Math 214A. 1 Affine Varieties. Yuyu Zhu. August 31, Closed Sets. These notes are based on lectures given by Prof. Merkurjev in Winter 2015.

Transcription

1 Math 214A Yuyu Zhu August 31, 2016 These notes are based on lectures given by Prof. Merkurjev in Winter Affine Varieties 1.1 Closed Sets Let k be an algebraically closed field. Denote A n = A n k = {a = (a 1,, a n ) a i k} e.g. A 0 =point, A 1 = k. Let A n = k[t 1,, T n ] be the k-algebra of polynomials, where for f A n, a A n, we have f(a) k. Now let S A n a subset, let Z(S) = {a A n f(a) = 0 for all f S} A n. A subset Z A n is called closed if Z = Z(S) for some S A n Example 1. For n = 1, Z A n closed Z is finite or Z = A 1. Example If S S A n, then Z(S) Z(S ). 2. S A n, I an ideal in A n generated by S. Then Z(S) = Z(I) = Z( I), where I = {g A n g m I for some m > 0}. 3. (Hilbert Basis Theorem) Every ideal in A n is finitely generated. Writing J = f 1,, f k, then Z(S) = Z(J) = Z({f 1,, f k }). 4. Z( S α ) = Z(S α ), with S α A n. This proves that intersection of closed sets is closed. 5. Z(S T ) = Z(S) Z(T ). So the union of finitely many closed sets is closed. Hence we get the Zariski topology on A n with closed sets Z(S). Going back to A 1, the open sets are or cofinite sets. So every two nonempty open sets must intersect. If X A n subset, let I(X) = {f A n f(x) = 0 for all x X}. Then we have this correspondence between subsets of A n and subsets of A n given by Z and I operators. In fact, I(X) is a radical ideal in A n. Properties. 1. If X X A n, then I(X) I(X ). 2. IZ(S) S, and ZI(X) X. 3. IZI = I and ZIZ = Z. 1

2 Proof. IZI(X) = IZ(I(X)) I(X) and ZI(X) X. So applying I, we get IZI(X) I(X). So IZI = I. 4. ZI(X) = X the closure of X. Proof. X ZI(X) closed, so X ZI(X). On the other hand, suppose Z(S) = X, then ZI(X) ZI(X) = ZIZ(S) = Z(S) = X. 5. S A n, then IZ(S) = S. Proof. Note that IZ(S) S, so IZ(S) S. On the other hand, recall Hilbert Nullstellensatz: S A n and f A n such that f is zero on the set of all common zeros of polynomials in S, i.e. f IZ(S), then f m in S for some m > 0. So we get the correspondence between radical ideals in A n and closed subsets in A n with inverse bijections given by the restriction of Z and I. We have A n, 0 A n, and Max(A n ) {points}. Warning: we need k to be algebraically closed for this to work, otherwise Nullstellensatz does not apply. A subset X A n is called quasi-affine if X = Z U with Z closed and U open in A n. e.g. In A 1, X is quasi-affine if and only if X is closed or open. Properties 1. Closed and open subsets are quasi-affine. 2. Intersection of two quasi-affine subsets are quasi-affine. Every closed (open) subset of a quasi-affine set is quasi-affine. 3. A subset X is quasi-affine if and only if X is open in X. Proof. Note that X = U X quasi-affine. On the other hand, if X = Z U, then X Z implies XsubsetZ. So X = Z U X U X. Then X = X U, i.e. X is open in X. 1.2 Regular Functions Suppose X A n closed subset. A function f X k is regular if F A n such that f(x) = F (x) for every x X. Let k[x] denote the k-algebra of all regular functions on X. Hence we get a surjective map A n k[x] by F F X with kernel being I(X). So k[x] A n /I(X) reduced k-algebra (having no nontrivial nilpotent element). Let t i = T i X be the coordinate functions on X. So k[x] is generated by t i, hence it is finitely generated by coordinate functions. Proposition 3. Every commutative finitely generated k-algebra is isomorphic to k[x] for some closed X A n. Proof. Consider A = k[a 1,, a n ] and the evaluation map A n A via T i a i. Let J be the kernel of the map, hence a radical ideal. Then we have A A n /J. Let X = Z(J), then I(X) = IZ(J) = J as J is radical. Hence we have A = A n /I(X) k[x] Example k[a n ] = A n, as I(A n ) = 0 2. Suppose X = {a} a point in A n. Then k[x] = k since all functions on a point are constant. 2

3 3. Suppose that f A n irreducible. Then X = Z(f) A n a hypersurface. So we have k[x] = A n / f = A n / f a domain, as f is irreducible hence f is prime and hence radical. Suppose X A n is closed, then we have k[a n ] k[x] surjective. Now suppose S k[x], denote Z X (S) = {x X f(x) = 0 for all f S} X. If T k[a n ] and maps to S k[x], we have Z X (S) = Z A n(t ) X closed in X. If Y X a subset, denote I X (Y ) = {f k[x] f(y) = 0 for all y Y } a radical ideal in k[x]. This gives a correspondence between closed subsets in X and radical ideals in k[x] via I X ( ) and Z X (x). In fact, observe that clsoed sets in X are a subset of closed sets in A n. Previously, we have a correspondence between closed sets in A n and radical ideals in A n given by I and Z, and we have an inclusion of radical ideals in k[x] into the radical ideals in A n via J φ 1 (J) where φ A n k[x] surjective k[x] A 1 0 { } m a maximal ideal. Also if S k[x], then Z X (S) = S = k[x] S = k[x]. Local Definition. Suppose X A n closed and f k[x], define D X (f) = X Z X (f) = {x X f(x) 0} the set in X. If f, g k[x], then we have D X (fg) = D X (f) D X (g). Moreover, if U X open, then X U = Z X (S) is closed with S k[x]. So U = X Z X (S) = f S D X (f). S can be chosen to be finite, so principal open sets form a basis for Zariski topology on X. Example 5. Set X = A 1 {0}, T A 1 = k[a 1 ] and t = T X. Suppose t(x) 0, x X, then 1 t X k should be regular. If X A n is quasi-affine, f X k and x X, then f is regular at x if there exists a neighborhood U of x and G, H A n such that H(y) 0 for any y U and f(y) = G(y) for all y. First, there H(y) exists G, H A n such that H(x) 0, f = H in a neighborhood of x. If X is closed, there exists g, h k[x], G with h(x) 0, f = g in a neighborhood of x. h Proposition 6. Let X be a closed subset, f X k. Then f is regular iff f is regular at every x X. Proof. If f is regular, then f = F X, F A n. So f = F. Hence f is regular at every x X. 1 On the other hand, x X, there exists a neighborhood U x of X such that f = gx h x in U x for some g x, h x k[x]. So g x = fh x in U x. As X U x is closed, there exists l x k[x] such that l x (x) 0 but l x X Ux 0. So we have g x l x = fh x l x holds in U x and X U x hence on X. Since (h x l x )(x) 0, we have f = gxlx h xl X in U x. Replace g x, h x by g x l x, h x l x respectively. Then g x = fh x in k[x] and f = gx h x in U x. Let S = {h x x X}. h x 0 implies that Z(S) = implies that S = k[x]. So we can write So f is regular. 1 = h x m x, m x k[x] almost all zero x X f = fh x m x = g x m x k[x] x X x X So if X A n is quasi-affine subset, a function f X k is regular if f is regular at every point. Write k[x] the k-algebra of all regular functions on X, and is reduced, but not finitely generated in general. 3

