LECTURES ON ALGEBRAIC GEOMETRY MATH 202A

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1 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A KIYOSHI IGUSA BRANDEIS UNIVERSITY Contents Introduction 1 Overfield 1 1. A ne varieties Lecture 1: Weak Nullstellensatz Lecture 2: Noether s normalization theorem Nullstellensatz Noetherian induction Generic points Morphisms Algebraically closed case Lecture 6: Review First look at schemes Projective varieties Lecture 7. Basic properties of projective space Segre embedding Irreducible algebraic sets in P n 24 References 25 Introduction These are lecture notes for Math 202a: Algebraic Geometry. The basic goal is to do varieties over a nonalgebraically closed field and simultaneously lay the groundwork for schemes (reduced of finite type). The lectures will expand on the very terse notes for the course and give full details. Overfield We will assume that k is a field and is an overfield. This is a field extension of k which has infinite transcendence degree and is algebraically closed. =k(x 1,X 2,X 3, ) We also want to allow the number of variables to be more than countably infinite so that k = Q, =C is an example. Date: September25,

2 2 KIYOSHI IGUSA BRANDEIS UNIVERSITY I went over the basic properties of the overfield starting with clarifying the definition of transcendence degree. Definition Suppose that K is a field extension of k or any integral domain which contains k. We say that x 1,,x n 2 K, n 1, are algebraically independent if they do not satisfy any polynomial equations over k. This means: given any polynomial f(x 1,,X n ) 2 k[x 1,,X n ], if f(x 1,,x n )=0thenf is zero as a polynomial. When n =1wesay that x 1 is transcendental over k. The transcendence degree of K over k is defined to be the largest number d so that K contains d algebraically independent elements x 1,,x d over k. (d can be infinite.) One of the basic theorems is the following. The Noether Normalization Theorem is a refined version of this basic theorem. Theorem ) If K is a field extension of k generated by x 1,,x n then a maximal algebraically independent subset of this set of generators is a transcendence basis. 2) Any two transcendence bases for K over k have the same number of elements. A basic property of the overfield is the following. Theorem For any finitely generated field extension L = k(x 1,,x n ) of k, there is an embedding : L,! over k. (over k means is the identity on k) Proof. Choose a transcendence basis x 1,,x k for L over k. Let (x i ) = X i for i = 1,,k. Since L is an algebraic extension of k(x 1,,x k ) and is algebraically closed, the embedding : k(x 1,,x k )! extends to an embedding : L!. The class had no problem deriving the following consequence. Corollary Any finitely generated domain over k embeds in L. 1. Affine varieties One of the basic objects of study will be subsets of C n defined by polynomials equations with coe cients in Q. These are a ne varieties over C but they also give a ne schemes over Q. More generally we take k to be any field and to be an overfield : an algebraically closed field extension of k of infinite transcendence degree as discussed above. 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 ] Lecture 1: Weak Nullstellensatz. Lemma Let a 2 n. Then the set of all f(x) 2 k[x] so that f(a) =0is a prime ideal which we denote (a) k[x]. (I.e., is a mapping of sets : n! Spec(k[X]).) For now, Spec(R) is the set of prime ideals in any commutative ring R. Students were well aware of the fact that an ideal a in any ring R (always commutative) is prime i R/a is a domain and a is a maximal ideal i R/a is a field. Proof. This is a prime ideal since it is the kernel of the k-algebra homomorphism ev a : k[x]! which sends f(x) tof(a). The image of this homomorphism is a subring of and thus a domain. I also use the notation ' a = ev a : evaluation at a.

3 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 3 In my lecture I emphasized the fact that f(x) as a function of x 2 n does not have easy to understand properties. It is a polynomial function. However, the function f 7! f(x) for any fixed x 2 n gives a homomorphism of k-algebras. (This means ring homomorphism which is the identity on k.) The lemmas being proved here use this nice interpretation of f(x) as a function of f. Lemma : n! Spec(k[X]) is surjective. We will give n a topology and Spec(k[X]) the quotient topology. Proof. Take any prime p k[x]. Then k[x]/p is a domain which contains k. Claim: There is an embedding ' : D = k[x]/p,!. Pf: D is finitely generated by the images x i of the variables X i : D = k[x 1,,x n ]. By renumbering the x i we may assume that x 1,,x k are algebraically independent. Embed D in its field of fractions Q(D) ={ a b a, b 2 D, b 6= 0}. ThenQ(D) =k(x 1,,x n ). Since x 1,,x k are algebraically independent, we have an isomorphism: L = k(x 1,,x k ) = k(x 1,,X k ) Since x k+1,,x n are algebraic over L, theembeddingl,! extends to an embedding Q(D),! and we get an embedding This proves the claim. ' : D = k[x]/p,! Q(D),! Let a i = '(X i ) 2. Let a =(a 1,,a n ) 2 n. Then f(a) ='(f) =0i f 2 p. So, (a) =p. So, : n! Spec(k[X]) is onto. Theorem (a) is a maximal ideal if and only if all coordinates a i 2 of a are algebraic over k. Proof. (We will go over this again after reviewing the Normalization Theorem.) Let ' a : k[x]! be given by evaluation at a so that (a) =ker' a.then' a (k[x]) = k[a 1,a 2,,a n ]. Since this is a subring of, it is an integral domain finitely generated over k. So, the normalization theorem applies and this is a field i 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 (a) where a 2 k n.( (a) =(X 1 a 1,,X n a n )) Lecture 2: 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 ]. Recall the definitions: If S is a subring of R an element a 2 R is called integral over S if there is a monic polynomial f(x) =X n + c 1 X n 1 + c 2 X n c n 2 S[X] so that f(a) = 0. This is equivalent to the statement that S[a] is finitely generated as an S-module. We say that R is an integral extension of S if S R and every element of R is integral over S. R is called integrally closed if the only integral extension of R is R itself.

