# R S. with the property that for every s S, φ(s) is a unit in R S, which is universal amongst all such rings. That is given any morphism

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1 8. Nullstellensatz We will need the notion of localisation, which is a straightforward generalisation of the notion of the field of fractions. Definition 8.1. Let R be a ring. We say that a subset S of R is multiplicatively closed if for every s 1 and s 2 in S, s 1 s 2 S, that is S S S. Definition-Lemma 8.2. Let R be a ring and let S be a multiplicatively closed subset, which contains 1 but not zero. The localisation of R at S, denoted R S, is a ring R S together with a ring homomorphism φ: R R S, with the property that for every s S, φ(s) is a unit in R S, which is universal amongst all such rings. That is given any morphism ψ : R R, with the property that ψ(s) is a unit, for every s S, there is a unique ring homomorphism R ψ T. φ R S Proof. This is almost identical to the construction of the field of fractions, and so we will skip most of the details. Formally we define R S to be the set of all pairs (r, s), where r R and s S, modulo the equivalence relation, (r 1, s 1 ) (r 2, s 2 ) iff s(r 1 s 2 r 2 s 1 ) for some s S. We denote an equivalence class by [r, s] (or more informally by r/s). Addition and multiplication are defined in the obvious way. Note that R is an integral domain, then S = R {0} is multiplicatively closed and the localisation is precisely the field of fractions. Note also that as we are not assuming that R is an integral domain, we need to throw in the extra factor of s, in the definition of the equivalence relation and the natural map R R S is not necessarily injective. Example 8.3. Suppose that p is a prime ideal in a ring R. Then S = R p is a multiplicatively closed subset of R. The localisation is denoted R p. It elements consist of all fractions r/f, where f / p. On 1

2 the other hand, suppose that f R is not nilpotent. Then the set of powers of f, S = { f n n N }, is a multiplicatively closed subset. The localisation consists of all elements of the form r/f n. For example, take R = Z and f = 2. Then R f = Z[1/2] Q consists of all fractions whose denominator is a power of two. Lemma 8.4. Let F be a field and let f F [x] be a polynomial. Then F [x] f is not a field. Proof. Suppose not. Clearly deg(f) > 0 so that 1+f 0. Therefore we may find g F [x] such that (1 + f) 1 = g f n, for some n. Multiplying out, we get that (1 + f) divides f n. So f n is congruent to 0 modulo (1 + f). On the other hand, f is congruent to 1 modulo (1 + f). The only possibility is that 1 + f is a unit, which is clearly impossible. Definition 8.5. Let R F be a subring of the field F. We say that c F is integral over S if and only if there is a monic polynomial m(x) = x n + a n 1 x n a 1 x + a 0 S[x], such that m(c) = 0. If R S F is an intermediary ring, we say that S is integral over R if every element of S is integral over R. The integral closure of R in F is the set of all elements integral over R. Lemma 8.6. Let R F be a subring of the field F. The following are equivalent: (1) c is integral over R, (2) R[c] is a finitely generated R-module, (3) there is an intermediary ring R[c] C S which is a finitely generated R-module. Proof. Suppose that c is integral over R. Pick m(x) R[x] monic such that m(c) = 0. If m(x) has degree d it is easy to see that 1, c, c 2,..., c d 1 generate R[c] as an R-module. Thus (1) implies (2). (2) implies (3) is clear. 2

