Math 572 { Lecture Notes Chapter 15 { Commutative Algebra and Algebraic Geometry Lecture 32 §15.1 Noetherian Rings and Affine Algebraic Sets In this chapter, R will usually be a commutative ring with 1 6= 0. Definition A commutative ring R is Noetherian or satisfies the ascending chain condi- tion if there are no infinite increasing chains of ideals in R. That is, whenever I1 ⊆ I2 ⊆ I3 ⊆ · · · is an increasing chain of ideals, there exists some m such that Ik = Im for all k ≥ m. Proposition 1 If I is an ideal in a Noetherian ring R, then R=I is Noetherian. Any homomor- phic image of a Noetherian ring is Noetherian. Proof. Any infinite ascending chain of ideals in R=I would correspond to an infinite ascending chain of ideals in R. The homomorphic image is isomorphic to R=I where I is the kernel of the homomorphism. 1 Theorem 2 The following are equivalent: (1) R is Noetherian. (2) Every nonempty set of ideals of R contains a maximal element under in- clusion. (3) Every ideal of R is finitely generated. Proof. (1))(2). Assume R is Noetherian. Let Σ by any nonempty collection of ideals, but assume Σ has no maximal element. Choose I1 2 Σ. Since I1 is not maximal, there exists I2 2 Σ such that I1 ( I2: Since I2 is not maximal, there exists I3 2 Σ such that I1 ( I2 ( I3: Continuing in this fashion produces an infinite ascending chain in Σ: I1 ( I2 ( I3 ( ··· ( Ik ( Ik+1 ( ··· : This contradicts the fact that R is Noetherian. Hence Σ must have a maximal element. (2))(3). Assume (2) holds. Let I be any ideal of R. If I is not finitely generated, there exist elements x1; x2; x3;::: 2 I such that I1 = (x1) ( I2 = (x1; x2) ( I3 = (x1; x2; x3) ( ··· Let Σ be the collection of ideals in the sequence above. By (2), Σ contains a 0 0 maximal element I . But then I = In for some n, which means that I was actually finitely generated, a contradiction. So, every ideal of R is finitely generated. (3))(1). Assume every ideal of R is finitely generated. Let I1 ⊆ I2 ⊆ I3 ⊆ · · · 2 be a chain of ideals, and let 1 [ I = Ii: i=1 Then I is an ideal, which by (3), is finitely generated I = (x1; x2; : : : ; xk) Then xi 2 Imi for some mi. Let m = max(m1; : : : ; mk). Then x1; : : : ; xk 2 In for all n ≥ m. Thus, In = I for all n ≥ m and the ascending chain of ideals is finite. Hence, R is Noetherian. Note In the previous proof, we did not every use the assumption of commutativity or that the ring contained 1. Hence, we could have defined a Noetherian ring as a ring (not necessarily commutative and not necessarily containing 1) satisfying the ascending chain condition. Then the previous theorem still holds. Example Every ideal in a PID is finitely generated (by a single element). Hence, any PID is Noetherian. Z, Z[i], F [x] for F a field are Noetherian. Z[x1; x2; x3;:::] is not Noetherian since the ideal (x1; x2; x3;:::) is not finitely generated. Z[x1; : : : ; xn] is Noetherian (by Hilbert's basis theorem). Example Noetherian ring may having infinitely long descending chains. For example, in Z 2 3 (2) ) (2 ) ) (2 ) ) ··· Theorem 3 (Hilbert's Basis Theorem) If R is a Noetherian ring, then so is R[x]. 3 Proof. Let I be any ideal in R[x]. We'll show that I is finitely generated, and so R[x] will be Noetherian. Let L ⊆ R be the set of all leading coefficients of polynomials in I. Thus, axm + lower order terms 2 I ) a 2 L: We'll show that L is an ideal of R. Since the zero polynomial is in I, 0 2 L. Let f = axd + ··· and g = bxe + ··· be polynomials in I. For any r 2 R, either ar − b = 0 or ar − b is the leading coefficient of rxef − xdg: This shows that L is an ideal in R. Since R is Noetherian, L is finitely generated: L = (a1; : : : ; an) ai 2 R: For each ai choose fi 2 I such that ei fi = aix + ··· Set N = max(e1; : : : ; en). For each d 2 f0; 1;:::;N − 1g, let Ld = \set of leading coefficients of elements of degree d in I along with 0". Let f = axd + ··· and g = bxd + ··· be polynomials of degree d in I. For any r 2 R, either ra − b = 0 or ra − b is the