4 > 4 KIYOSHI IGUSA BRANDEIS UNIVERSITY 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 Going-Up theorem. Lemma Let R be an integral extension of S. IfR is a field then S is a field. Proof. Take any a 6= 02 S. Then R contains b =1/a. Since R is integral over S, thereis a monic polynomial f(x) 2 S[X] so that f(b) = 0: Multiply by a n 1 to get: So, b =1/a 2 S and S is a field. b n + c 1 b n c n =0, c i 2 S 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 ]withd>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 2 R is not nilpotent. I.e., f n 6= 0 for all n 0. Then M = {1,f,f 2, } is a multiplicative set which means that M is closed under multiplication, contains 1 and does not contain 0. In general R M is defined to be the ring of all fractions R M = { a a 2 R, b 2 M} b with the usual addition and multiplication rule for fractions and the identity a b = ac bc for a all c 2 M. In particular, b = 0i thereisac 2 M so that ac = 0inR. When M = {1,f,f 2, } for f not nilpotent, the notation is: R f = R M which is given by inverting just f: a R f = f n where a f n =0i af m = 0 for some m. In particular 1 f 6= 0. Another description of R f is: R f = R[X]/(fX 1). We verified this in detail in class using the universal properties of R f and R[X]. The universal property of R f is the following. Let ' : R! S be any ring homomorphism which sends f to a unit in S (unit means invertible element). Then there is a unique ring homomorphism ' : R f! S making the following diagram commute: R ' / S R f ' Let S = k[x]/(xf 1). Then X = 1 f uniquely through R f. in S, so the homomorphism R,! R[X] S factors

5 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 5 The universal property of R[X] is the following. Given any ring homomorphism : R! S and any element x 2 S, there is a unique homomorphism : R[X]! S which is on R and sends X to x. In this case we have a homomorphism : R[X]! R f which sends each a 2 R to a 1 2 R f and sends X to 1 f 2 R f.then (Xf) = 1. So, Xf 1 is in the kernel of. So, there is an induced homomorphism R[X]/(Xf 1)! R f So we have mapping in both directions R f $ R[X]/(Xf identity maps by universality. 1) and the compositions are the Zariski topology. For any subset S of k[x], let V (S) be the set of all x 2 n so that f(x) = 0 for all f 2 S. Two extreme cases are: V (;) = n and V (1) = ;. Definition AsubsetZ n is called an algebraic set defined over k if Z = V (S) for some S k[x]. Example SL(2, ) is the algebraic subset of 4 defined over any k by SL(2, ) = {(a, b, c, d) 2 4 : ad bc =1} GL(2, ) is defined by the inequality ad bc 6= 0. This does NOT define an algebraic subset of 4. However, GL(2, ) is equivalent to the algebraic subset of 5 (defined over any k) given by GL(2, ) = {(a, b, c, d, e) 2 5 :(ad bc)e =1}. The point is: GL(2, ) does not satisfy the definition of an algebraic subset of 4. However, it is in bijection with a true algebraic subset of 5.Wewillusethistodefine the structure of a variety on GL(2, ) to be the one given by this structure on this algebraic subset of 5. 1 One of the consequences of this concept is that the function ad bc, which is not a polynomial function on GL(2, ) 4, is a polynomial function on the corresponding subset of 5. So, is a regular function on GL(2, ). 1 ad bc The basic property of algebraic sets is: Theorem The sets V (S) form the closed subsets of a topology on n.(thisiscalled the Zariski topology or the k-topology.) Proof. Recall that a collection of subsets C of a set X form the closed sets in a topology i they satisfy the following. The standard terminology is A is closed means A 2C. (1) ; is closed. (2) X is closed. (3) Any intersection of closed sets is closed: C closed ) T C is closed. (4) The union of two closed sets is closed: A, B closed ) A [ B is closed. The definition of the set V (S) can be parsed as follows: V (S) = \ V (f). f2s So V (S) is an intersection of the hypersurfaces V (f) given by single equations f(a) = 0. So, T V (S )=V([S ). This proves (3): The collection of sets {V (S)} is closed under arbitrary intersections. If S, T are two subsets of k[x] and ST = {fg : f 2 S, g 2 T } then V (ST)= \ V (fg)= \ (V (f) [ V (g)) = \ V (g) =V (S) [ V (T ). f2s,g2t f2s,g2t f2s V (f) [ \ g2t

