p = {x R x n = 0 for some n > 0}.
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1 16 JENIA TEVELEV 5. APPLICATIONS OF LOCALIZATION 5.1. Nilradical. The localization homomorphism φ : R S 1 R induces an injective map of spectra Spec S 1 R Spec R, p φ 1 (p). The image consists of prime ideals which do not intersect S. Localization is used when we want to focus on ideals disjoint from S. Here is a fairly typical example. PROPOSITION Let R be a commutative ring. The intersection of all its prime ideals is equal to the set of nilpotent elements (called the nilradical) of R: p = {x R x n = 0 for some n > 0}. p Spec R Proof. If p is a prime ideal and x n = 0 for some n then x n p and therefore x p. Now suppose that x is not nilpotent. Then S = {1, x, x 2,... } is a multiplicative system. Consider the localization S 1 R and any maximal (and hence prime) ideal I of it. Then p = R I is a prime ideal of R that does not intersect S, and therefore does not contain x Going-up Theorem. THEOREM (Going-up Theorem). Let A B be rings and suppose that B is integral over A. Then the pull-back map Spec B Spec A is surjective. Proof. We have to show that for any prime ideal p A, there exists a prime ideal q B such that p = q A. Let S A be a complement of p. We can view S as a multiplicative system not only in A but also in B. Localizing at S gives a commutative diagram B A α > S 1 B β > Ap (6) where the vertical arrows are inclusions. Notice that S 1 B is integral over A p : if b B is a root of a monic polynomial b n + a 1 b n a n = 0 then b/s for s S is a root of a monic polynomial (b/s) n + (a 1 /s)(b/s) n (a n /s n ) = 0. Suppose we can prove the theorem for A p S 1 B. Then there exists a prime ideal J of S 1 B such that J A p = S 1 p. Let q = α 1 (J). We claim that q A = p. This follows from commutativity of the diagram (6) and because β 1 (S 1 p) = p by Prop It remains to consider the case when A is a local ring with a maximal ideal m. Let m B be any maximal ideal. We claim that m A = m. In
2 NOTES ON COMMUTATIVE ALGEBRA 17 any case, m A m because m A is a proper ideal of A and m is its only maximal ideal. We have a commutative diagram B > B/m A > A/(m A) where the vertical arrows are inclusions. Notice that B/m is a field integral over A/(m A). But then A/(m A) is a field (see the proof of Lemma 2.2.2). So m A is a maximal ideal, and therefore m A = m. EXAMPLE Consider the embedding f : Z Z[i]. Since i is a root of a monic polynomial T = 0, this is an integral extension. These rings are PIDs, and non-zero prime ideals correspond to primes (irreducible elements) in Z and in Z[i], respectively (up-to association). Suppose γ Z[i] is prime and let p Z be a prime such that (p) = (γ) Z. We have a commutative diagram Z/(p) > Z[i]/(p) > Z[i]/(γ) The field extension Z/(p) Z[i]/(γ) is generated by the image of i in Z[i]/(γ), and consequently has degree 1 or 2 depending on whether 1 is a square in Z/(p) or not. If 1 is not a square then the map Z[i]/(p) Z[i]/(γ) is an isomorphism (as both sets have p 2 elements and this map is surjective). Therefore, in this case (p) = (γ) and p = γ up to association. If 1 is a square modulo p then Z[i]/(γ) has p elements. The number of elements in Z[i]/(γ) is the area of the square with sides γ and iγ, therefore, γ = p. It follows that γγ = p. Since Z[i] is a PID, it follows that there are exactly two possibilities for γ, unless γ and γ are associate, i.e. if γ/γ is a unit in Z[i]. There are just 4 units, ±1 and ±i, and it is easy to see that γ and γ are associate if and only if γ = 1 + i (up to association). Finally, it is very easy to see (using the fact that F p is cyclic) that 1 is not a square modulo p if and only if p 3 mod 4. So the full picture of the pull-back map of spectra is as follows:
