Commutative Algebra We will mostly follow \Introduction to Commutative Algebra" by Atiyah and MacDonald. All rings will be commutative (with identity). The examples we have encountered of commutative rings all come from the following: 1) Z. 2) k where k is a field. 3) A quotient A=I where A is a ring and I is an ideal of A. 4) A polynomial ring A[x1;:::; xn] is a commutative ring. We construct another important example, the power series ring. Let A be a commutative ring. Let Ai = A for i 2 N. Set Q R = i2 Ai , which is an A-module. Define a product on R by N P (ai )(bj ) = (ck ) where ck = i+j=k ai bj . This makes R into a ring. Let x be the sequence (ai ) where a1 = 1 and ai = 0 if i 6= 1. Then i x is the sequence (bj ) where bj = 1 if j = i, bj = 0 otherwise. Every element f 2 R has a unique expansion f = P α xi with i2N i P j P k P αi 2 A. If g = βj x 2 R then fg = γk x where γk = αi βj . We write R = A[[x]]. Now that we know it exists, we forget this complicated notation, and use the usual notation, representing an P i element f 2 R by a series f = αi x . We iterate the construction, giving us power series rings A[[x1; x2]] = (A[[x1]])[[x2]] and A[[x1;:::; xn]] for all n. Every element f 2 A[[x1;:::; xn]] has a unique expression f = P α xi1 ··· xin with α 2 A. i1;:::;in2N i1;:::;in 1 n i1;:::;in Lemma 1. Suppose that k is a field. Then f = P a xi1 ··· xin 2 R = k[[x ;:::; x ]] is a unit iff i1;:::;in2N i1;:::;in 1 n 1 n a0;:::;0 6= 0. Proof. Suppose that the constant term a0;:::;0 of f is nonzero. Let −1 I be the ideal I = (x1;:::; xn). Let g1 = a0;:::;0. Then fg1 − 1 2 I . We will inductively construct for all r > 0 gr 2 R such that r r−1 fgr − 1 2 I and gr − gr−1 2 I . Suppose that we have constructed g . Let fg − 1 = P b xi1 ··· xin . Let r r i1+:::+in≥r i1;:::;in 1 n g = g + P c xi1 ··· xin , where r+1 r i1+···+in=r i1;:::;in 1 n −1 ci1;:::;in = −a0;:::;0bi1;:::;in . Then fg − 1 = (fg − 1) + f (P c xi1 ··· xin ) r+1 r i1+···+in=r i1;:::;in 1 n = P b xi1 ··· xin i1+:::+in=r i1;:::;in 1 n +a (P c xi1 ··· xin ) 0 i1+···+in=r i1;:::;in 1 n +terms of order larger than r in x1;:::; xn = 2 I r+1: The gn have been constructed so that there exists a series g 2 R r+1 r such that g − gr 2 I for all r. We have that fg − 1 2 I for all r so fg − 1 = 0. Thus fg = 1 and so f is a unit in R. Verify that a unit has a nonzero constant term, to complete the proof. Suppose I ; J are ideals in a ring A. The ideal IJ is the ideal generated by the set of products P ffg j f 2 I and g 2 Jg. The ideal IJ consists of the sums fi gi such that fi 2 I and fi 2 J. The intersection I \ J of two ideals is an ideal. The sum of two ideals, I + J = ff + g j f 2 I and g 2 Jg is an ideal. Theorem 2. Every nonzero ring A has at least one maximal ideal. Proof. Let Σ be the set of all ideals in A which are not equal to A. Order Σ by inclusion. Σ is not empty since 0 2 Σ. To apply Zorn's lemma, we must show that every chain in Σ has an upper bound in Σ. Let fai g be a chain in Σ, so that for each ai , aj , either ai ⊂ aj or aj ⊂ ai . Let a = [ai . Then a is an ideal (verify this) and 1 2 a because 1 62 ai for all i. Hence a 2 Σ and a is an upper bound of the chain fai g. Thus by Zorn's lemma, Σ has a maximal element, which is thus a maximal ideal. Corollary 3. Every nonzero ring A has a prime ideal. A maximal ideal is a prime ideal. Corollary 4. If a 6= A is an ideal in a ring A then there exists a maximal ideal of A which contains