4 Example X = A 1 {0} and t coordinate function on X, so 1 t k[x]. This means that k[t, t 1 ] k[x], but not generated by t. (In fact, they are equal.) 2. Let X = Z(T 1 T 2 T 3 T 4 ) Z(T 2, T 4 ) A 4 a quasi-affine subset. Let t i = T i X. Let f = t1 t 4 = t3 t 2 and for all x X, t 2 (x) 0 or t 4 (x) 0. So X = U 2 U 4, where U i = {t i 0}. Then f is regular. And we have f = t1 t 4 Some Properties: defined on U 4 and f = t3 t 2 defined on U 2 1. Let Y be a quasi-affine subset of a quasi-affine set X, f k[x], then f Y k[y ] 2. Let X be a quasi-affine set, f X k and X = α U α open cover. Then we have f k[x] f Uα k[u α ] for each α. The converse follows from the local argument of regularity. 3. If f X k is regular at x X, then f is regular in a neighborhood of x. 4. Let f k[x] be such that f(x) 0 for all x X, then 1 f ink[x]. 5. Proof. Let x X. Then in a neighborhood of x, f(y) = G(y) for polynomials G, H. f(x) 0 H(y) implies that G(x) 0. So G is not zero in a neighborhood of x, i.e. f = G and G(y) 0 y. Then H 1 f = H G is regular at x. Lemma 8. Let X = α u α be an open cover of a topological space X, Y X a subset. Then Y is closed (in X) if and only if Y U α is closed in U α for every α. Proof. The forward direction is immediate. For the converse, let s prove that X Y is open. Take x X Y, then x U α for some α. There is a closed subset Z α X such that Y U α = Z α U α. Then U = (X Z α ) U α is open in X. Since x Y, x Z α. Then x (X Z α ) U α. We claim that U Y =. Suppose y U Y, then y U α Y = Z α U α Z α, a contradiction. So U is a neighborhood of x, with U X Y. Let f k[x]. Then Z X (f) = {x X f(x) = 0} is closed in X. Proof. Suppose x X, then in a neighborhood U x of x, f = Gx H x Z X (f) U x = Z Ux (f) = Z Ux (G x ) = Z X (G x ) U x is closed in U x. As X = x X U x, Z x (f) is then closed in X by the lemma. Let f k[x], then f X k = A 1 is continuous. with G x, H x polynomials. So Proof. Suppose Z A 1 is closed. Want to show that f 1 (Z) is closed in X. Note that the only closed sets in A 1 is finite or A 1. If Z = A 1, then f 1 (Z) = X closed in X. Suppose Z is finite, then f 1 (Z) = z Z f 1 (z) = z Z Z X (f z) finite union of closed sets is closed. Let X be q quasi-affine set. Then we get maps between closed subsets of X and radical ideals in k[x] given by I and Z. Let Z X closed, then Z = Z X (S). Then (ZI)(Z) = ZIZ X (S) = Z(S) = Z. So ZI is the identity, I is injective and Z is surjective. Therefore we have more radical ideals in k[x] than the closed subsets of X. Example 9. Let X = A 2 (0, 0), then we will see that k[x] = k[a 2 ] = A 2. If J 1 = 1 the unit ideal in k[x], and J 2 = t 1, t 2. Then Z(J 1 ) =, but Z(J 2 ) = as we have removed the only zero (0, 0). So the map is not always injective. 4

5 1.3 Regular Maps Let X A n be a quasi-affine set and f X A m by x (f 1 (x), f 2 (x),, f m (x)) where f i X k. We say that f is regular if f i k[x] for all i. Example Every f X k = A 1 a regular function is a regular map. 2. The inclusion map X A n is regular. Now let X A n and Y A m quasi-affine sets. We say a map f X Y is regular if the composition X f Y A m is regular. To give a regular map f X Y is to give f 1,, f m k[x] such that (f 1 (x),, f m (x)) Y for all x X. Example Inclusions are regular. 2. Let f X Y be regular, and X X be a quasi-affine subset. Then f X X Y is regular. Now suppose Im(f) Y is a quasi-affine subset of Y, then f X Y is regular. 3. (Image of quasi-affine map is not necessarily quasi-affine) Consider the regular map f A 2 A 2 defined by f(x, y) = (xy, y). Let X = im(f). Then X = A 2 /(L/(0, 0)), where L is the x-axis. So X = A 2 and A 2 /X = L/(0, 0). If X is quasi-affine, then X is open in A 2. This implies that L/(0, 0) is closed in A 2, a contradiction. So X is not quasi-affine. Proposition 12. Regular maps are continuous. Proof. Suppose f X k A m is regular and Z Y closed. Then there exists S A m such that Z = Z Y (S). Then f 1 (Z) = G S f 1 (Z Y (G)) where f 1 (Z Y (G)) = {x X G(f 1 (x),, f m (x)) = G(f 1,, f m )(x) = 0} is closed in X. Corollary 13. If f X Y is a regular map, Y Y is a quasi-affine subset, then f 1 (Y ) is a quasi-affine subset of X. Proof. Suppose Y = Z U with Z closed and U open in Y. Then f 1 (Y ) = f 1 (Z) f 1 (U) quasi-affine in X. Proposition 14. Let X and Y be two quasi-affine sets, f X Y. Then f is regular if and only if for all h k[y ], the composition h f k[x]. Proof. : Suppose t i be the coordinate functions on Y, and f = (f 1, f 2,, f m ). Then t i f = f i regular, so f is regular. : Suppose h k[y ]. Is h f regular? Let x X, then y = f(x) Y. In a neighborhood of y, h = G H where G, H are polynomials and H 0. Then on f 1 (U) a neighborhood of x, we have h f = G(f1,f2,,fm) H(f 1,f 2,,f m) on f 1 (U), with numerator and denominator both regular. So h f is regular at x. Hence h f is regular. Proposition 15. If X f Y g Z are regular maps, then g f is also regular. For f X Y regular, the induced map f k[y ] X by h h f is a homomorphism of k-algebras. Now X f Y g Z, the induced map (gf) = f g k[z] k[x], i.e. the functor is contravariant. 1.4 Products We identify A n A m = A n+m. NoW let X = Z A n(s) closed in A n, then A n+m X Y = Z A n+m(s T ). Now suppose X A n, Y A m quasi-affine sets, then X = X 1 /X 2, Y = Y 1 /Y 2, with X i, Y i closed. Then X Y = (X 1 Y 1 )/(X 1 Y 2 X 2 Y 1 ) quasi-affine in A n+m. We have two natural projections p X Y X 5

6 and q X Y Y, with corresponding k-algebra homomorphism p k[x] k[x Y ] and q k[y ] k[x Y ]. So we have the map φ k[x] k k[y ] k[x Y ] via f g ((x, y) (f(x), f(y))) Proposition 16. The map φ is injective. Proof. Let {f i } I basis for k[x]. Every element in k[x] k k[y ] can be uniquely written as h = i I f i g i for g i k[y ], with terms almost all zero. Now suppose 0 = φ(h)(x, y) = f i (x)g i (y) x X, y Y i I Then for all y Y, f i g i (y) = 0 in k[x]. So g i (y) = 0 for all i and y Y. Hence g i = 0 for all i and therefore h = 0. Corollary 17. φ is an isomorphism if X and Y are closed. Proof. If X and Y are closed, then k[x Y ] is generated by coordinate functions. Take X A n quasi-affine set. Take X X X = {(x, x) x X} = Z(t i t i ) closed in X X. This is called the diagonal of X. Let d X X X by x (x, x) the diagonal map. Then Im(d) = X. Moreover, this map gives an isomorphism between X and X. Remark The Zariski topology on X Y is not the product topology. The diagonal X is closed in X X (for product topology) if and only if X is Hausdorff. But the Zariski topology is not Hausdorff as any two non-empty open sets intersect. Consider f X Y A m and f X Y A m regular maps of quasi-affine sets. Then f f X X Y Y is called the product of maps. If f = (f 1,..., f m ) and f = (f 1,..., f m), then f f = (f 1 p,..., f m p, f 1 q,, f m q) where p, q projection map from X Y onto X and Y respectively. Now suppose f X Y regular map, then f 1 Y X Y Y Y by (x, y) (f(x), y). Then X Y (f 1 Y ) 1 ( Y ) = {(x, y) X Y f(x) = y} = Γ f. This is isomorphic to X under p Γ f X and X Γ f x (x, f(x)). Proposition 18. (Universal Property of Product) Let X, Y, Z be quasi-affine sets and f Z X and g Z Y two regular maps. Then there exists a unique regular map h Z X Y such that p h = f and q h = g, where p, q the usual projection maps. Proof. Assuming h exists, then p(h(z)) = f(z) and q(h(z)) = g(z), so necessarily h(z) = (f(z), g(z)). This gives uniqueness. Now let h Z d Z Z f g X Y compositions of regular maps so regular. This proves the existence. Proposition 19. Let f, g X Y two regular maps. If f and g agree on a dense subset of X, then f = g. Proof. Let h X d X X f g Y Y Y Then h 1 ( Y ) X is closed and contains a dense subset of X. So h 1 ( Y ) = X, i.e. f = g. 6