3 18 JENIA TEVELEV Before giving examples of a more geometric nature, let s develop the language of algebraic sets a bit further. 6. BASIC ALGEBRAIC GEOMETRY Let R be the algebra of polynomials in n variables over an algebraically closed field k. We view R as the ring of functions on the affine space A n k. Weak Nullstellensatz provides a bijection between A n and the set of maximal ideals in R. A n is endowed with Zariski topology with closed sets V (I) = {x A n f(x) = 0 for any f I} for any ideal I R (and strong Nullstellensatz provides a bijection between the set of closed subsets of A n and the set of radical ideals of R) Irreducible algebraic sets. DEFINITION A subset of a topological space is called irreducible if it is not a union of two proper closed subsets. What are the irreducible subsets of R n in the usual Euclidean topology? Only points. (Why?) Not so in Zariski topology! LEMMA The set of irreducible subsets of A n is in bijection with Spec R via the usual correspondence Y I(Y ). Proof. Suppose Y A n is a reducible subset, Y = Y 1 Y 2. Let I (resp. I 1 and I 2 ) be the ideals of all polynomials vanishing along Y (resp. Y 1 and Y 2 ). Since Y 1 and Y 2 are proper subsets of Y, we have I I 1 and I I 2 by the Nullstellensatz. So there exist f I 1 \ I and g I 2 \ I. However, clearly f g I(Y ). This shows that I is not prime. Conversely, if I = I(Y ) is not prime then we can find f, g R \ I such that fg I. This means that, for any y Y, either f(y) = 0 or g(y) = 0. It follows that Y can be decomposed as (Y V (f)) (Y V (g)). What s going on here? We see that in the geometric situation, points of A n correspond to maximal ideals of R = k[x 1,..., x n ] and other prime ideals of R correspond to irreducible subsets of A n which are not points. EXAMPLE The only prime ideal in k[x] that is not maximal is (0), i.e. Spec k[x] = A 1 k {η}, where η = (0). One can visualize η as a sort of a fuzzy point. EXAMPLE According to the homework, the prime ideals of k[x, y] are maximal ideals (x a, y b); ideals (f), where f k[x, y] is an irreducible polynomial; (0). The corresponding closed algebraic sets are points (a, b) A 2 ; algebraic curves (f = 0), where f k[x, y] is irreducible; A 2. Draw a picture of Spec k[x, y].
4 NOTES ON COMMUTATIVE ALGEBRA 19 REMARK One can endow Spec R with Zariski topology for any ring R. Namely, for any subset I R, we declare V (I) := {p Spec R p I} to be a closed subset in Zariski topology (see the homework for details). If R = k[x 1,..., x n ] then A n with Zariski topology can be identified with the set of maximal ideals MaxSpec R Spec R with induced topology. Fuzzy points discussed above then have a remarkable property, rare in traditional topological contexts: they are not closed! For example, consider (0) Spec k[x]. This point belongs to only one Zariski closed subset, namely V ((0)). It follows that (0) = Spec k[x]. What are the closures of points in Spec k[x, y]? 6.2. Irreducible components. DEFINITION Let Y A n be a closed algebraic set. Maximal (by inclusion) irreducible subsets of Y are called irreducible components of Y. THEOREM A closed algebraic set Y has finitely many irreducible components Y 1,..., Y r and we have Y = Y 1... Y r. Proof. It suffices to prove that Y can be written as a finite union of irreducible subsets Z 1,..., Z r. Indeed, by throwing away subsets contained in other subsets we can also assume that Z i Z j for i j. If Z Y is any subset then Z = (Z Z 1 )... (Z Z r ), and so if Z is irreducible then it must be contained in one of the Z i s. It follows that Z i s are exactly irreducible components of Y. Let s prove the claim. If Y is irreducible then there is nothing to do. If Y = Y 1 Y 2 then we can start breaking Y 1 and Y 2 further into unions of proper closed subsets. We claim that this process eventually stops and produces the decomposition of Y as a finite union of irreducible subsets. Indeed, if the process doesn t terminate then we will produce a nested chain of closed subsets Y = Y 1 Y 2 Y 3... where Y i Y i+1. But then I(Y 1 ) I(Y 2 ) I(Y 3 )... is an increasing chain of ideals and I(Y i ) I(Y i+1 ) by Nullstellensatz. But k[x 1,..., x n ] is a Noetherian ring (Hilbert s basis theorem). REMARK What does this theorem tell us on the algebraic side? By Nullstellensatz, a closed algebraic set Y A n corresponds to a radical ideal I k[x 1,..., x n ]. Irreducible subsets contained in Y correspond to prime ideals p I. Irreducible components Y 1,..., Y r of Y correspond to minimal (by inclusion) prime ideals p 1,..., p r containing I. We claim that in fact I = p 1... p r. Indeed, if f p i for any i then f vanishes along each Y i, and so f I.