6 6 KIYOSHI IGUSA BRANDEIS UNIVERSITY This proves (4): The collection of sets {V (S)} is closed under finite unions. Since V (;) = n and V (1) = ;, the collection of sets {V (S)} satisfies the axioms for the closed sets in a topology on n as listed above. For any subset Y of n we take the induced topology. Thus a subset A Y is closed i A = Y \ V (S) for some S. Exercise If Y = V (S) is a closed subset of n show that the (relatively) open sets Y f = {x 2 Y : f(x) 6= 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 2 U then f i (y) 6= 0 for some i. Soy 2 Y fi U. So, the Y f are basic open sets in Y Nullstellensatz. 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 2 k[x] so that some power of f lies in a. The following statement is clear: Proposition For any subset S k[x]: Proof. What is clear is that V (S) =V ((S)) = V (r(s)). V (S) V ((S)) V (r(s)) since S (S) r(s). So, we need to show that V (S) V ((S)) V (r(s)). (1) Proof that V (S) V ((S)): The ideal I =(S) generated by the set S in the ring k[x] is the set of all linear combinations of elements of S: o (S) =nx f g f 2 S, g 2 k[x]. If a 2 V (S) thenf (a) = 0 for all f 2 S. But this implies P f (a)g (a) = 0. So, a 2 V ((S)) which shows that V (S) =V ((S)). (2) V ((S)) V (r(s)): (Lecture 3 starts here.) I did a complicated proof of this in class but, after class, Mac showed me a quick proof: f(a) =0, f k (a) =0. Why does this prove (2)? (After some discussion we decided this is obvious.) For any subset Z n let I(Z) be the set of all f(x) 2 k[x] so that f(z) = 0 for all z 2 Z. Sincef(z) =0i f 2 (z), this is equivalent to: I(Z) = \ (z) z2z Therefore, I(Z) is an ideal in k[x]. Also, being an intersection of prime ideals, it is a radical ideal. (Recall that, if an ideal I is contained in a prime ideal p then r(i) p. Since each (z) =kerev z : k[x]! is a prime ideal, r(i(z)) (z) for each z. So, r(i(z)) T (z) =I(Z). So, r(i(z)) = I(Z). So, I(Z) isaradical ideal, i.e., an ideal equal to its radical.) Hilbert s Nullstellensatz states: Theorem For any ideal a in k[x], I(V (a)) = r(a).

7 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 7 Proof. (from the red book [4]) It is clear that a I(V (a)) and therefore r(a) r(i(v (a))) = I(V (a)). So, suppose g is not in r(a). Then we will show that g /2 I(V (a)) by finding a point in x 2 V (a) so that g(x) 6= 0. (And x 2 V (a) means f(x) = 0 for all f 2 a.) Step 1. Since g is not in r(a), g is not nilpotent in k[x]/a. So, we can invert g in k[x]/a without making the ring trivial: k[x] 6= 0. a 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].] Step 2a. Reformulate the equation: k[x] 0 6= a g = k[x] g (a) g = R/(1 X n+1g) (a) = R (a, 1 X n+1 g) where R = k[x][x n+1 ]=k[x 1,,X n+1 ] and (a) is the ideal in R generated by a k[x] =k[x 1,,X n ]. The first equality follows from the universal properties: the ring homomorphism k[x]! k[x] g /(a) has a in its kernel. So, it induces a homomorphism k[x]/a! k[x] g /(a). This sends g to a unit. So, there is a unique induced homomorphism (k[x]/a) g! k[x] g /(a). Similarly, we get a map the other way and, by the universal properties defining these maps, both compositions are the identity. The second equality comes from the equation k[x] g = k[x][xn+1 ]/(X n+1 g 1) discussed earlier. The third equality is clear. Step 2b. Let J =(a, 1 X n+1 g). Since this R/J 6= 0, J is a proper ideal in R. So, J is contained in some maximal ideal m. Recall that : n+1! Spec(R) issurjective. So, there is a point a 2 n+1 with (a) =kerev a = m. By the weak Nullstellensatz, all coordinates a i of a are algebraic over k. Step 3: Reinterpretation: Let x =(a 1,,a n ) 2 n. (So, a =(x, a n+1 ).) a J m = (a) implies that f(a) = 0 for all f 2 a. But f(a) = f(x) since f 2 a k[x] is a polynomial in X 1,,X n. So, the point x 2 n lies in V (a). But 1 X n+1 g is also in J m =kerev a. So, 1 = a n+1 g(x) making g(x) =1/a n+1 6=0 as required to prove the theorem. Since a i are algebraic over k, (x) =ker' 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 N = r(0). Recall that the Jacobson radical of any ring is by definition the intersection of all maximal ideals. There is a related theorem which says that the nilradical (r(0)) is the intersection of all prime ideals. For finitely generated algebras over any field these are the same and this is the key fact that allows us to study varieties by looking only at the k points in the variety. Example Let k = R and a =(X 2 + Y 2 ). Then k = C is algebraically closed but is much bigger. The algebraic set V (a) is the set of all (x, y) 2 2 so that x 2 + y 2 = 0.