7 1.5 Affine and Quasi-affine Varieties Take X = Z(xy 1) A 2 closed and Y = A 1 /{0} = D A 1(t) = {x A 1 t(x) 0} A 1 is not closed. But they are isomorphic under the map X Y by (x, y) x and Y X by t (t, 1 ) which is regular since t t 0. So closed-ness is not preserve. Now let X be a closed set, f k[x] and D X (f) = {x X f(x) 0} X principal open set. Define Z = Z(f t 1) X A 1 closed, where t is the coordinate function. I claim that D X (f) Z. The maps 1 are D X (f) Z z (x, f(x) ) and Z D X(f) (x, t) x. Now k[d X (f)] k[z] = k[x A 1 ]/ ft 1. But k[x A 1 ] k[x] k k[a 1 ] = k[x] k k[t] = k[x][t]. Thus k[x A 1 ]/ ft 1 = k[x][t]/ ft 1. Also k[x][t]/(ft 1) k[x] (f) localization of k[x] with respect to {f n } Then k[x] is reduced then k[x] (f) is reduced. Hence ft 1 = ft 1. So k[d X (f)] k[x] (f). Example k[a 1 /{0} = k[d A 1(t)] = k[t] (t) = k[t, t 1 ]. 2. k[a 2 /{(0, 0)}] = k[t 1, t 2 ]. If U X dense, then the restriction map k[x] k[u] is injective. If f U X the inclusion map, then the restriction map is f. It is injective as if the maps agree on U dense subset, they agree on X. Let U i = D A 2(T i ) A 2 for i = 1, 2. Then U 1 U 2 X = A 2 /{(0, 0)} A 2. Similarly, we have k[a 2 ] k[x] k[u i ] k[u 1 U 2 ] But k[u 1 ] = k[t 1, T 2, T 1 1 ] and k[u 2 ] = k[t 1, T 2, T 1 2 ]. And k[u 1 U 2 ] = D A 2(T 1 T 2 ) = k[t 1, T 2, T 1 1, T 1 2 ]. So we actually get equality in above. A quasi-affine variety is a quasi-affine set. An affine variety is a quasi-affine variety that is isomorphic to a closed set. If X is an affine variety, then D X (f) is also an affine variety. Example 21. A 2 /{(0, 0)} is not affine. 1. The map Z from radical ideals in k[x] = k[t 1, t 2 ] to closed subsets in X is not injective: Z(k[t 1, t 2 ]) = Z( t 1, t 2 ) =. 2. f X A 2 is not an isomorphism. Apply f k[a 2 ] k[x] isomorphism as they are both polynomial rings in two variables. But the functor AffVar(k) op CAlg(k) is full and faithful. 3. g A 1 /{0} X is closed, but g k[x] k[a 1 /{0}] = k[t, t 1 ] is not surjective. Theorem 22. Let X be a quasi-affine variety, x X. Then x has an affine neighborhood. Proof. Suppose X = Z 1 /Z 2 A n quasi-affine subset, with Z 1, Z 2 closed in A n. So x Z 2, then there exists F A n F Z2 = 0 but F (x) 0. Let f = F Z1 k[z 1 ]. Then x D Z1 (f) = Z 1 /Z(f) Z 1 /Z 2 = X. So D Z1 (f) is an affine neighborhood of x in X Corollary 23. Every quasi-affine variety admits an affine open cover. 1.6 Irreducible Variety Lemma 24. Let X be a topological space. Then the following are equivalent 1. Every subsequence of closed sets Z 1 Z 2 is stable, i.e. n such that Z n = Z m for all m n 2. Every non-empty set of closed subsets of X has a minimal element. We say X is Noetherian if X satisfies 1 and 2. Example 25. A n is Noetherian since there is one-to-one correspondence (which reverses the inclusion) between closed subsets of A n and radical ideals in A n which is a Noetherian ring. So every decreasing sequence of closed subsets in A n stabilizes. 7

8 Lemma 26. Every subspace of a Noetherian space is Noetherian. Corollary 27. Every subset of A n is Noetherian. Lemma 28. Every Noetherian space is compact. Proof. Let Z α X a collection of closed subsets. Let A = {Z α1 Z αn } set of all finite intersection of closed subsets. So there exists Z α1 Z αn the minimal element in A. Claim 29. This intersection is empty. Supposet not. There exists x Z αi, i = 1,, n. There exists α n+1 x Z α1. Then x n+1 i=1 Z αi < n i=1 Z αi a contradiction. Hence we prove the claim and finish the proof Proposition 30. Every quasi-affine variety admits a finite affine open cover. Lemma 31. Let X be a topological space. Then the following are equivalent. 1. If Z i < X, i = 1, 2 are closed, then Z 1 Z 2 X. 2. If U i X, i = 1, 2 non-empty open subsets, then U 1 U Every non-empty subset of X is dense in X. Proof. We know that : Suppose U X open, then U (X/Ū) =. So X/Ū =, i.e. X = Ū. 3 2: U 1, U 2 X non-empty open sets. If U 1 U 2 =, then U 1 X/U 2 closed. So Ū1 X/U 2 X. So U 1 is not dense in X. A topological space satisfying 1-3, is called irreducible. Irreducibility Test A X irreducible if A Z 1 Z 2, Z 1, Z 2 closed in X, then either A Z 1 or A Z 2. Proposition 32. A quasi-affine variety is irreducible k[x] is a domain. Proof. : Suppose f, g k[x] with fg = 0. Then X = Z(0) = Z(fg) = Z(f) Z(g) closed sets. So Z(f) = X or Z(g) = X, i.e. f = 0 or g = 0. : Suppose X = Z 1 Z 2, Z i closed. Suppose Z 1 X and Z 2 X. Then there exist f i k[x] f i 0 but f i Zi = 0 for i = 1, 2. Then f 1 f 2 is trivial on X = Z 1 Z 2, i.e. f 1 f 2 = 0, contradiction. Example A n is irreducible. 2. A 1 /{0} is irreducible as k[a 1 /{0}] = k[t, t 1 ] a domain. 3. If f A n is irreducible, then Z(f) A n is irreducible as: k[z(f)] = A n /(f) a domain as (f) is prime. 4. Let X = Z(T 1 ) Z(T 2 ) = Z(T 1 T 2 ) A 2. Then k[x] = k[t 1, T 2 ]/(T 1 T 2 ) which is not a domain as t 1 t 2 0 on the axes. Note that Z(T i ) A 1, hence both axes are irreducible, so X is a union of irreducible parts. Actually, these components are unique. 5. Suppose X is an affine variety. We have a corresponding between closed subsets in X and radical ideals in k[x]. So if Y is closed and I the corresponding ideal then k[y ] k[x]/i. This implies correspondence between irreducible closed subsets and prime ideals in k[x]. Theorem 34. Let X be a Noetherian topological space. Then there are irreducible closed subsets X i X, i = 1,, n such that X = X i and X i / X j for i j and these subsets are unique. Proof. Let Y be the smallest closed subset of X that has no decomposition. Then Y is not irreducible. So Y = Y Y, Y, Y closed proper subset of Y. By minimality of Y, Y and Y have decomposition, so does Y = Y Y, a contradiction. This prove the existence part. Now suppose X 1 X 2 X h = X i X 2 X l X i By irreducibility test, X i X j for some j. Similarly, there is some k such that X j X k. So we have i = k, X i = X j, the decomposition is unique. 8

