheaves We want to get to the basic definition of varieties and schemes. An affine variety is a irreducible closed subset of Ω n together with a sheaf of regular functions and similarly for projective varieties. Affine and projective schemes are quotient spaces of these Presheaf. Definition Let X be any topological space. Then a presheaf on X is a functor F from the category of open subsets of X and inclusion maps to another category C. By the definition of a functor this means that F assigns to each open subset U X an object F (U) C and to every pair of open subsets U 1 U 2 a restriction morphisms res U 2 U 1 : F (U 2 ) F (U 1 ) in C which satisfies two conditions. (1) If U 1 = U 2 then the restriction map is the identity: F (U 2 ) = F (U 1 ) (2) If U 1 U 2 U 3 then the following diagram commutes: res U 3 U 1 F (U 1 ) F (U 3 ) res U 3 U 2 F (U 2 ) res U 3 U 2 In other words: res U 2 U 1 res U 3 U 2 = res U 3 U 1. We will usually take the category C to be a category of additive groups or a category of rings. The condition F ( ) = 0 will be added later as a requirement for F to be a sheaf. Example (1) Let A be any additive group. Then the constant presheaf is defined to be F (U) = A if U is nonempty and F ( ) = 0. For any inclusion U 1 U 2 we define the restriction map to be the identity map A A if U 1 and the zero mapping if U 1 =. (2) Let X be a smooth (C ) manifold and, for every open U X, let F (U) = C (U) be the ring of all C functions f : U R. Restriction is given by restriction of functions. (3) Let X Ω n be an irreducible closed subset and, for every open U X, let F (U) be the ring of all functions f : U Ω which have the property that there exists a polynomial f(x) k[x] so that f(x) = f(x) for all x U. Again restriction is given by restriction. (4) Let X Ω n be an irreducible closed subset. Let G(U) be the set of all f : U Ω so that there exist g, h k[x] for which, for all x U we have: h(x) 0 and f(x) = g(x)/h(x). Definition Given a presheaf F on X with values in a category C which has direct limits (such as the category of additive groups or the category of kalgebras) and any x X, we define the stalk F x of F at x to be the inverse limit of all F (U) where U is an open neighborhood of x: F x := dir lim x U F (U) Since open neighborhoods of x form an inverse system, {F (U)} form a direct system in C. So, this makes sense.
18 18 KIYOSHI IGUSA BRANDEIS UNIVERSITY The elements of F (U) are called sections of F over U since we imagine that F is something like a fiber bundle over X and that F (U) is the set of sections of this bundle over U. I.e., suppose that p : E X is any fixed continuous mapping. Then a section of E over U X is a mapping s : U E so that p s = id U. We usually consider continuous sections. If p is a smooth mapping of smooth manifolds, we might also consider smooth sections. An element of F x is called the germ of a section at x. By definition, germs are equivalence classes of pairs: (U, s) where x U X and s F (U). This is called a local section defined in a neighborhood of x. Two local sections (U, s), (V, t) are equivalent if there exists a third local section (W, r) of F at x so that x W U V and the restriction maps res U W : F (U) F (W ) and resv W : F (V ) F (W ) send both s, t to r. The stalk F x is the set of equivalence classes of local sections [(U, s)] which we abbreviate as [s]. Usually we say s is a section defined in a nbh of x and [s] is the germ of s at x. Example In the last example G(U) = {f : U Ω : ( g, h)( x U)f(x) = g(x)/h(x)} we have that G x is the ring of all f = g/h so that h(x) 0. This is the ring Γ(X) = k[x]/p localized at the maximal ideal m x of all functions f Γ(X) so that f(x) = 0. (Recall that a ring can be localized at any prime ideal and the result R p will be a local ring, i.e., it has a unique maximal ideal.) 3.2. Sheaf. Definition A presheaf F is defined to be a sheaf if the following sequence is exact for any collection {U α } of open sets in X where U = U α : 0 F (U) α F (U α ) α,β F (U α U β ) Exactness means two things: (1) The restriction maps res U U α : F (U) F (U α ) give a monomorphism F (U) α F (U α ) In other words, any two sections γ, γ of F over U whose restrictions to every U α agree must be equal. (2) If γ α F (U α ) is a family of sections which agree on all overlaps: γ α Uα U β = γ β Uα U β for all α, β, then there exists a unique γ F (U) so that γ α = γ Uα Proposition F ( ) has exactly one element. Proof. Let U =. Then U is covered by the empty collection of open sets U α (i.e., there is no α). Then the statement restriction of γ, γ F (U) to any U α agree is vacuously true. So, γ = γ and F (U) has only one element. The constant presheaf is not a sheaf in general. Take X be the union of two disjoint open sets U 1, U 2 in R 2. Then we have 0 F (X) F (U 1 ) F (U 2 ) F (U 1 U 2 )