8 8 KIYOSHI IGUSA BRANDEIS UNIVERSITY R 2 C 2 (1,i) V (a) \ R 2 V (a) \ C 2 V (a) \ R 2 is just one point. But V (a) \ C 2 is two lines given by y = ix and y = X 2 + y 2 =(Y ix)(y + ix). (x) R[X, Y ] is a maximal ideal i x 2 C 2. For example, (1,i)={f 2 R[X, Y ]:f(1,i)=0} =(X 1,Y i) \ R[X, Y ]= (X 1,Y 2 + 1) where ( ) needs proof. This is a maximal ideal since is a field. Corollary states that R[X, Y ] (1,i) = R[Y ] (Y 2 + 1) = C J = I(V (a) \ C 2 )=r(a) = Nullst. I(V (a)) R[X, Y ] since V (a) \ C 2 is the intersection of all maximal ideals containing a. Jacobson radical of R[X, Y ]/a. Question: Show geometrically that V (a) is irreducible in this case. ix since Thus J/a is the Corollary There is an order reversing bijection between the set of radical ideals in k[x] and the set of Zariski closed subsets of n defined over k. Proof. This follows from the formulas that we proved. where ( ) is the Nullstellensatz. radical ideals! closed subsets of n start with a = r(a)! V (a) I(V (a)) = r(a) =a V (a) start with any closed set: I(Z) =I(V (S)) = r(s) Z = V (S) =V (r(s)) I(Z)! V (I(Z)) = V (r(s)) Exercise Show that the closure (in the Q-topology) of the point (e, e 2 ) 2 C 2 is the parabola {(z,z 2 ) z 2 C} 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: Given I 1 I 2 I 3 Then eventually I n = I n+1 = for some n. Therefore, the closed subsets of n satisfy the DCC: any descending chain of closed subsets is eventually constant: Z 1 Z 2 Z 3 implies that, for some n, Z n = Z n+1 =. This allows for the following argument called Noetherian induction: For every statement about closed subsets of n which is not true, there is a minimal counterexample.

9 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 9 Proposition Every closed subset of 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. So, Y is not the union of finitely many irreducible closed subsets. In particular, Y is not irreducible. So, Y is the union of two proper closed subsets Y = Y 1 [ Y 2. By minimality of Y, each of Y 1,Y 2 is a union of finitely many irreducible closed subsets. So Y = Y 1 [ Y 2 = [ Y j 1 [ [ Y j 2 where each Y j i is irreducible. So, Y is a finite union of irreducible closed sets. The proof of the next proposition uses the following argument. If X is irreducible and X S Y i then X Y i for some i. The reason is: X = X \ [ Y i = [ X \ Y i Since X is irreducible, it must be equal to one of these: X = X \ Y i which implies X Y i. Proposition The decomposition of a closed set Y as a union of irreducible closed subsets Y = S Y j is unique. The unique pieces Y j of a closed set Y are the components of Y. Proof. If an irreducible X S Y i then X is contained in one of the Y i. So, if S X i = S Y j then X 1 Y j for some j. By the same argument, Y j X k for some k. But, X 1 X k implies that X 1 = X k and k = 1 (otherwise the decomposition X = S X i has redundant summands). So, X 1 = Y j. Similarly, X 2 = Y k where j 6= k (since X 1 6= X 2 ). Do this for all the X i to prove the theorem. (Any Y j which are left over are redundant summands.) Lecture 4 starts here: Theorem The prime ideal correspond to the irreducible subsets under the correspondence of Corollary Proof. One direction is clear: Y not irreducible ) I(Y ) not prime: Y = Y 1 [ Y 2, Y i ( Y ) I(Y )=I(Y 1 ) \ I(Y 2 ), I(Y ) ( I(Y i ) But a prime ideal cannot be written as an intersection of two larger ideals. [Pf: If p = I 1 \I 2 and p ( I i then there are a 2 I 1 \p,b2 I 2 \p. Butab 2 I 1 I 2 I 1 \ I 2 = p. Theneithera or b is in p, a contradiction.] So, I(Y ) is not prime. Equivalently, I(Y )prime) Y irreducible. Conversely, suppose I(Y ) is not prime. Then, by the lemma below, I(Y ) is an intersection of prime ideals: I(Y )=p 1 \ p 2 \ \p m, I(Y ) ( p i. Then Y is a union of the corresponding closed sets (which are irreducible by the first part of the proof): Y = V (p 1 ) [ V (p 2 ) [ [V (p m ), V(p i ) ( Y. So, I(Y ) not prime ) Y not irreducible. So, Y irreducible, I(Y )prime. Lemma In any Noetherian ring A, any radical ideal is the intersection of finitely many prime ideals.