9 We call X i s irreducible components of X. Example 35. Let X = Z(y 2 xz, z 2 y 3 ) A 3. Then on X, we have z 3 (z x 3 ) = z 4 x 3 z 3 = z 4 y 6 = y 6 y 6 = 0. So X = Z X (z) Z X (z x 3 ). Observe that Z X (z) = Z A 3(z, y 2 xz, z 2 y 3 ) = Z A 3(y, z) A 1 irreducible On Z X (z x 3 ), note that x 4 = xz = y 2, so we have x 4 (y x 2 ) = x 4 y x 6 = y 3 x 6 = y 3 z 2 = 0. So Z X (z x 3 ) = Z X (z x 3, x) Z X (z x 3, y x 2 ). But Z X (z x 3, x) = Z A 3(x, y, z) = {0} which is in Z X (z) already. Now k[z X (z x 3, y x 2 )] = k[x, y, z]/(z x 3, y x 2 ) = k[x]. So Z X (z x 3, y x 2 ) A 1 irreducible. So X = Z A 3(y, z) Z A 3(z x 3, y x 2 ). The first is embedded as the x-axis, and the second is a line under the parametrization x = t, y = t 2, z = t 3. The first set contains (1, 0, 0) and the second set contains (1, 1, 1). Proposition 36. Let A be a subset of topological space X. Then A is irreducible if and only if Ā is irreducible. Corollary 37. Let X be an irreducible topological space, U X a non-empty open subset, then U is also irreducible. Proposition 38. If X and Y are irreducible quasi-affine varieties, then so is X Y. Proof. Fix y Y, then X X y X Y closed. Suppose X Y = Z 1 Z 2 with Z i X Y closed, then X y Z 1 Z 2. So X y Z 1 or X y Z 2. Now let Y i = {y Y (x, y) Z i, x X}. Then Y = Y 1 Y 2. I claim that Y i is closed in Y. For all x X, (x y) Z i = x Y x, with Y x Y closed as intersection of two closed sets. Then we have Y = {y Y (x, y) Z i, x X} = Y i x X so Y i is closed as intersection of closed sets. As Y is irreducible, then Y = Y i. Hence X Y = Z i. Corollary 39. Let A and B be two commutative k-algebras. If A and B are domains, then so is A k B. 1.7 Rational Functions and Maps Let X be a irreducible quasi-affine variety. Consider the set of pairs (U, f) where U X nonempty open subset and f k[u] (f is a regular function on U.) We say that (U 1, f 1 ) (U 2, f 2 ) if f 1 U1 U 2 = f 2 U1 U 2, i.e. agree on the intersection of domian. Note that since X is irreducible, U 1 U 2. Check this is an equivalence relation. Denote by k(x) the set of equivalence classes of pairs, [U, f]. If U U open, then (U, f) (U, f ) where f U = f. Thus [U, f] = [U, f ], so we can shrink open sets and replace the functions by their restrictions. These equivalence classes are called rational functions. We define the operation as follows: [U 1, f 1 ] + [U 2, f 2 ] = [U 1 U 2, f 1 U1 U 2 + f 2 U1 U 2 ] a [U, f] = [U, af], a k This makes k(x) a commutative k-algebra. Note that the zero element is 0 = [U, 0] = [X, 0]. So if [U, f] is nonzero in k(x), then f 0. Note that [D X (f), 1 f ] [U, f] = [D X(f), 1] the identity element in k(x) Hence, k(x) is a field. In fact a field extension of k. We have the natural embedding k[x] k(x) by f [X, f] We say X is irreducible iff k[x] is a domain. Recall that X is irreducible, if X is non-empty and not a union of two proper closed subsets. We know every X is a unique union of irreducible varieties. Now let X be irreducible, we embed k[x] k(x). 9

10 Proposition 40. k(x) is a quotient field of k[x]. Proof. Suppose [U, f] k(x) with U X open and f k[u]. Let x X. In a neighborhood U of x, f = G H with G, H polynomials and H 0 in U. Now let g = G U, h = H U. So on U, f U = g h. Hence [U, f] = [U, f U ] = [U,g] [U,h] = [X,g] [X,h] We write f for [X, f] for f k[x] and [X, f] k(x). So every f k(x) can be written as f = g h with g, h k[x] with h 0. Remark 41. If X A n, then k(x) is generated by the coordinate functions t i. In particular, k(x)/k is finitely generated as a field. Let X be irreducible and f k(x). So f = [U, g] with U X open and g k[u]. Let Ũ = U where the union is taken over all U X open such that f = [U, g] for some g k[u]. Then (U) is open in X. If f = [U, g ], then g, g agree on U U. So there exists h k[u U ] such that h U = g, h U = g. So there exists a unique g k[ũ] such that f = [Ũ, g]. We call Ũ domain of definition of f and we say that f is regular at every x Ũ. Example 42. Let X = Z(x 2 + y 2 1) A 2 closed subset. Then k[x] = k[x, y]/(x 2 + y 2 1). Observe that f = 1 y x k(x) is defined at all (x, y) with x 0. As on k[x], we have x2 = 1 y 2 = (1 + y)(1 y), f = 1 y x = x is defined at (0, 1) but not (0, 1). So f is regular at X/{(0, 1)}. 1+y Suppose X, Y quasi-affine varieties, with X irreducible. Consider pairs with 0 U X open and f U Y regular map. Recall that (U, f) (U, f ) if f, f agree on U U. A rational map X Y is an equivalence class [U, f] of a pair (U, f). Observe that a rational function on X is a rational map X A 1. Now suppose Y A m and [U, f] X Y, then f = (f 1,, f m ) with f i k[u]. We can view f i as rational functions on X. So to give a rational map X Y is the same as to give rational functions f 1,, f m k(x) such that (f 1 (x),, f m (x)) Y for every x where all f i s are defined. Every rational map X Y has the domain of definition Ũ X. Let f X Y defined on Ũ X, then f = [Ũ, g] where g Ũ Y is regular and define the image Im(f) = Im( g) Y. We say that f is dominant if Im(f) is dense in Y. Suppose g f X Y Z If Im(f) Ũg =, we cannot define g f. But if f is dominant, then Im(f) Y dense, so Im(f) Ũg and we can define g f as follows: Now take g f = [f 1 (Ũg) Ũf, h] where h(x) = g(f(x)) g f X Y A 1 where g k(x) and f dominant. Then we have g f k(x). So we have a induced map f k(y ) k(x) by g g f, which is a homomorphism of fields over k. Moreover (f f) = f f. So we have a functor from the opposite category of irreducible quasi-affine varieties and dominant rational maps to category of field extensions of k. It is given by X k(x) and f f. Example 43. Let X be irreducible and U X a nonempty open set, so U is irreducible. We have the inclusion map U X which is dominant because the image is U and every nonempty set is dense in a irreducible variety. I claim this is an isomorphism in the above category. We define an inverse map X U by [U, id U ]. So U X are birationally isomorphic. Therefore k(x) k(u). Proposition 44. This functor is full and faithful. Proof. Suppose X, Y irreducible varieties. Let α k(y ) k(x) be a homomorphism over k. We want to show that there exists f X Y such that α = f. Now replace Y by a nonempty open subvariety. 10