19 But these are: which cannot be exact. Exercise Show that is exact iff is exact in the usual sense. NOTES ON ALGEBRAIC GEOMETRY MATH 202A 19 0 Z Z Z 0 0 A f B g h C 0 A f B g h C Another classical example of a presheaf which is not a sheaf is the following. Let M n be a smooth manifold and let m >> n. Let E be the presheaf on M given by E(U) = {smooth embeddings f : U R m } This is a presheaf since the restriction of any embedding U R m to an open subset V U is still an embedding. However, this is not a sheaf since a family of compatible local embeddings might have a selfintersection. What we want to do is to replace the presheaf by a sheaf which is locally the same as the presheaf. In this example the resulting sheaf is the sheaf of local immersions: E (U) = {smooth immersions f : U R m } Immersions are mappings which are embeddings when restricted to some open covering of U: f : U R m is an immersion if there is an open covering U α of U so that f Uα is an embedding for every α. Sheafification of a presheaf is a generalization of this construction Sheafification. We could try to construct a sheaf F out of a presheaf F by considering all open coverings of every open set U and pasting together F (U α )s. But this is messy. Instead we use the stalks. A section of F over U will be defined to be a section of the stalk bundle of F which looks locally like sections of F. By the stalk bundle of F we mean the set of all pairs (x, g) where x X and g F x. (g for germ ). Call this E(F ) = {(x, g)}. For every open subset U X we define F (U) to be the set of all sections σ of the stalk bundle (i.e., σ(x) F x for all x U) satisfying the following local compatibility condition. (*) For every x U there is a local section (V, s) at x (Recall this means x V and s F (V )) which represents the section σ(y) F y for all y V. I.e., σ(y) is the germ of s at y for every y V. We think of this as saying that the section σ is continuous. Example Consider the presheaf G(U) of functions on U which are given by fractions f(x) = g(x)/h(x). In class I showed that g, h Γ(X) = k[x]/p are unique up to multiplication by a common factor. The stalk is G x is the local ring of all fractions g/h where h(x) 0. What is the sheafification G (U)? (1) A section σ of the stalk bundle over U gives a mapping U Ω by x σ(x) (evaluated at x). So, at each x U there might be different polynomials g(x), h(x) so that σ(x) = [g/h] (the germ of g/h at x). (2) The local condition says that there are particular polynomials g(x), h(x) so that σ(y) is the germ of g/h at y.
20 20 KIYOSHI IGUSA BRANDEIS UNIVERSITY In other words, a section of G over U is a function f : U Ω which is locally given by rational functions. So, G (U) = Γ(U), the ring of regular function on U. Theorem F is a sheaf and any morphism of F into a sheaf factors uniquely through F. Proof. This is intuitively obvious since the sections of F are locally defined. Suppose that γ, γ F (U) and γ Uα = γ Uα for all α. By definition γ, γ are special sections of the stalk bundle E(F ). They are determined by their value at each point. Since U = U α, the condition implies γ x = γ x for all x U. So, γ = γ. For exactness at the next point: if sections of U α agree on overlaps, they define a (not necessarily continuous) section of U. But continuity on each U α implies continuity on U. Let o X, the structure sheaf of X be defined by o X (U) = G (U) the ring of regular functions on U. Theorem Let U = X f be the set of points in X so that f 0. Then Γ(X f ) = Γ(X) f. We need the following lemma first. Lemma Let K be the field of fractions of the domain Γ(X) = k[x]/p. For every x X we have G x K and a section σ of the stalk bundle of G satisfies the local compatibility condition (*) if and only if σ(x), σ(y) are the same element of K for all x, y X. In other words, G X(U) = G x K x U This follows from the fact that any two fractions f/g, h/k which give the same function U Ω on any open set U are the same element of K. Proof of Theorem By the lemma we have: o X (X f ) = x X f G x K We also have R f K is the subring generated by R = Γ(X) and 1 f. So, elements have the form g/f m which is a function on X f. So, R f o X (X f ). We need to prove the reverse inclusion. So, take any F o X (X f ). Let This is an ideal in R. F = B = {h R : hf R} g f m f m F R f m B So, we want to show that f m B for some m. In other words, f r(b). For all x X f, F o X (X f ) K. So, F = g/h where h(x) 0. In other words h B. So, x / V (B). Since this holds for all x X f we have X f V (B) =. So, V (B) V (f). By the Nullstellensatz, this implies r(f) r(b). So, f r(b) as needed. Taking the case f = 1 (so that X f = X) we get: Corollary Γ(X) = o X (X). In other words, function on X which are given locally by fractions f/g can be given globally as a single polynomial function. Theorem Let x X. Then o x = K if and only if x is generic.