10 10 KIYOSHI IGUSA BRANDEIS UNIVERSITY The proof uses the Primary Decomposition Theorem which says that, in any Noetherian ring R, any ideal a is a finite intersection of primary ideals: a = q 1 \ q 2 \ \q m. Recall that an ideal q in any ring R is primary if it satisfies the condition that a, b 2 q, a/2 q implies that b n 2 q for some n. In other words, b 2 r(q). The important property for us is: The radical of any primary ideal is prime. Proof: Suppose that a, b 2 R, ab 2 r(q) but a/2 r(q). Then (ab) n = a n b n 2 q for some n. Buta/2 r(q) means that a n /2 q. Sinceq is primary, this implies (b n ) m 2 q for some m. Butthenb 2 r(q). So, r(q) is a prime ideal. Proof of Lemma By the Primary Decomposition Theorem, any ideal in a Noetherian ring is a finite intersection of primary ideals: a = T q i.sinceaisaradical ideal we have \ a = r(a) =r qi = \ r(q i ). Proof of the equality = : The inclusion ( ) is clear since T q i q i implies r( T q i ) r(q i ) for each i. So, suppose a 2 T r(q i ). Then a 2 r(q i ) for each i. Then, for each i there is an integer n i so that a n i 2 q i. Let n be the maximum of these n i. Then a n 2 T q i = a. So, a 2 r(a) and a = T r(q i ). The lemma follows since each r(q i )isprime. 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 2 A so that ab 2 a but no power of a or b lies in a. For each n 1letI n =(a : a n ) be the set of all c 2 A so that a n c 2 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 /2 a these ideals properly contain a. We will show that J \K = a giving a contradiction (showing that a is primary). Let c 2 J \ K. Since c 2 J, c = a n x + y where y 2 a. Since ak a, ac 2 a. So, a n+1 x 2 a. So, x 2 I n+1 = I n. So, a n x 2 a. So, c 2 a Generic points. Theorem The closure of any point y 2 n is an irreducible algebraic set defined over k. Conversely, for any irreducible algebraic subset Y n, there is an element y 2 Y (called a generic point of Y )sothaty = y. Proof. The first statement is easy: Suppose that y = Y 1 [ Y 2.Theny must be in one of the two sets. Say, y 2 Y 1.SinceY i are closed, we get y Y 1.ButY 1 cannot be bigger than y. So, Y 1 = y. So, y is not the union of two smaller closed subsets. 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 2 n so that (y) =p. Then y = V (p) =Y. We went through the proof of this in class: For any ideal a we have an equivalence: y 2 V (a), (8f 2 a)f(y) =0, a p = (y) =kerev y

11 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 11 But the closure of y is the intersection of all closed sets containing y and all closed sets have the form V (a) for some ideal a: y = \ V (a) =V (p) =Y. y2v (a) V (a) =\ a p Corollary Two point x, y 2 n map to the same prime ideal (x) = (y) if and only if x = y. Exercise (1) Show that x y i x = y is an equivalence relation on n. (2) Show that x 2 y i (x) (y). (3) Show that x is a finite set i (x) is a maximal ideal. (4) When is x 2 n a closed point? (What if k is not perfect?) I went through an example of generic points, but the example was a little too advanced. We will discuss this more later. Definition The Krull dimension d of any ring R is the maximum length of a chain of prime ideals. p 0 ( p 1 ( ( p d For example, the dimension of Z is d = 1 since the longest tower is: 0 ( (p) for any prime number p. Later we will discuss the theorem that the Krull dimension of k[x 1,,X n ]isn. For example, when n = 2, the longest chain of prime ideals looks like this: 0 ( p ( m where m is a maximal ideal. The prime ideal p in the middle is not zero and not maximal. It is called a height 1 prime. (0 has height 0, maximal ideals have height 2 in k[x 1,X 2 ].) For such a chain of prime ideals we have the corresponding irreducible closed subsets: V (0) ) V (p) ) V (m). Each of these has a generic point. Conversely, for any point y 2 2, y is one of these irreducible closed sets by Theorem So, 2 has three kinds of points: (0) y is a finite set. Equivalently, (y) is a maximal ideal. (This is not obvious and requires proof.) (1) y = V (p) isacurve in 2. (2) y = 2. Equivalently (y) = 0. The case y = V (p) occurs when the coordinates of y are algebraically dependent on each other but one of them is not algebraic over k. For example, y =(T 1,T1 3 ). (Recall that is the algebraic closure of k(t 1,T 2, ), the field of rational functions on infinitely many variables.) Exercise Show that the closure of the point y =(T 1,T 3 1 )istheset y = V ((X 2 X 3 1)) = {(x, x 3 ) x 2 }.

