# Extension theorems for homomorphisms

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1 Algebraic Geometry Fall 2009 Extension theorems for homomorphisms In this note, we prove some extension theorems for homomorphisms from rings to algebraically closed fields. The prototype is the following result: Theorem 1 (Extension theorem for algebraic extensions). If L/K is an algebraic extension of fields, then any embedding σ of K into an algebraically closed field k can be extended to an embedding σ : L k. This was proved in class using Zorn s Lemma. Note that some hypothesis akin to k is algebraically closed is needed for such a statement, since e.g. if L = K(α) and f α (x) is the minimal polynomial of α over K, then σ must map α to a root of f α (x) in k. Since f α can be any monic polynomial in K[x], we need K to be algebraically closed in k. A simple way to guarantee this is to require that k itself be algebraically closed. We now prove an important generalization of Theorem 1 from algebraic extensions of fields to integral extensions of rings: Theorem 2 (Extension theorem for integral extensions). Let A B be rings with B integral over A. Then every homomorphism φ of A to an algebraically closed field k can be extended to a homomorphism φ : B k. Proof. Let p = ker(φ), and let S = A\p. Since φ factors through the canonical homomorphism A S 1 A (by the universal property of localization), S 1 B is integral over S 1 A (by problem 3(a) on page 2 of Homework 1), and the diagram B S 1 B A S 1 A is commutative (again by the universal property of localization), we are reduced to the case where A is a local ring and ker(φ) is the unique maximal ideal m of A (by replacing A with A p = S 1 A and B with S 1 B). By the Going-Up Theorem (Theorem 3 below), there is a maximal ideal M of B lying over m (i.e., such that M A = m). Since B/A is integral, B/M is an algebraic extension of A/m. By Theorem 1, we can extend φ : A/m k to a homomorphism φ : B/M k. 1

2 2 Composing with the natural projection maps A A/m and B B/M gives the desired extension of φ: In the proof, we used: B B/M A A/m φ k = φ k Theorem 3 (Going-Up Theorem). If A B are rings with B integral over A, and p is a prime ideal of A, then there exists a prime ideal q lying over p, i.e., such that q A = p. Furthermore, q is maximal iff p is maximal. For this, we need a lemma: Lemma 1. If A B are rings with B integral over A, q is a prime ideal of B, and p = q A, then q is maximal iff p is maximal. Proof. Since B/q is integral over A/p, and both are integral domains, we are reduced to proving the following statement: If A B are integral domains with B integral over A, then A is a field iff B is a field. To see this, suppose first that A is a field. If y B is nonzero, let y n + a n 1 y n a 0 be an integral dependence of minimal degree, with a i A. As B is a domain, a 0 0, so y 1 = a 1 0 (y n a 1 ) B. Thus B is a field. Conversely, if B is a field and x A is nonzero, then x 1 B, hence is integral over A, so we have x m + a m 1x m a 0 = 0 with a i A. Thus x 1 = (a m 1 + a m 2x + + a 0x m 1 ) A, so A is a field. Proof of Theorem 3. Let S = A\p. As before, we have a commutative diagram β B S 1 B α A S 1 A with S 1 B is integral over S 1 A = A p. Let M be any maximal ideal of S 1 B. Then m = M A p is a maximal ideal in the local ring A p by

3 Lemma 1, hence equals pa p. Under the correspondence between prime ideals of S 1 B and prime ideals of B disjoint from S, if q = β 1 (M) is the prime ideal of B corresponding to M, one sees from the commutativity of the above diagram that q A = α 1 (m) = α 1 (pa p ) = p. One can use the Extension Theorem for Integral Extensions to give an insightful proof of the Algebraic Nullstellensatz: Theorem 4 (Algebraic Nullstellensatz). If L/K is a field extension and L is finitely generated as a K-algebra, then L is a finite extension of K. Proof. Let x 1,..., x n be generators for L as a K-algebra. If L/K is algebraic, then (since L/K is finitely generated) it must be a finite extension, and we re done. So let s assume, for the sake of contradiction, that L/K is transcendental. Let t 1,..., t r L be a transcendence basis for L/K, i.e., K[t 1,..., t r ] is isomorphic to a polynomial ring in r variables, and x 1,..., x n are algebraic over K(t 1,..., t r ). (Convince yourself that such a basis always exists.) Let p i (X) be the minimal polynomial of x i over K(t 1,..., t r ). Multiplying by a suitable nonzero element h of K[t 1,..., t r ] to clear denominators, we obtain a polynomial q i (X) = h p i (X) K[t 1,..., t r ][X] with q i (x i ) = 0. Let f = f(t 1,..., t r ) K[t 1,..., t r ] be the product of the leading coefficients of the q i (X). Then each x i is integral over the ring K[t 1,..., t r ] f. Let K be an algebraic closure of K. Since K is infinite and f 0, there exist y 1,..., y r K such that f(y 1,..., y r ) 0. Consider the unique homomorphism φ : K[t 1,..., t r ] K which is the identity on K, and for which φ(t i ) = y i. Since φ(f) 0, the universal property of localization shows that φ extends uniquely to a homomorphism φ : K[t 1,..., t r ] f K. By Theorem 2, we can extend φ to a homomorphism φ : K[t 1,..., t r ] f [x 1,..., x n ] K which is the identity on K. But K[t 1,..., t r ] f [x 1,..., x n ] = L. Since L is a field, φ must be injective, so L can be embedded as a subfield of K containing K. Thus L/K is algebraic, contradiction. We now consider the following related question: given a subring A of a field K, and a homomorphism φ from A to an algebraically closed field k, when is it possible to extend φ to a larger subring of K? If A is not integrally closed in K, then by Theorem 2 we can extend φ to the integral closure A of A in K. Can we extend even further? The following result is the key to understanding the answer to this question: 3