a. Proof. Let π : A ! A=a be the canonical homomorphism. Let m be a maximal ideal of A=a. Let n = π−1(m). The kernel of the surjection A ! (A=a)=m is n so A=n =∼ (A=a)=m which is a field, so n is a maximal ideal of A. We have that a = π−1(0) ⊂ n. Corollary 5. Every non unit of A is contained in a maximal ideal. Proof. The ideal generated by a non unit is not equal to A. A local ring is a ring which has a unique maximal ideal. Every field is a local ring. Proposition 6. Let A be a ring and m an ideal of A such that m 6= A and every x 2 A n m is a unit in A. Then A is a local ring and m is its maximal ideal. Proof. Every ideal which is not equal to A consists of non units, and so is contained in m. Hence m is the unique maximal ideal of A. A power series ring R = k[[x1;:::; xn]] over a field k is a local ring. In Lemma 1 we showed that f 2 R is a unit if and only if its constant term is nonzero which is equivalent to f 62 I = (x1;:::; xn). Thus by Proposition 6, R is a local ring with maximal ideal I . An element x in a ring A is said to be nilpotent if xn = 0 for some n > 0. Proposition 7. The set Nil(A) of all nilpotent elements in a ring A is an ideal and A=Nil(A) has no nonzero nilpotent elements. Prove this. To prove that Nil(A) is an ideal, use the binomial theorem, which is valid in any commutative ring. The set of nilpotent elements Nil(A) in A is called the nilradical of A. Proposition 8. The nilradical of a ring A is the intersection of all prime ideals of A. Proof. Let N = Nil(A) and N0 be the intersection of all prime ideals of A. If f 2 A is nilpotent and if P is a prime ideal, then f n = 0 2 P for some n > 0 so f 2 P since P is a prime ideal. Thus N ⊂ N0. Suppose that f 2 A is not nilpotent. Let Σ be the set of ideals a of A with the property that f n 62 a for all n > 0. The set Σ is ordered by inclusion. Σ is non empty since 0 2 Σ. Further, A 62 Σ. Suppose that fai g is a chain in Σ. Then a = [ai is an ideal in A which does not contain any positive power of f so a 2 Σ is an upper bound of fai g. By Zorn's Lemma, Σ has a maximal element P. We will show that P is a prime ideal. Let x; y 62 P. Then the ideals P + (x) and P + (y) strictly contain P and thus do not belong to Σ, so f m 2 P + (x) and f n 2 P + (y) for some positive m; n. Thus f m+n 2 (P + (x))(P + (y)) = P + (xy). Hence the ideal P + (xy) is not in Σ and thus xy 62 P, and so P is a prime ideal. Hence we have a prime ideal P such that f 62 P so f 62 N0, so N0 ⊂ N. The Jacobson radical J(A) is defined to be the intersection of all maximal ideals of A. Proposition 9. x 2 J(A) iff 1 − xy is a unit in A for all y 2 A. ) Suppose x 2 J(A) and 1 − xy is not a unit for some y 2 A. By Corollary 5, it belongs to some maximal ideal m of A. But x 2 J(A) ⊂ m implies xy 2 m implies 1 2 m, a contradiction. ( Suppose x 62 m for some maximal ideal m of A. Then m and x generate the unit ideal A so that u + xy = 1 for some u 2 m and some y 2 A. Hence 1 − xy 2 m and is thus not a unit. Let I and J be ideals in a ring A. Then IJ ⊂ I \ J but in general these two ideals are not equal. A simple example where equality fails is I = (x; y) and J = (x; y − 1) in the polynomial ring R = k[x; y] where k is a field. x 2 I \ J but x 62 IJ. Going back to the general case of ideals I and J in a ring A, we always have (I + J)(I \ J) = I (I \ J) + J(I \ J) ⊂ IJ: Hence IJ = I \ J if I + J = A. Two ideals I and J in a ring A are said to be coprime if I + J = A.