12 12 KIYOSHI IGUSA BRANDEIS UNIVERSITY 1.6. Morphisms. Let Y n be an irreducible closed subset defined over k. Then a regular function f : Y! is any mapping which is the restriction to Y of a mapping n! given by a polynomial in k[x 1,,X n ]. The coordinate ring (Y ) of Y is the ring of all regular functions Y!. If two polynomials give the same regular function on Y then their di erence is zero on Y and conversely. So, (Y ) = k[x]/i(y ). To clarify this: By definition, we have an epimorphism: I(Y ),! k[x] (Y ) with kernel I(Y ) since every regular function is defined to be given by a polynomial. To set the notation, we denote polynomials by f(x 1,,X n ) and the corresponding regular function on Y by f(x 1,,x n ) using lower case letters to denote the functions x i : Y,! n! which are projection to the ith coordinate. For example, the polynomial f = XY 3 5X gives the function f(x, y) =xy 3 5x. Since Y is irreducible, I(Y ) is a prime ideal and (Y ) is an integral domain. If y 2 Y is a general point then f : Y! isuniquelydeterminedbyf(y). So, (Y ) is isomorphic to the image of y under k[x]. Given Y,Z irreducible closed subsets of n, m resp, a morphism Y! Z is a mapping f : Y! Z given by polynomial equations. Thus, there exist m polynomials f i 2 k[x 1,,X n ] so that f(y) =(f i (y)) for all y 2 Y. There is an induced map: f : (Z)! (Y ) given by composition of polynomial mappings: f (g : Z! ) = g f : Y! Z!. For any g 2 (Z) k[t 1,,T m ], f (g) =g f : Y! is given on each y 2 Y n by f (g)(y) =g(f(y)) = g(f 1 (y),,f m (y)). Theorem Hom(Y,Z) = Hom k alg ( (Z), (Y )). Proof. Easy part: Given f : Y! Z, we get f : (Z)! (Y ) as described above. Hard part: Given any ' : (Z)! (Y ) we need to construct a polynomial mapping f ' : Y! Z. The ring (Z) is generated by the images of the generators T i of k[t 1,,T m ] which are the projection maps t j : Z,! m! tothejth coordinate. (The coordinates of any point z 2 Z are z j = t j (z).) So, the homomorphism ' is determined by the images '(t j ):Y! of these projection maps. We define f ' : Y! Z to be the mapping given by f ' (y) =('(t 1 )(y),,'(t m )(y)) 2 m for all y 2 Y. We need to verify that this is an element of Z. We use the fact that Z = V (I(Z)). In other words, a point x 2 m lies in Z i g(x) = 0 for all g(t ) 2 I(Z) k[t ]. When we say that ' : (Z)! (Y ) is a homomorphism of k-algebras we mean: ' (linear combination of products of functions j : Z! ) = the same linear combination of products of the functions '( j ):Y!. In other words, for any polynomial g( j ) of (any) functions '(g( j )) = g('( j )) : Y!. j : Z! wehave:

13 Now take LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 13 j = t j, the projection maps Z,! m!. Then: '(g(t 1,,t m )) = g('(t 1 ),,'(t m )) : Y!. For any g(t ) 2 I(Z) the function g(t )=g(t 1,,t m ) = 0 since this is the image of g(t ) in (Z). So, '(g(t )) = g('(t )) = 0 : Y!. So, for any y 2 Y : g('(t 1 )(y),,'(t m )(y)) = 0. Since this holds for all g(t ) 2 I(Z) we get: ('(t 1 )(y),,'(t m )(y)) 2 Z as claimed. Lecture 5 starts here. We are left with two more easy steps: Verify that f ' = ' and f ' = f if ' = ' f. The map (f ' ) : (Z)! (Y ) is given on the generators t j : Z! of (Z) by (f ' ) (t j )=t j f ' = t j (f ' )=t j ('(t 1 ),,'(t m )) = '(t j ). In other words, f'(t j )='(t j ). Since the t j generate (Z), f' = '. Conversely, if we start with f : Y! Z and ' = f then f ' =('(t 1 ),,'(t m )) = (f (t 1 ),,f (t m )) = (t 1 f,,t m f). This is equal to f since it is the function whose jth coordinate is t j of f. So, f ' = f. Proposition The ideal I(f(Y ))/I(Z) is equal to ker f f,thejthcoordinate Proof. Since f (g) = g f, g 2 ker f i g f = 0 i g(f(y )) = 0 i g 2 I(f(Y )). But the polynomial representing a function Z! liesini(f(y )) i the function lies in I(f(Y ))/I(Z). Proposition Any morphism f : Y! Z is continuous in the Zariski topology. Proof. First, suppose that Y = n, Z = m. Then we need to show that the inverse image of any close set V (a) m is closed in n. But, for any y 2 n, y 2 f 1 (V (a)), f(y) 2 V (a), g(f(y)) = 0 8g 2 a, y 2 V ({g f g 2 a}) So, f 1 (V (a)) = V ({g f g 2 a}) is closed. In the general case a morphism f : Y! Z if the restriction of a morphism F : n! m. Let C Z be closed. Then C is closed in m. So, F 1 (C) is closed in n and f 1 (C) = F 1 (C) \ Y is closed in Y Algebraically closed case. Let s compare the definitions and the last theorem to the standard setup. Assume in this subsection that k = 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 Suppose k = k and y 2 n. Then y = {y} (y is a closed point) i y 2 k n. Proof. (() Suppose a 2 k n.then (a) = ker k[x] eva! k = (X 1 a 1,,X n a n )=m a (Z). By the weak Nullstellensatz, these are all of the maximal ideals in k[x].