11 We may assume Y is affine. So Y A m is closed. Let t 1,, t m be the coordinate functions on Y and t i k[y ] k(y ). Let g i = α(t i ) k(x). All the g i s are defined on U X open. So the restrictions g i k[u] k(u) = k(x). So we have k(y ) α k(x) t i k[y ] β k[u] g i Since Y is affine, there exists a regular map h U Y such that h = β and h (t i ) = g i. So h defines a rational map f X Y given by [U, h]. So f k(y ) k(x) = k(u) t i g i = α(t i ). This shows that f and α agree on the generators t i, so f = α. So the functor is full. Now given f, g X Y such that f = g, we know that f = (f 1,, f m ), g = (g 1,, g m ) if Y A m. Then f i = f (t i ) = g (t i ) = g i. So f = g, i.e. the functor is faithful. Now to show f is dominant, it suffices to show h is dominant. Simply note that Im(h) = h(z(0)) = Z(f ) 1 (0)) = Z(ker(β)) = Z(0) = Y and therefore h is dominant. k(x) is always finitely generated. Corollary 45. The functor is an equivalence if the target is the category of finitely generated extensions of k. Proof. Suppose K/k finitely generated field extension. We want a X quasi-affine irreducible vareity such that k(x) K. Suppose K = k(f 1,, f n ) and let A = k[f 1,, f n ] K finitely generated domain. So there exists an irreducible affine variety X such that A k[x]. So k(x) = qf k[x] = qf A = K. We write X Y if they are birationally isomorphic to each other. Note that X Y implies that X Y. However, when U X open, then U X, but it is not necessarily true that U X. For example, A 1 /{0} A 1 but their corresponding coordinate rings are k[t, t 1 ] and k[t], which are not isomorphic as isomorphism send constant to constant. Example 46. Let X = (x 3 y 2 ) A 2. We have the following maps A 1 t (t2,t 3 ) X (x,y) y x A 2 Then X A 1 but X / A 1. Proposition 47. Two irreducible varieties X and Y are birationally isomorphic if and only if there exists non-empty open sets U X, V Y such that U V. Proof. X U V Y Suppose X Y by f U Y and g V X, for U, V open in X, Y respectively. Then we get the maps X f 1 (V ) f V g X Y g 1 (U ) g U f Y Both maps are identities, so we have g(f(x)) = x for all x f 1 (V ) and f(g(y)) = y for all y g 1 (U ). But X U =f 1 (g 1 (U )) f 1 (V ) Y V =g 1 (f 1 (V )) g 1 (U ) If x U f 1 (V ), then g(f(x)) = x. Applying f yields f(x) = f(g(f(x))) V. So f(g(f(x))) = f(x) V. Similarly if y V, then g(y) U and f(g(y)) = y. Hence we get inverse isomorphism f and g between U and V. Proposition 48. Every irreducible variety is birationally isomorphic to a hyperplane in A n. 11

12 Proof. Let X be such variety. Consider K = k(x)/k finitely generated field extension. Let E = k(t 1,, t m ), over which K is separable. So K = E(α) for some α K. Let f E[t] be the minimal polynomial of α. So we can find g k[t 1,, t m, t] = R[t] where R = k[t 1,, t m ] a UFD and g has relatively prime coefficients in R. So f is primitive. If g is irreducible in E[t] and primitive in R[t] and since R is a UFD, g is irreducible in k[t 1,, t m, t]. So Y = Z A m+1(g) A m+1 is irreducible, and k(y ) = qfk[y ] = qfk[t 1,, t m, t]/(g) = qfe[t]/(f) = K = k(x). Note that k(a n ) = k(t 1,, T n )/k is purely transcendental. An irreducible variety X is called rational if X A n for some n k(x)/k purely transcendental there are non-empty open subsets U X and V A n such that U V. Example 49. Suppose char(f ) 2 and X = Z(x 2 + y 2 1) A 2. So k(x) = qfk[x] = qfk[x, y]/(x 2 + y 2 1) = k(x, 1 x 2 ). X A 1 by the map t ( t, t2 1 ) and (x, y) 1+y. Note that quadratic is 1+t 2 1+t 2 x always rational. 1.8 Local Ring of a Subvariety Suppose X a quasi-affine variety, and Y X a closed irreducible variety. An neighborhood of Y in X is an open set U X such that U Y. Note that if U 1, U 2 two neighborhood of Y, so is U 1 U 2 as Y is irreducible. Example When Y = {y}, then the neighborhood is just the usual neighborhood. 2. Y = X then every non-empty open subset is a neighborhood. Consider the pair (U, f) where U is a neighborhood of Y and f k[u]. We define an equivalence relation: (U 1, f 1 ) (U 2, f 2 ) if there exists a neighborhood W of Y in X such that W U 1 U 2 and f 1, f 2 agree on W. The set of classes [U, f] forms a k-algebra, denoted O X,Y Example 51. Suppose X is irreducible, then O X,X = k(x). But in general, O X,Y is not a field. Proposition 52. Now define the map: Define m X,Y = ker(π), an ideal in O X,Y. Then: 1. The map π is surjective. π O X,Y k[y ] [U, f] [U Y, f U Y ] 2. O X,Y is a local ring with unique maximal ideal and residue filed k(y ). Proof. Suppose g k(y ), so g = [W, h] with W = U Y Y, U X open, and h k[w ]. Let y W so there exists an affine neighborhood U of y in U. So y U Y, then U is a neighborhood of Y in X. Let W = U Y closed in U, so W U closed and affine. The map k[u ] restr. k[w ] is surjective, so there exists f k[u ] such that f W = h W. Hence π([u, f ]) = [W, h W ] = [W, h] = g For the second claim, note that k(y ) is a field and m X,Y maximal ideal in O X,Y. Let [U, f] O/m X,Y. Then f U Y is not zero. So D U (f) Y, i.e. D U (f) is a neighborhood of Y in X. Moreover, [D U (f), f][d U (f), f 1 ] = 1, where [U, f] = [D U (f), f]. So [U, f] 1 = [D U (f), f 1 ] Consider g Y X X g(y ) = Y a dominant regular map and Y X closed and irreducible. This induces the map g O X,Y O X,Y [U, f ] [g 1 (U ), g f ] as U X a neighborhood of Y in X and f k[u ] so g f k[g 1 (U )]. Since g is dominant, g(y ) U. Hence g 1 Y. So g 1 (U ) is a neighborhood of Y in X. So we have the following maps: O X,Y g O X,Y k(y ) (g Y ) k(y ) 12