4 4 Theorem 5. Let A be a subring of a field K, and let x K. If φ is a homomorphism from A to an algebraically closed field k, then φ can be extended to a homomorphism from either A[x] or A[x 1 ] to k. Proof. By first extending φ to a homomorphism from A p to k, with p = ker(φ), we may assume that A is a local ring with maximal ideal m = ker(φ). If x is integral over A, then φ has an extension to A[x] by Theorem 2. Now assume that x is not integral over A, and let B = A[x 1 ]. We claim that mb B. Assuming this, it follows that mb is contained in some maximal ideal M of B. As M A contains m and m is maximal, it follows that M A = m. Let y be the image in B/M of x 1 B. Since B/M = (A/m)[y], we can extend the given map ψ : A/m k to ψ : B/M k. Indeed, if y is algebraic over A/m, then this follows from Theorem 1, while if y is transcendental then we can define ψ by sending y to any element of k that we like. (In fact, y must be algebraic over A/m by the Algebraic Nullstellensatz, but we don t need to use this.) Composing with the natural projection B B/M gives the desired extension. To prove the claim, note that otherwise we can write 1 = a 0 + a 1 x a n x n with a i m. Multiplying by x n, we get (1 a 0 )x n + b n 1 x n b 0 = 0 with b i A. As a 0 m, 1 a 0 A, and dividing by (1 a 0 ) 1 we see that x is integral over A, a contradiction. Corollary 1. Let A be a subring of a field K, and let φ be a homomorphism from A to an algebraically closed field k. If R is a maximal subring of K for which φ extends to a homomorphism Φ : R k (such subrings exist by Zorn s lemma), then R is a valuation ring of K: for every x K, either x R or x 1 R. Proof. Let Σ be the set of pairs (C, ψ) where C is a subring of K and ψ : C k is a homomorphism extending φ. By Zorn s lemma, there is a maximal element (R, Φ) of Σ, and R is a valuation ring by Theorem 5. Note that the ring R given in Corollary 1 is a local ring with maximal ideal m equal to ker(φ): otherwise we could extend Φ even further to R p, where p = ker(φ). (Exercise: R p R unless R is already a local ring with maximal ideal p. Another exercise: Show that every valuation ring is a local ring.) Combining this observation with Corollary 1, we

5 obtain the following result (which is how the Extension Theorem is formulated in [Bump]): Theorem 6 (The Extension Theorem). Let A be a subring of a field K, and let φ : A k be a homomorphism from A to an algebraically closed field k. Then there exists a valuation ring R of K containing A, and a homomorphism Φ : R k extending φ, such that ker(φ) is the unique maximal ideal of R. One application of the Extension Theorem is the fact that if A is a subring of a field K, then the integral closure of A in K is the intersection of all valuation rings of K which contain A (Problem 6 on Homework 1). The Extension Theorem can also be used to prove the extension theorem for integral extensions and the Going-Up Theorem, though in these notes we have gone the other way, and used those results to prove the Extension Theorem. See [Bump] for an alternative, more direct proof of the Extension Theorem. The main advantages to the proofs given in these notes are: (a) the statement of the Extension Theorem is more motivated, and (b) our approach to the extension theorem for integral extensions and the Going-Up Theorem does not require the assumption made in [Bump] that A and B are integral domains. 5

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