14 14 KIYOSHI IGUSA BRANDEIS UNIVERSITY Recall from the proof of Theorem that y = V ( (y)). So, a = V (m a )={y 2 n f(y) =08f 2 m a } Since X i a i 2 m a this implies (X i a i )(y) =y i a i = 0. So, y i = a i for all i and y = a. So, a = {a}. ()) Conversely, suppose that y = {y}. Ify/2 k n then (y) =p is not maximal (by weak Nullstellensatz). So, there is a maximal ideal m a p. Then {a} = V (m a ) V (p) =y. So, a 2 y. Since a 6= y, y has at least two element. This contradicts the assumption y = {y}. Theorem If k = k and Y = V (p) is an irreducible closed subset in n corresponding to the prime ideal p k[x 1,,X n ] then Y 0 = Y \ k n = {a 2 k n p m a } Proof. For all a 2 k n we have: p m a, V (p) =Y V (m a ) ={a}, a 2 Y. Lemma If Y n is an irreducible algebraic set then Y = Y 0. Proof. Let p = I(Y ). By the theorem above, Y 0 = Y \ k n is the set of all points a 2 k n so that p m a. Since the Jacobson radical of k[x]/p is equal to its nilradical, we have: I(Y 0 )= \ \ m a = m a = r(p) =p a2y 0 m a p (The Jacobson radical of k[x]/p is T m a /p and the nilradical of k[x]/p is r(p)/p. Weproved that these are equal, so the numerators are equal.) So, Y 0 = V (I(Y 0 )) = V (p) =Y. When k = k, we define (Y 0 ) to be the ring of all functions f : Y 0! k which are given by polynomials f(x) 2 k[x]. Thus, by definition, we have an epimorphism k[x] (Y 0 ). The kernel if I(Y 0 )=I(Y). So, (Y 0 )= k[x] I(Y 0 ) = k[x] = (Y ). I(Y ) So, any polynomial function f : Y 0! k extends uniquely to a regular function f : 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 ( (Z), (Y )) = Hom k alg ( (Z 0 ), (Y 0 )). We already know that Hom(Y,Z) = Hom k alg ( (Z), (Y )) = Hom k alg ( (Z 0 ), (Y 0 )). So, we just need to check that Hom(Y 0,Z 0 ) = Hom(Y,Z). This implies the standard bijection: Hom(Y 0,Z 0 ) = Hom k alg ( (Z 0 ), (Y 0 )). Proof. Any morphism f : Y! Z sends Y 0 into Z 0 since evaluation of a polynomial with coe cients in k at a point with coordinates all in k will give elements of k. So f(y 0 ) k m \ Z = Z 0. So, we get a restriction map Hom(Y,Z)! Hom(Y 0,Z 0 ). Conversely, any morphism f : Y 0! Z 0 has coordinates f j : Y 0! k. These extend uniquely to regular functions f j : Y!. Together these give a morphism f : Y! m.

15 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 15 We need to verify that f(y ) Z. Equivalently, we need to show Y f 1 (Z). But f is continuous and Z is closed. So, f 1 (Z) is a closed set and Y 0 f 1 (Z 0 ) f 1 (Z). Taking the closure of both sides we get Y = Y 0 f 1 (Z). So, f(y ) Z. If we extend a function Y 0! Z 0 to Y then restrict again to Y 0 we get back the same function. Also, the extension f is unique. So, restricting and extending must give back the same function Y! Z. So, Hom(Y 0,Z 0 ) = Hom(Y,Z). The proof of the following is easy and we skipped it. I emphasized instead the two key properties of open sets: (1) If U is a nonempty open subset of an irreducible set Y then U = Y. The reason is: Y = U [ Y \U. SinceY is irreducible and Y 6= Y \U, wemusthavey = U. (2) If U, V are nonempty open subsets of Y then U \ V is nonempty. The proof is: If U \ V = ; then Y =(Y \U) [ (Y \V ), a contradiction. 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. For the next corollary, I emphasized the statement. Definition If U is an open subset of Y, a regular function on U is a mapping f : U! which is given locally by rational functions. This means that, for all y 2 U there is an open neighborhood V of y in U and polynomial functions g, h : V! so that h(x) 6= 0 for all x 2 V and f(x) = g(x) h(x) for all x 2 V. The polynomials g, h may be di erent in di erent neighborhoods V as long as the functions f = g/h agree on the intersections of these V s. When k = k, a regular function f : U 0 = U \ k n! k is defined to be a function which is given locally by rational functions f = g/h. The corollary below says that the ring of regular function f : U 0! k is equal to the ring of regular functions U!. In other words, f extends uniquely to f : U!. We will come back to the discussion of regular functions on open sets after we look at projective varieties. Corollary For any open subset U of Y,anymappingU\ Y 0! k given locally as a rational function with coe cients in k extends uniquely to a regular function U! and all regular functions U! take U \ Y 0 into k. Proof. Let f : U \ Y 0! k and, for each x 2 U \ Y 0, suppose that there is a basic open neighborhood V = Y g U of x and h 2 Y so that f(y) =h(y)/g(y) m for all y 2 V \ Y 0. Then h/g m is one extension of f to V.SinceV\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 a ne variety and of morphisms of a ne varieties by restricting to just the closed points.