13 where the vertical maps are π and π respectively. Moreover, g (m X,Y ) m X,Y so g is a local homomorphism of local rings. Remark 53. Suppose U X open, then if U a neighborhood of Y in X, U Y Y closed and irreducible. Then we have O X,Y OU,U Y i.e. we can replace X by an affine U to calculate the local ring. Now take Y X irreducible closed subset, what is O X,Y? We have a map α k[x] O X,Y f [X, f] We have m X,Y O X,Y maximal ideal and α 1 (m X,Y ) = {f k[x] f Y = 0} = I(Y ) prime ideal in k[x]. If g k[x]/i(y ), α(g) O X,Y. By the universal property of localization, α extends to a unique ring homomorphism This makes sense as α(g) O X,Y. Proposition 54. ᾱ is an isomorphism. ᾱ k[x] I(Y ) O X,Y f g α(f) α(g) Proof. Suppose α( f g ) = 0. Then 0 = [X, f] O X,Y. So f U = 0 for a neighborhood U of Y in X. Need: f g = 0 in k[x] I(Y ). Pick y Y U, then y / X/U X closed. Then there exists h k[x] such that h X/U = 0 but h(y) 0. So h / I(Y ). So fh = 0, and thus f h = 0 in k[x] I(Y ). This proves injectivity. Now take [U, f] O X,Y, where U is a neighborhood of Y in X and f k[u]. Let y Y U. In a neighborhood W of y in U, f = g where g, h k[x] and h(y) 0. Then we have h [U, f] = [W, f W ] = [W, g W ] [X, g] = [W, h W ] [X, h] = α(g) α(h) = ᾱ( g h ) Im(ᾱ) Example Let {0} = Y X = Z(xy) A 2. X is a union of two axes, so it is reducible. But k[x] = k[x, y]/(xy) x, y /(xy) = I(Y ). Then O X,Y = k[x] I(Y ) not a domain, since xy = 0 but x, ȳ 0. This reflects the fact that there are two irreducible component passing through Y. 2. Let Y = {(0, 0)} X = Z(y 2 x 2 x 3 ) A 2. Then k[x] = k[x, y]/(y 2 x 3 x 2 ) is a domain as (y 2 x 3 x 2 ) is irreducible. So X is irreducible and I(Y ) = (x, y)/(y 2 x 3 x 2 ). So O X,Y = k[x] I(Y ) is a domain. This reflects the fact that there are one irreducible component passing through Y. Now take Y X closed with X, Y both irreducible Then k[x] O X,Y k(x) quotient field of O X,Y. So O X,Y = {f k(x) f is regular in a neighborhood of Y X} = {f k(x) f is regular at a point y Y }. If Y = {y}, then O X,y = O X,y = {f k(x) f is regular at y}. Now suppose Y Z X with Y, Z irreducible and closed. We have that ker(o X,Y O X,Z k(z)) is a prime ideal in O X,Y. This ideal is just {[U, f] f Z = 0} = I(Z). So the closed irreducible subsets in X containing Y is in correspondence with prime ideals in O X,Y by Z I(Z). Proposition 56. This map is a bijection. Proof. Idea: Let U X be a neighborhood of Y in X. Consider the collection of closed irreducible subsets U containing U Y. This is the same as the closed irreducible subsets Z in X containing Y. This corresponds to prime ideals in O U,U Y, but O X,Y = O U,U Y. Now choose U affine. We may assume that X is affine. Then the collection of closed irreducible Y Z X is in correspondence with prime ideals p k[x] with p I(Y ) in correspondence with prime ideals in k[x] I(Y ) = O X,Y. The composition of these two maps is checked to be the same as the map above. 13

14 Moreover, this map is inclusion reversing, so in particular Y corresponds to m X,Y. The irreducible components of X containing Y correspond to minimal prime ideals in O X,Y. There is only one irreducible component of X containing Y if and only if O X,Y is a domain. Now suppose X, Y quasi-affine varieties and X, Y A n. Consider X X {0} A n+1 and Y Y {1} A n+1. Then write X + Y, the disjoint union of X and Y, or sum of varieties, to be X + Y = X {0} Y {1} Moreover, k[x + Y ] = k[x] k[y ] product of k-algebras. 2 Quasi-projective Varieties 2.1 Quasi-projective Subsets Recall that if we have a map from compact space to Hausdorff space, the image is always closed. However the projection from X = Z(xy 1) A 1 has image that is open but not closed. If X = X { }, then we can map to the origin. These varieties are called projective. So projective varieties are true compact. We will construct a full and faithful functor QAVar QPVar. A subset Y A n+1 is called cone if y Y, c k, then cy Y. A polynomial F k[t 0, t 1,, t n ] is called homogeneous of degree d if F (ct) = c d F (t) for all c k. An ideal I A n+1 = k[t 0, t 1,, t n ] is called homogeneous if F I implies that the degree i homogeneous component F i I (F = i 0 F i ). Remark An ideal I k[t 0, t 1,, t n ] is homogeneous if and only if I is generated by homogeneous polynomials. 2. If I is homogeneous, then so is I. Lemma 58. A closed subset Y A n+1 is a cone if and only if I(Y ) A n+1 is homogeneous. Proof. : Suppose Y is a cone and F = i 0 F i I(Y ). Then 0 = F (cy) = i 0 c i F i (y) for every c k, y Y. So F i (y) = 0 for all i, y, i.e. F i I(Y ). : Suppose y Y, c k. I(Y ) is generated by a set of homogeneous polynomials. If F is a homogeneous generator, then F (cy) = c d F (y) = 0 for some d. So for all G I(Y ), we have G(cy) = 0. So cy Z(I(Y )) = Y. So Y is a cone. So we have this one-to-one correspondence between closed cones in A n+1 and radical homogeneous ideals in A n+1 by I and Z. Under which we have A n+1 and {0} t 0, t 1,, t n. The unit group k acts on A n+1 /{0} via c (a 0,, a n ) = (ca 0,, ca n ). Then P n k = P n = (A n+1 /{0})/k projective space. The elements are classes [a 0 a 1 a n ] with not all a i s are zero. Moreover, [a 0 a 1 a n ] = [b 0 b 1 b n ] if and only if c k b i = ca i i if and only if a i b j = a j b i i, j. So we have the map π A n+1 /{0} P n+1 (a 0, a 1,, a n ) [a 0 a 1 a n ] If X P n, then π 1 (X) is a cone in P n. We say that X P n+1 is closed (resp. open) if π 1 (X) A n+1 /{0} is closed (resp. open). Now suppose S A n+1 a set of homogeneous polynomials, then Z(S) = {a P n F (a) = 0 for all F S} P n. All closed sets take this form. If I A n+1 homogeneous ideal, then Z(I) = Z(S) where S is the set of all homogeneous polynomials in I. If F is a homogeneous polynomial in A n+1, then D(F ) = D P n(f ) = {a P n F (a) 0}. So we get a one-to-one correspondence between closed sets in P n and radical homogeneous ideals in A n+1 different from t 0, t 1,, t n via the map I and Z. Moreover, A n+1 and P n 0. Theorem 59. (Projective Nullstellensatz) Let I A n+1 be a homogeneous ideal. Then Z(I) t 0, t 1,, t n k I for some k > 0. 14