5 20 JENIA TEVELEV 6.3. Affine algebraic sets. Regular functions. DEFINITION A closed algebraic set X A n is also called an affine algebraic set, especially if we want to ignore its embedding into A n. Likewise, an irreducible algebraic set X A n is called an affine algebraic variety. DEFINITION Let Y A n k be a closed algebraic set. The algebra of regular functions on Y is the algebra of restrictions of polynomial functions O(Y ) = k[x 1,..., x n ]/I(Y ). This algebra is also known as the coordinate algebra of Y. Notice that Y is irreducible if and only if O(Y ) is a domain. DEFINITION A ring R is called reduced if its nilradical is equal to 0. LEMMA O(Y ) is a finitely generated reduced k-algebra. Moreover, any finitely generated reduced k-algebra is isomorphic to the algebra of the form O(Y ). Proof. Since I(Y ) is a radical ideal, the nilradical of O(Y ) is the zero ideal. Now let A be any finitely generated reduced k-algebra. Then we can write A k[x 1,..., x n ]/I. We claim that I = I. Indeed, if x I then x + I is a nilpotent in A, hence zero. It follows that A O(X), where X = V (I). We can extend Nullstellensatz and other previously discussed results from A n to any affine algebraic set Y as follows: COROLLARY (a) There is a bijection between the set of points of Y and MaxSpec O(Y ) which sends a point to the ideal of functions vanishing at this point. (b) There is a bijection between the set of closed algebraic subsets of Y and the set of radical ideals of MaxSpec O(Y ) which sends a subset to the ideal of functions vanishing along this subset. (c) There is a bijection (defined as in part (b)) between the set of irreducible algebraic subsets of Y and Spec O(Y ). (d) Closed algebraic subsets of Y are closed subsets of a topology on Y called Zariski topology. This is a topology induced by the inclusion Y A n. Proof. This follows from analogous results about A n. For example, points of Y (viewed as a subset of A n ) correspond to maximal ideals of k[x 1,..., x n ] containing I. By the first isomorphism theorem these ideals correspond to maximal ideals of O(Y ) = k[x 1,..., x n ]/I. Interpreting properties of Y in terms of the algebra O(Y ) is a coordinatefree way of thinking about Y independent of the embedding Y A n Morphisms of affine algebraic sets. Given closed algebraic sets Y 1 A n x 1,...,x n and Y 2 A m y 1,...,y m (subscripts indicate variables), how should we define maps from Y 1 to Y 2? DEFINITION A map α : Y 1 Y 2 is called a regular morphism if α is the restriction of a polynomial map A n x 1,...,x n A m y 1,...,y m, y i = f i (x 1,..., x n ),
6 NOTES ON COMMUTATIVE ALGEBRA 21 i.e. if there exists m polynomials f 1,..., f m in variables x 1,..., x n such that α(a 1,..., a n ) = (f 1 (a 1,..., a n ),..., f m (a 1,..., a n )) for any point (a 1,..., a n ) Y 1. EXAMPLE The map t (t 2, t 3 ) is a morphism from A 1 to the cusp V (y 2 x 3 ) A 2. EXAMPLE A morphism from A 1 t to the parabola X = V (y x 2 ) A 2, t (t, t 2 ), and a morphism X A 1, (x, y) x, are inverses of each other. DEFINITION For any regular morphism α : Y 1 Y 2, a pull-back homomorphism α : O(Y 2 ) O(Y 1 ) is defined as follows. If f O(Y 2 ) and a Y 1 then we simply define (α f)(a) = f(α(a)). To show that α f O(Y 1 ), we have to check that it is the