16 16 KIYOSHI IGUSA BRANDEIS UNIVERSITY 1.8. Lecture 6: Review. Before discussing projective space, let us briefly review the concepts. (1) k is any field and = k(t 1,T 2, ) is a very large field containing k. (2) There is an order reversing 1-1 correspondence between radical ideals (ideals a = r(a)) in k[x] and closed subsets Z of n given by V (a) :={y 2 n : f(y) =08f 2 a} a = I(Z) :={f 2 k[x] :f(z) =0} (3) Under this bijection, the prime ideals p (k[x] correspond to the irreducible subsets Y n (those not equal to a union of two proper closed subsets). (4) Every irreducible set Y contains an element y called a generic point so that y = Y. I(Y )=p = I(y) = (y) =ker(ev y : k[x]! ) (If y 2 Y is not generic then y = Z ( Y is an irreducible proper subset.) (5) If Y n,z m are irreducible, a morphism f : Y! Z is a function given by m polynomial equations in n variables. (6) Hom(Y,Z) = Hom k alg ( (Z), (Y )) where (Y )= k[x] I(Y ) is the ring of all polynomial functions f : Y!. (7) For any open subset U Y,(Y irreducible), a regular function on U is defined to be a mapping f : U! so that f is given locally by rational functions f(x) = g(x) h(x). These functions form a ring (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 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: apple T1 T 2 T 3 T 4 This is the general 2 2 matrix whose entries are general elements of k. 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 di erent dummy variables. Using the concept that dummy variables are equivalent we will make the transition from varieties to schemes. Definition For any (commutative) ring R let Spec(R) denote the topological space whose elements are the prime ideals p ( 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/2 p} for any f 2 R. Recall that a collection of subsets of X form a basis for a topology on X if the intersection of any two basic open sets is a union of basic open sets. In this definition

17 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 17 we have X f \ X g = X fg since f,g /2 p, fg /2 p. So the sets X f form a basis for some topology on Spec(R). Two special cases: X 0 = ;,X 1 = X. If we think of p as representing the ring homomorphism ' p : R R/p then f/2 p, ' p (f) 6= 0. When R = k[x 1,,X n ] and p = (y) =kerev y then R/p and (1.1) [p] 2 X f, ' p (f) =f(y) 6= 0, y/2 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 )wherey n is irreducible.) Lemma Let R be a finitely generated domain over any field k. Let f 2 R be represented by a polynomial f(x). Let X = Spec(R). Then, for any y 2 Y, (y) =p 2 X f, y 2 Y f. (Recall that Y f = {y 2 Y : f(y) 6= 0}.) In other words, 1 (X f )=Y f Proof. This follows from (1.1): : Y! Spec(R). y 2 Y f, y/2 V (f), [p] 2 X f. Theorem There is a continuous epimorphism : Y X = Spec(R) =Spec(k[X]/I(Y )) where sending each point y 2 Y to the prime ideal (y) =kerev y : k[x]!. Furthermore, 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 (V Spec(R) is open i 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 = (x) for some x 2 n and I(Y ) p i x 2 y = Y. Therefore, : Y X = Spec(R) is surjective. The lemma shows that is continuous. The lemma also shows that is open since 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 di erence between di erent generic points of the same Y? (These are all elements of 1 ( (y)).) Theorem For any y 2 n there is a 1-1 correspondence between 1 ( (y)) (the set of generic points of y = Y ) and the set of all k-algebra monomorphisms ' : R = k[x]/ (y),!.

18 18 KIYOSHI IGUSA BRANDEIS UNIVERSITY Proof. Let p = (y). Then x 2 1 ( (y)) i (x) =p. For each such x we have ' = ev x : k[x]/p,! Conversely, given any ' : k[x]/p,!, we get back the point x since its coordinates are given by x i = '(X i ). Since ev x (X i )=x i, this gives a bijection x $ '. 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 R-algebra homomorphism R,!, 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 R-algebra monomorphisms C,! (one is the inclusion map C,! and the other is the composition C! C,! where the first map is complex conjugation). So, 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 = V (X 2 X1 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) 2. 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, B) =(X 2 + ix 1,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 2 C lies on the unit circle. The point (A, B) =(z,1/z) is,in(x 1,X 2 ) coordinates, given by 1 1 (X 1,X 2 )= z, 1 z + 1 2i z 2 z with complex conjugate: 1 1 2i z z, 1 z z

19 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 19 giving A =1/z. Another way to see this: When we conjugate X 1 and X 2,(A, B) is replaced with (A 0,B 0 )=(X 2 + ix 1, X 2 ix 1 )=(B,A). In other we conjugate A, B and switch them. So, the new A is A 0 = B =1/z. The mapping z 7! 1/z interchanges the outside of the circle and the inside minus the origin. (In polar coordinates z = re i, 1/z = 1 r e.) These two points are identified in Spec(R). By dimension theory (to be discussed later) there is only 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. As I said in class, the unique generic point [0] 2 Spec(R) is usually drawn as a cloud which is smeared over the whole space since the closure of the point [0] is the whole space. Spec(R) = [ (generic point)