15 Proof. Note that Z P n(i) = if and only if Z A n+1(i) {0} = Z A n+1( t 0, t 1,, t n ) if and only if t 0, t 1,, t n I Note that P n = π(a n+1 /0) is Noetherian and irreducible as image of irreducible. Every subset of P n is compact and admits decomposition into union of irreducible closed subsets. A quasi-projective subsets in P n has the form U Z where U P n open and Z P n closed. A subset X P n is quasi-projective if and only if X is open in X. 2.2 Regular Functions Let X P n be a quasi-projective set and x X. A function f X k is called regular at x if there are homogeneous polynomials F and G of the same degree such that G(x) 0 and f(y) = F (y) for all y in G(y) the neighborhood of x (so G(y) 0 for all y in the neighborhood). We say that f is regular if f is regular at all x X. Let k[x] denote the k-algebra of all regular functions on X. Note that there is no global definition for regular functions. Example 60. Claim: k[p n ] = k. To see this, suppose f k[p n ] not constant. Then there exists x P n such that f = F G in a neighborhood U of x and F, G are coprime, homogeneous non-constant polynomials of the same degree. Then we have Z(G) / Z(F ) as F / G. Then there exists x P n such that G(x ) = 0 and F (x ) 0. In a neighborhood U of x, we write f = F fraction of homogeneous polynomials of the same degree. So G in the dense set U U (as P n is irreducible), we have F G = F. Then F G = F G holds as equality of G polynomials. So 0 F (x )G (x ) = F (x )G(x ) = 0, a contradiction. So such constant f does not exists. Some properties: 1. If f k[x] and Y X quasi-projective set, then f Y k[y ]. 2. Suppose X = α U α open cover and f X k. Then f k[x] if and only if f Uα k[u α ] for all α. 3. If f X k is regular at x X, then f is regular in a neighborhood of x. 4. If f k[x], then {x X f(x) = 0} is closed in X. So f is continuous as a map X A If f k[x] and f(x) 0 for all x X, then 1 f k[x]. We say that f X Y P m is regular at x X if there exists a neighborhood U of x and f 0, f 1,, f m k[u] such that f i is nowhere zero on U for some i, with the property that [f(x )] = [f 0 (x ), f 1 (x ),, f m (x )] for all x U. We say f is regular if f is regular at every x X. As before, f is regular if and only if X f Y P m is regular. We can write f i = Gi H at a neighborhood of x, where G i, H are homogeneous polynomials of the same degree d. We don t subscript H as we can find a common denominator. Then f = [ G 0 H G 1 H G m H ] = [G 0 G 1 G m ] Example Suppose f X Y is regular and X X and Y Y with f(x ) Y, then f X X Y is also regular. 2. X Y then X Y is regular. Example Let L 1, L 2,, L k be linear forms on P n, linearly independent. Let Z = Z(L 1, L 2,, L k ) P n. Define f P n /Z P k 1 x [L 1 (x) L 2 (x) L k (x)] called projection with center Z. If k = n, then Z is a point, so if we take L i = t i the coordinate function, then Z = {[1 0 0]}. Thus we get a projection P n /{ } P n 1. 15

16 2. Let m, n be integers. Consider the monomial t α0 0 tα1 1 tαn n with α i = m. Then there are ( n+m n ) such monomials. We use the multi-index notation α = (α 0, α 1,, α n ). Introduce new variables S α. Define v n,m P n P dn,m [t 0 t 1 t n ] [S α ] = t α = t α0 0 tα1 1 tαn n where d n,m = ( n+m n ) 1. For example, if α = (0,, m,, 0) where m is in the ith tuple then [S α ] = t m i. If t i 0, then S α 0. This is Veronese map. Let Z = Z( S α S β S γ S δ α + β = γ + δ ) P dn,m. Observe that Im(v n,m ) Z. We want to show that v n,m gives an isomorphism between P n and Z. Let β be a multi-index with β = n 1 and let β i = (β 0, β 1,, β i + 1,, β n ). Construct g Z P n by [S α ] [S β1,, S βn ]. This independent of choice of β as another β and β i would satisfy S β i S j = S βj S βi for all i, j and [S α ] Z. So [S β1 S βn ] = [S β 1 S β n ] whenever it is defined. And g is the regular map inverting v n,m. For example, when n = 1, m = 2, we have three monomials t 2 0, t 0 t 1, t 2 1. Then v 1,2 P 1 P 2 [t 0 t 1 ] [S 0 S 1 S 2 ] = [t 2 0 t 0 t 1 t 2 1] Then Z = Im(v 1,2 ) = Z(S 2 1 S 0 S 2 ) P 2 conic given by S 2 1 S 0 S 2, so P 1 Z. The map g Z P 1 sends [S 0 S 1 S 2 ] [S 0 S 1 ] = [S 1 S 2 ]. Proposition 63. A regular map f X Y is continuous. Proof. It suffices to show that f 1 (Z(G)) is closed in X, where G is homogeneous polynomial. Now let x f 1 (Z(G)). In a neighborhood U of x, f = [f 0 f 1 f n ], f i k[u] are nowhere zero on U. Then f 1 (Z(G)) U = {y U G(f 0, f 1,, f n )(y) = 0} closed in U. Then f 1 (Z(G)) is locally closed, so it is closed. Corollary 64. If f X Y is regular and Y is quasi-projective, then f 1 (Y ) is also quasi-projective. Lemma 65. Let f X Y be a regular map, g k[y ]. Then g f k[x]. Proof. Let x X and y = f(x). Suppose g = F in a neighborhood W of y in Y. In a neighborhood U of G x in X, f = [f 0 f 1 f n ] with f i k[u], one is nowhere zero on U. In U f 1 (W ) a neighborhood of x g(f(x)) = F (f 0, f 1,, f n )(x) G(f 0, f 1,, f n )(x) = F (f 1 0, f 1,, f n )(x) G(f 0, f 1,, f n )(x) which is regular on U f 1 (W ). Hence is regular in a neighborhood of x, hence is regular. So f X Y regular gives a map f k[y ] k[x] g g f is a k-algebra homomorphism. Proposition 66. Let f X Y be a map of quasi-projective sets. Then f is regular if and only if for every open V Y and every g k[v ] the composition g (f f 1 (Y )) k[f 1 (Y )]. Proof. : If f f 1 (Y ) f 1 (Y ) V is regular then g (f f 1 (Y )) is regular. : Let x X and y = f(x) = [y 0 y 1 y m ] with y k 0. We may assume that y k = 1. Let s 0,, s m be the coordinates of P m. Then V = D Y (s k ) is a neighborhood of y in Y. Write f = [f 0 f 1 f m ] in a neighborhood of x in X. Dividing all by f k we may assume that f k 1 on U. Need: all f i s are regular at x. Take g = si s k k[v ]. Therefore, g (f f 1 (Y )) is regular. In f 1 (V ) U, g f = f i so is regular at x. Corollary 67. The composition of two regular maps is regular. We will construct a fully faithful functor θ QAffVar(k) QProjSets(k) We have A n with coordinates T 1,, T n and P n with coordinates S 0,, S n. We define U = D P n(s 0 ) P n open. Now we construct a map A n g U by (T 1,, T n ) [1 T 1 T n ]. We get an inverse map by [S 0 S 1 S n ] ( S1 S 0,, Sn S 0 ). I claim that g is a homomorphism. Take F k[t 1,, T n ] and G = F ( S1 S 0,, Sn S 0 ) s k 0 for sufficiently large k so that G is a polynomial. 16

17 Then g(z A n)(f ) = Z U (f). Conversely, given G homogeneous in k[s 0, S 1,, S n ], define F = G(1, T 1,, T n ) k[t 1,, T n ]. Then g 1 (Z U (G)) = Z A n(f ). If V A n is an open set and h g(v ) k is a function, then h is regular if and only if h (g V ) V k is regular. Locally, we can write h = K on g(v ) where K, L are homogeneous L of the same degree. Then h g = K(1, T 1,, T n ) L(1, T 1,, T n ) is regular. If h g = F H, where F, H k[t 1,, T n ], then h = F ( S1 S 0,, Sn S 0 ) S0 k H( S1 S 0,, Sn S 0 ) S0 k is regular. Then θ is the assignment given by X A n g(x) U P n. If f X Y, with X A n and Y A m, we have g n (X) U n P n and g m (Y ) U m P m. Then we have We claim that θ is full and faithful. We have θ(f) g n (X) g 1 n X f Y gm g m (Y ) f X gn g n (X) θ(f) g m (Y ) g 1 m Y So QAffVar(k) is full subcategory of QProjSets(k) = QProjVar(k). Definition 1. A quasi-projective variety is a quasi-projective set. A quasi-affine variety is a quasiprojective variety that is isomorphic to a quasi-affine set. A projective variety is isomorphic to a closed set in P n. Some properties: 1. Every quasi-projective variety admits an open affine cover. Proof. We have P n = n i=0 D P n(s i ), but D P n(s i ) A n, so is affine. In general, if X P n is quasi-projective, then X = n (D P n(s i ) X) i=0 and each member of this intersection is quasi-affine variety, and every quasi-affine variety admits an open affine finite cover. 2. A map f = (f 1, f 2,, f n ) X A n is regular if and only if all f i are regular functions. 3. k[d P n(s i )] = k[ S0 S i,, Si 1 S i L = a i S i for a i k. Then with a i S i L = 1., Si+1 S i,, Sn S i ]. To generalize, let L be a nonzero linear form on S i, so k[d P n(l)] = k[ S 0 L,, S n L ] Let F A n+1 be a homogeneous polynomial with deg(f ) > 0. Then D P n(f ) = P n /Z(F ). If F is linear, then D P n(f ) A n is affine. Proposition 68. The variety D P n(f ) is always affine. Proof. Consider the Veronese embedding v n,m P n P dn,m where d n,m = ( n+m n ) 1. Then Z = Im(v n,m) P dn,m is closed as the image of embedding. Recall our notations that if α = (α 0, α 1,, α n ), then α = n i=0 α i and S α = S α0 0 Sα1 1 Sαn n. We have proved that Z P n. Now let F = a α S α, for a α k α =m 17