leading coefficient of rf − g Hence, Ld is also an ideal of R, and so is finitely generated. Ld = (bd1; bd2; : : : ; bdnd ): For each bdi choose fdi 2 I such that d fdi = bdix + ··· : 4 Let I0 be the ideal in R[x] defined by 0 I = ff1; : : : ; fng [ ffdi : 0 ≤ d < N; 1 ≤ i ≤ ndg : By construction I0 ⊆ I: We'll show I0 = I and so I is finitely generated. If I 6= I0, there exists some nonzero polynomial f 2 I n I0 of minimal degree m. Write f = axm + ··· where deg(f) = m and a is the leading coefficient. Case 1. Suppose m ≥ N. Since a 2 L a = r1a1 + ··· + rnan for some ri 2 R. Then m−e1 m−en 0 g = r1x f1 + ··· rnx fn 2 I has degree m and leading coefficient a. Then f − g 2 I n I0 has degree < m. By the minimality of m, f − g = 0 and so f = g 2 I0, a contradiction. Case 2. Suppose m = d < N. In this case, a 2 Ld. Write a = r1bd1 + ··· + rnd bdnd ; ri 2 R: Then 0 g = r1fd1 + ··· + rnd fdnd 2 I had degree m = d and has the same leading coefficient as f. As before, we have a contradiction. Hence, I = I0. This proves that R[x] is Noetherian. Corollary If R is Noetherian, R[x1; : : : ; xn] is Noetherian. If k is a field k[x1; : : : ; xn] is Noetherian. 5 Lecture 33 Lecture 34 Definition Let k be a field. A ring R is called a k-algebra if k is contained in the center of R and the identity of k is the identity of R. Definition (1) The ring R is a finitely generated k-algebra if R is generated as a ring by k together with some finite set r1; r2; : : : ; rn of elements of R. (2) Let R and S be k-algebras. A map : R ! S is a k-algebra homomor- phism if is a ring homomorphism that is the identity on k. Corollary 5 The ring R is a finitely generated k-algebra if and only if there is some surjective k-algebra homomorphism φ: k[x1; x2; : : : ; xn] ! R: Hence, any finitely generated k-algebra is Noetherian. Proof. If R is generated as a k-algebra by r1; : : : ; rn, we define the map φ: k[x1; : : : ; xn] ! R by φ(xi) = ri and φ(a) = a for all a 2 k. Then φ extends uniquely to a surjective homomorphism. Conversely, given a surjective homomorphism, the image of x1; : : : ; xn generates R as a k-algebra, proving that R is finitely generated. Then R is isomorphic to the quotients of the Noetherian ring k[x1; : : : ; xn] modulo the kernel, which is again Noetherian. Affine Algebraic Sets 6 An is the set of n-tuples of the field k is called affine n-space. (Apply the forgetful functor to the vector space kn to get affine space.) n Polynomials f 2 k[x1; : : : ; xn] may be regarded as functions f : A ! k by evalua- tion: f :(a1; : : : ; an) 7! f(a1; : : : ; an) 2 k: This gives a ring of functions, denoted k[An], called the coordinate ring of An. Definition If S ⊆ k[An], then n Z(S) := f(a1; : : : ; an) 2 A : f(a1; : : : ; an) = 0 for all f 2 Sg is called an affine algebraic set. V = Z(S) is called the locus of S in An. Examples 1. If n = 1 and f(x) 2 k[x], then Z(f) is the set of roots of f. n 2. One point subsets f(a1; : : : ; an)g of A are Z(x1 − a1; : : : ; xn − an). 3. If f 2 k[x1; : : : ; xn] is a nonconstant polynomial, Z(f) is called a hyper- surface. In R2, Z(y−x2) is the parabola y = x2, Z(xy−1) is the hyperbola y = 1=x. In R3, Z(x2 + y2=4 + z2=9 − 1) is an ellipsoid. 7 Basic Properties of Z Let S; T ⊆ k[An]. (1) If S ⊆ T , then Z(T ) ⊆ Z(S). (2) Z(I) = Z(S) where I = (S) is the ideal generated by S. (3) Z(S) \Z(T ) = Z(S [ T ). (4) For ideals I;J ⊆ k[An], Z(I) [Z(J) = Z(IJ). (5) Z(0) = An and Z(1) = ;. By (2), we may consider the map n n Z : fideals of k[A ]g ! faffine algebraic sets in A g: Since k[x1; : : : ; xn] is Noetherian, an ideal I = (f1; : : : ; fq) is finitely generated, and from (3) Z(I) = Z(f1) \Z(f2) \···\Z(fq): So, each affine algebraic set is the intersection of finitely many hypersurfaces. Definition For any A ⊆ An, I(A) := ff 2 k[x1; : : : ; xn]: f(a1; : : : ; an) = 0 for all (a1; : : : ; an) 2 Ag: I(A) is an ideal and is the largest ideal that vanishes on A. 8 Basic Properties of I Let A; B ⊆ An (6) If A ⊆ B, then I(B) ⊆ I(A).