21 NOTES ON ALGEBRAIC GEOMETRY MATH 202A 21 Proof. Suppose x is generic and let f/g be any element of K. Then g 0 implies that g(x) 0 since, otherwise, x V (g) X contradicting the fact that x = X. So, f/g o x. Conversely, suppose that x is not generic. Then x = Y X. Then Y = V (p). Let f 0 p. Then 1/f / o x. So, o x K. Proposition Suppose that X Ω n, Y Ω m are irreducible closed subsets and f : X Y is a continuous function. Then the following are equivalent. (1) f is a morphism (i.e., f is given by m regular functions on X) (2) g Γ(Y, o Y ), g f Γ(X, o X ). I.e., f : Γ(Y, o Y ) Γ(X, o X ). (3) U Y open, f : Γ(U, o Y ) Γ(f 1 U, o X ). (4) x X, the composition of any map germ in o f(x) with f is a map germ in o x. Although the statements have slightly different meanings, the proof is the same as in [4].
22 22 KIYOSHI IGUSA BRANDEIS UNIVERSITY 4. Affine varieties and affine schemes We are now ready to give the definition of an affine variety. Then we will paste affine varieties together to get prevarieties. Affine schemes and preschemes will be seen to be quotient spaces with natural structure sheaves on them Affine varieties. Definition An affine variety over (k, Ω) is a topological space X together with a sheaf o X of Ωvalued functions on X so that the pair (X, o X ) is isomorphic to an irreducible subset Σ of Ω n for some n together with its structure sheaf o Σ : (X, o X ) = (Σ, o Σ ). Affine nspace A n is defined to be (X, o X ) where X = Ω n. Elements of Γ(U, o X ) will be called regular functions on U. When we say that o X is a sheaf of Ωvalued functions on X we mean that it is a subsheaf of the sheaf of functions X Ω. Equivalently, for all open U X, Γ(U, o X ) is a subset of the set of functions U Ω and, for all V U, the restriction map res U V : Γ(U, o X) Γ(V, o X ) is given by restriction of functions. Note that any mapping f : X Y induces a mapping on the sheaves of Ωvalued functions on X, Y : f : F (Y ) F (X). The last Proposition gives us the definition of a morphism of affine varieties. Definition A morphism of affine varieties (X, o X ) (Y, o Y ) is a continuous function f : X Y which satisfies any of the following equivalent conditions. (1) There are isomorphisms g : (X, o X ) = (Σ, o Σ ) and h : (Y, o Y ) = (Σ, o Σ ) so that the composition h f g 1 : Σ Σ is a morphism (i.e., a mapping given by polynomial equations). (2) f sends Γ(Y, o Y ) to Γ(X, o X ). (3) For every open subset U Y, f : F (U) F (f 1 U) sends Γ(U, o Y ) to Γ(f 1 U, o X ). (4) For every x X, f (o f(x) ) o x. (The composition of any regular map germ Y, f(x) Ω with f : X, x Y, f(x) is regular.) It follows immediately that Hom k alg (Γ(Y ), Γ(X)) = Mor((X, o X ), (Y, o Y )) The structure sheaf o X helps us to distinguish between affine varieties which, as topological spaces, are equivalent, i.e., homeomorphic, but not isomorphic as varieties. Example Suppose that k is algebraically closed. Then the only closed subsets of A 1 are finite subsets of k and the whole space A 1. Let f : A 1 A 2 be the morphism given by f(x) = (x 2, x 3 ). The image of f is Σ = {(s, t) A 2 : s 3 = t 2 }. Then f induces a mapping A 1 Σ which is a homeomorphism (it is a continuous bijection which sends finite sets to finite sets). However, the structure sheaves are different since R = Γ(Σ) = k[s, T ]/(S 3 T 2 ) has a maximal ideal m = (S, T ) (corresponding to 0 Σ) so that m/m 2 is 2dimensional as vector space over k = R/m. (Whereas, every maximal ideal M in Γ(A 1 ) = k[x] has the form M = (X a), a k. So, M/M 2 = k is one dimensional.) Therefore, the structure sheaf of Σ contains the information that 0 is different from the closed points of A 1 : 0 is a singular point of Σ. Exercise Let (X, o X ) be an affine variety. Show that any section of o X over any open set U is a continuous function U Ω. Equivalently, for any f Γ(X, o X ), X f = {x X : f(x) 0} is an open subset of X.
V (f) :={[x] 2 P n ( ) f(x) =0}. If (x) ( x) thenf( x) =
20 KIYOSHI IGUSA BRANDEIS UNIVERSITY 2. Projective varieties For any field F, the standard definition of projective space P n (F ) is that it is the set of one dimensional F vector subspaces of F n+.
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