restriction of a polynomial function on A n. But f itself is the restriction of a polynomial function f k[y 1,..., y m ]. Then α (f) is the restriction of the function f(f 1 (x 1,..., x n ),..., f m (x 1,..., x n )), which is obviously a polynomial in n variables. EXAMPLE For the map α : A 1 X = V (y 2 x 3 ) A 2, t (t 2, t 3 ), we have O(X) = k[x, y]/(x 3 y 2 ), and α : k[x, y]/(x 3 y 2 ) k[t], x t 2, y t 3. The image of α is k[t 2, t 3 ] k[t]. The field of fractions of k[t 2, t 3 ] is equal to k(t) and k[t] is the integral closure of k[t 2, t 3 ]. Geometrically, we say that A 1 is the normalization of the cuspidal curve X. EXAMPLE Regular morphisms between A 1 and X = V (y x 2 ) A 2 from Example induce isomorphisms O(X) = k[x, y]/(y x 2 ) k[t] = O(A 1 ), (x t, y t 2 ), (t x). LEMMA Let Mor(Y 1, Y 2 ) be the set of regular morphisms Y 1 Y 2. Then Mor(Y 1, Y 2 ) Hom Rings (O(Y 2 ), O(Y 1 )), α α. In other words, a regular morphism of affine algebraic sets is completely determined by its pull-back homomorphism, and any homomorphism of algebras of regular functions arises as the pull-back for some regular morphism. REMARK In the language of category theory, the category of affine algebraic sets and regular morphisms is equivalent to the category of finitely generated k-algebras (with a trivial nilradical) and homomorphisms.
7 22 JENIA TEVELEV Proof. If α : Y 1 Y 2 is a morphism and α(a 1,..., a n ) = (b 1,..., b m ) for some (a 1,..., a n ) Y 1 then b i = α (y i )(a 1,..., a n ). So α determines α. Let F : O(Y 2 ) O(Y 1 ) be any homomorphism. We have to realize it as a pull-back homomorphism for some morphism of algebraic sets. We can construct a homomorphism F that fits into the commutative diagram of homomorphisms O(Y 2 ) f f k[y 1,..., y m ] F > O(Y 1 ) F y i f i > k[x 1,..., x n ] by choosing, for each i = 1,..., m, a representative f i k[x 1,..., x n ] of a coset F (ȳ i ). Consider a regular morphism α : A n A m, (a 1,..., a n ) (f 1 (a 1,..., a n ),..., f m (a 1,..., a n )). We claim that α(y 1 ) Y 2. It suffices to check that any polynomial f I(Y 2 ) vanishes on α(y 1 ), i.e. that f(f 1 (x 1,..., x n ),..., f m (x 1,..., x n ) I(Y 1 ). But modulo I(Y 1 ) this polynomial is equal to F ( f)=0. For example, we can think about the embedding of a closed algebraic set X A n as a regular morphism X A n. The corresponding pull-back homomorphism is just the canonical projection O(A n ) = k[x 1,..., x n ] k[x 1,..., x n ]/I(X) = O(X). And vice versa, the following proposition shows that any realization of an affine algebraic set X as a closed algebraic subset of the affine space A m is equivalent to a choice of generators of the algebra O(X). PROPOSITION Given a choice of generators ȳ 1,..., ȳ m of O(X), consider a homomorphism k[y 1,..., y m ] α O(X) which sends y i to ȳ i. Let α : X A m be the corresponding map. Then α(x) A m is a closed algebraic set and α : X α(x) is an isomorphism of affine algebraic sets. Proof. By definition, X is a closed algebraic subset in some affine space A n. Let Y = V (Ker α ) A m. By the first isomorphism theorem, we have O(X) O(Y ). By Lemma 6.4.7, this shows that X and Y are isomorphic. Since the map X Y is our map α, we see that α(x) = Y is closed.
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