18 Recall that v n,m ([S 0 S 1 S n ]) = [S α ]. If T α are the coordinates of P dn,m, then this is T α = S α. Let L = α =m a α T α, which is linear on P dn,m. Then and moreover, v n,m (D P n(f )) = D Z (L) D P n(f ) D z (L) = Z D P dn,m (L) Since F is closed, this intersection is closed in D P dn,m (L) A dn,m. So it is affine. Example 69. Let X = Z(s 1 s 2 s 2 0) P 2. We get a map Let T 1 = s1 s 0 X P 1 [s 0 s 1 s 2 ] [s 0 s 1 ] = [s 2 s 0 ] and T 2 = s2 s 0, so T 1 T 2 = 1. Then D X (s 0 ) = Z A 2(T 1 T 2 1). Also Z X (s 0 ) = X/D X (s 0 ) = {[0 1 0], [0 0 1]} call these points x 1, x 2. Now let the coordinates of P 1 be w 0, w 1. Then D P 1(w 0 ) = A 1. The coordinate on the affine line is u = w1 w 0. We have the map [1 T 1 T 2 ] [1 T 1 ] so the map X P 1 restricts to the map D X (s 0 ) A 1 = D P 1(w 0 ). This gives the projection map from the hyperbola to the affine line. But P 1 /D P 1(w 0 ) = {[0 1]}, call this point. So if f X P 1 is the map, we have f(x 1 ) = and f(x 2 ) = 0 A Rational Maps an Functions Let X be irreducible quasi-projective variety. For a pair (U, f) with 0 U X open, f k[u], we say (U, f) (U, f ) if f, f agree on U U. Denote by F (X) the field of rational functions on X, which are set of equivalence classes of [U, f]. As before, we get a correspondece between the opposite category of irreducible quasi-projective varieties with morphisms dominant rational maps and the category of finitely generated extensions of k, by the assignment X F (X) and f f. The category of irreducible quasi-affine varieties with dominant rational maps is an equivalence with both the above categories, so the functor in the previous paragraph is actually an equivalence. 2.4 Local Ring of a Subvariety Let X be a quasi-affine variety, and Y X closed and irreducible subvariety. Consider (U, f) the pair, with U a neighborhood of Y in X (i.e. U Y ), and f k[u]. Denote by O X,Y the local ring of equivalence classes. Moreover, we may reduce to the affine or quasi-affine case for the study of local objects as O X,Y O U,Y U. 2.5 Products of Quasi-projective Varieties The main point is that P n P n / P n+m, although A n A m A n+m. Let S i P n, T j P m for 0 i n, 0 j m. Consider also P k where k = (n + 1)(m + 1) 1 with coordinates w ij. Define a map φ P n P m P k ([S i ], [T j ]) [w ij ] where w ij = S i T j. This is well-defined. Now write V = Z(w ij w kl w il w kj ) P k, which is closed. These equations can be rewritten as S i S j S k S l S i S l S k S j = 0. So Im(φ) V. 18

19 Lemma 70. The map φ P n P m V is a bijection. Proof. Construct ψ V P n P m [w ij ] ([w 0l, w 1l,, w nl ], [w k0, w k1,, w km ]), where 0 k n, 0 l m and the map is independent of k and l. Then ψ = φ 1. Define the product P n P m = V P (n+1)(m+1) 1. For example, P 1 P 1 P 3. The map φ P n P m P k is called the Segre embedding. We have canonical projections to P n and P m, which are regular. Let F r (S, T ) be a homogeneous polynomials in S and T of possible different degree. Then Z(F r (S, T )) P n P m P k is closed. Since w ij = S i T j, and if F r homogeneous in S and T of the same degree, then F r (S, T ) = G r (W ) by replacing the S i T j with w ij. Suppose that deg S (F r ) = a and deg T (F r ) = b with a < b, then Z(F r (S, T )) = Z(Si b a F r (S, T )). Consider P n P m. Take X = Z(S) P n and Y = Z(T ) P m closed. Then X Y = Z(S T ) P n P m is closed. If X P n and Y P m are quasi-projective, say X = Z 1 /Z 1 and Y = Z 2 /Z 2, then X Y = Z 1 Z 2 /(Z 1 Z 2 Z 1 Z 2 ) P n P m is quasi-projective. Similarly, if X P n and Y P m are open, then X Y P n P m is open by taking Z 1 = P n and Z 2 = P m. In fact, X Y is the categorical product in QProjVar(k). If X, Y are quasi-affine, then they are quasi-projective so in either category, X Y is well-defined. Some properties: 1. If X and Y are affine (resp. projective), then so is X Y. 2. If X X, Y Y closed (resp. open), then so is X Y X Y. 3. If f X X and g Y Y are regular, then so is f g X Y X Y. The component of f g are: X Y p X f X regular and X Y q Y g Y regular. 4. If X and Y are irreducible, then so is X Y. 5. The natural map k[x] k k[y ] φ X,Y k[x Y ] is an isomorphism. Proof. We know that when X and Y are affine. The prrof that φ X,Y is injective is the same. Choose basis {g i } a basis for k[y ]. Then every element in k[x] k[y ] can be uniquely written as i f i g i with f i k[x]. I claim that if X = U α an open affine over and φ Uα,Y for all α, then so is φ X,Y. To see this, take h k[x Y ]. Now h Uα Y Im(φ Uα,Y ) implies that there exists a unique f α,i k[u α ] such that for all α. But h Uα Y (x, y) = i h (Uα U β ) Y (x, y) = i f α,i (x)g i (y) = φ Uα,Y ( f α,i g i ) i f α,i Uα Uβ (x)g i (y) = f β,i Uα Uβ (x)g i (y) i By uniqueness, f α,i Uα Uβ = f β,i Uα Uβ so there exists unique f i k[x] such that f i Uα = f α,i. Then φ X,Y ( i f i g i ) Uα Y = φ Uα,Y ( i f i Uα g i ) = φ Uα,Y ( f α,i g i ) = h Uα Y i Then φ X,Y ( i f i g i ) = h. Hence we prove the claim. For any X and Y affine, X = U α open affine cover. Then φ Uα,Y is an isomorphism. So φ X,Y is also an isomorphism by the claim. Now for arbitrary X, Y, take Y = W β be an open affine cover. Then φ X,Wβ is an isomorphism. Then φ X,Y is an isomorphism. 6. Suppose X P n and Y P m. Then a function on X Y is regular at (x, y) is equal to F (S,T ) G(S,T ) is a neighborhood U of (x, y), where F and G are homogeneous in S and T of the same degree with G 0 everywhere in U. 19