3 Now suppose that C is a finitely generated R-module. Multiplication by c defines an R-linear map φ: C C. Pick generators c 1, c 2,..., c k for the R-module C. Then we may find A = (a ij ) M k (R) such that φ(c i ) = a ij c j. Then m(x) = det(a λi) R[x] is a monic polynomial and m(φ) = 0, by Cayley-Hamilton. But then m(c) = m(φ(1)) = 0. Hence (3) implies (1). Lemma 8.7. Let R F be a subring of the field F. If S = R[r 1, r 2,..., r k ] where each r 1, r 2,..., r k is integral over S then S is integral over R. Proof. By (8.6) it suffices to prove that S is a finitely generated R- module. By induction on k we may assume that S = R[r 1, r 2,..., r k 1 ] is finitely generated R-module. As S is a finitely generated S -module (r k is integral over S as it is integral over R) it follows that S is a finitely generated R-module. We will need the following result later: Lemma 8.8. Let R F be a subring of the field F. The integral closure S or R in F is a ring. Proof. Let a and b be in S. It suffices to prove that a ± b and ab are in S. But a ± b and ab belong to R[a, b] and this is finitely generated over R by (8.7). Lemma 8.9. Let E be a field and let R be a subring. If E is integral over R then R is a field. Proof. Pick a R and let b E be the inverse. As E is integral over R, we may find r 1, r 2,..., r n R such that b n + r 1 b n r n = 0. Multiply both sides by a n 1 and solve for b to get b = r 1 a r 2 a 2 r n a n 1 A. Lemma Let E/F be a field extension. If E is finitely generated as an F -algebra then E/F is algebraic. 3

4 Proof. By assumption E = F [f 1, f 2,..., f m ]. We proceed by induction on m. Let f = f m. By induction E = F (f)[f 1, f 2,..., f m 1 ] is algebraic over F (f). Let m i (x) F (f)[x] be the minimal polynomial of f i. Clearing denominators, we may assume that m i (x) F [f][x]. Let a i be the leading coefficient of m i (x) and let a be the product of the a i. Then (1/a i )m i (x) F [f] a [x] is a monic polynomial, so that f i is integral over F [f] a. By (8.9) F [f] a is a field. But then f is algebraic over F by (8.4). Theorem 8.11 (Weak Nullstellensatz). Let K be an algebraically closed field. Then an ideal m R = K[x 1, x 2,..., x n ] is maximal if and only if it has the form m p = x 1 a 1, x 2 a 2,..., x n a n, for some point p = (a 1, a 2,..., a n ) K n. Proof. Let m R be an ideal and let L = R/m. Then m is maximal if and only if L = R/m is a field and L = K if and only if m = m p for some point p. So we may assume that L is a field and we want to prove that L = K. But L is a finitely generated algebra over K (generated by the images of x 1, x 2,..., x n ) so that by (8.10) L/K is algebraic. As K is algebraically closed, L = K. Corollary 8.12 (Weak Nullstellensatz). Let K be an algebraically closed field. If f 1, f 2,..., f m R = K[x 1, x 2,..., x n ] is a sequence of polynomials then either (1) f 1, f 2,..., f m have a common zero, or (2) there are polynomials g 1, g 2,..., g m K[x 1, x 2,..., x n ] such that f 1 g 1 + f 2 g f m g m = 1. Proof. Let I = f 1, f 2,..., f m R be the ideal generated by the polynomials f 1, f 2,..., f m. Note that (1) holds if and only if I is contained in one of the ideals m p for some p = (a 1, a 2,..., a n ) K n. Indeed, in this case f 1, f 2,..., f n all vanish at p. On the other hand, note that (2) holds if and only if I = R. So suppose that I R. Pick a maximal ideal m containing I. By (8.11) we may find p K n such that m = m p. Theorem 8.13 (Strong Nullstellensatz). Let K be an algebraically closed field. 4

5 If f 1, f 2,..., f m, g R = K[x 1, x 2,..., x n ] is a sequence of polynomials then either (1) f 1, f 2,..., f m have a common zero, at a point where the polynomial g is not equal to zero, or (2) there are polynomials g 1, g 2,..., g m K[x 1, x 2,..., x n ] such that f 1 g 1 + f 2 g f m g m = g r, for some natural number r. Proof. We use the trick of Rabinowitsch. Let S = R[y] = K[x 1, x 2,..., x n, y], where y is an indeterminate and consider the polynomials f 1, f 2,..., f m, yg 1. If (1) does not hold then these equations don t have any solutions at all. By the weak Nullstellensatz (8.12) we may find polynomials g 1, g 2,..., g m, h S such that f 1 g 1 + f 2 g f m g m + h(yg 1) = 1. Let z = 1/y. Clearing denominators by multiplying through some large power z r of z, and relabelling, we get Now set z = g. f 1 g 1 + f 2 g f m g m + h(g z) = z r. 5

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