# Integral Extensions. Chapter Integral Elements Definitions and Comments Lemma

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1 Chapter 2 Integral Extensions 2.1 Integral Elements Definitions and Comments Let R be a subring of the ring S, and let α S. We say that α is integral over R if α isarootofamonic polynomial with coefficients in R. IfR is a field and S an extension field of R, then α is integral over R iff α is algebraic over R, so we are generalizing a familiar notion. If α is a complex number that is integral over Z, then α is said to be an algebraic integer For example, if d is any integer, then d is an algebraic integer, because it is a root of x 2 d. Notice that 2/3 is a root of the polynomial f(x) =3x 2, but f is not monic, so we cannot conclude that 2/3 is an algebraic integer. In a first course in algebraic number theory, one proves that a rational number that is an algebraic integer must belong to Z, so 2/3 is not an algebraic integer. There are several conditions equivalent to integrality of α over R, and a key step is the following result, sometimes called the determinant trick Lemma Let R, S and α be as above, and recall that a module is faithful if its annihilator is 0. Let M be a finitely generated R-module that is faithful as an R[α]-module. Let I be an ideal of R such that αm IM. Then α is a root of a monic polynomial with coefficients in I. Proof. let x 1,...,x n generate M over R. Then αx i IM, so we may write αx i = n j=1 c ijx j with c ij I. Thus n (δ ij α c ij )x j =0, 1 i n. j=1 In matrix form, we have Ax = 0, where A is a matrix with entries α c ii on the main diagonal, and c ij elsewhere. Multiplying on the left by the adjoint matrix, we get x i = 0 for all i, where is the determinant of A. But then annihilates all of M, so = 0. Expanding the determinant yields the desired monic polynomial. 1

2 2 CHAPTER 2. INTEGRAL EXTENSIONS Remark If αm IM, then in particular, α stabilizes M, in other words, αm M Theorem Let R be a subring of S, with α S. The following conditions are equivalent: (1) α is integral over R; (2) R[α] is a finitely generated R-module; (3) R[α] is contained in a subring R of S that is a finitely generated R-module; (4) There is a faithful R[α]-module M that is finitely generated as an R-module. Proof. (1) implies (2): If α is a root of a monic polynomial over R of degree n, then α n and all higher powers of α can be expressed as linear combinations of lower powers of α. Thus 1,α,α 2,...,α n 1 generate R[α] over R. (2) implies (3): Take R = R[α]. (3) implies (4): Take M = R.Ify R[α] and ym = 0, then y = y1 =0. (4) implies (1): Apply (2.1.2) with I = R. We are going to prove a transitivity property for integral extensions, and the following result will be helpful Lemma Let R be a subring of S, with α 1,...,α n S. If α 1 is integral over R, α 2 is integral over R[α 1 ],..., and α n is integral over R[α 1,...,α n 1 ], then R[α 1,...,α n ] is a finitely generated R-module. Proof. The n = 1 case follows from (2.1.4), part (2). Going from n 1ton amounts to proving that if A, B and C are rings, with C a finitely generated B-module and B a finitely generated A-module, then C is a finitely generated A-module. This follows by a brief computation: C = r By j,b= j=1 s Ax k, so C = k= Transitivity of Integral Extensions r j=1 k=1 s Ay j x k. Let A, B and C be subrings of R. IfC is integral over B, that is, every element of C is integral over B, and B is integral over A, then C is integral over A. Proof. Let x C, with x n + b n 1 x n b 1 x + b 0 = 0. Then x is integral over A[b 0,...,b n 1 ]. Each b i is integral over A, hence over A[b 0,...,b i 1 ]. By (2.1.5), A[b 0,...,b n 1,x] is a finitely generated A-module. By (2.1.4), part (3), x is integral over A.

3 2.1. INTEGRAL ELEMENTS Definitions and Comments If R is a subring of S, the integral closure of R in S is the set R c of elements of S that are integral over R. Note that R R c because each a R isarootofx a. We say that R is integrally closed in S if R c = R. If we simply say that R is integrally closed without reference to S, we assume that R is an integral domain with fraction field K, and R is integrally closed in K. If the elements x and y of S are integral over R, then just as in the proof of (2.1.6), it follows from (2.1.5) that R[x, y] is a finitely generated R-module. Since x + y, x y and xy belong to this module, they are integral over R by (2.1.4), part (3). The important conclusion is that R c is a subring of S containing R. If we take the integral closure of the integral closure, we get nothing new Proposition The integral closure R c of R in S is integrally closed in S. Proof. By definition, R c (R c ) c. Thus let x (R c ) c, so that x is integral over R c.asin the proof of (2.1.6), x is integral over R. Thusx R c. We can identify a large class of integrally closed rings Proposition If R is a UFD, then R is integrally closed. Proof. Let x belong to the fraction field K of R. Write x = a/b where a, b R and a and b are relatively prime. If x is integral over R, there is an equation of the form (a/b) n + a n 1 (a/b) n a 1 (a/b)+a 0 =0 with a i R. Multiplying by b n, we have a n + bc = 0, with c R. Thus b divides a n, which cannot happen for relatively prime a and b unless b has no prime factors at all, in other words, b is a unit. But then x = ab 1 R. A domain that is an integral extension of a field must be a field, as the next result shows Proposition Let R be a subring of the integral domain S, with S integral over R. Then R is a field if and only if S is a field. Proof. Assume that S is a field, and let a be a nonzero element of R. Since a 1 S, there is an equation of the form (a 1 ) n + c n 1 (a 1 ) n c 1 a 1 + c 0 =0 with c i R. Multiply the equation by a n 1 to get a 1 = (c n c 1 a n 2 + c 0 a n 1 ) R.

4 4 CHAPTER 2. INTEGRAL EXTENSIONS Now assume that R is a field, and let b be a nonzero element of S. By (2.1.4) part (2), R[b] is a finite-dimensional vector space over R. Let f be the R-linear transformation on this vector space given by multiplication by b, in other words, f(z) =bz, z R[b]. Since R[b] is a subring of S, it is an integral domain. Thus if bz = 0 (with b 0 by choice of b), we have z = 0 and f is injective. But any linear transformation on a finite-dimensional vector space is injective iff it is surjective. Therefore if b S and b 0, there is an element c R[b] S such that bc = 1. Consequently, S is a field Preview Let S be integral over the subring R. We will analyze in great detail the relation between prime ideals of R and those of S. Suppose that Q is a prime ideal of S, and let P = Q R. (We say that Q lies over P.) Then P is a prime ideal of R, because it is the preimage of Q under the inclusion map from R into S. The map a + P a + Q is a well-defined injection of R/P into S/Q, because P = Q R. Thus we can regard R/P as a subring of S/Q. Moreover, S/Q is integral over R/P. To see this, let b + Q S/Q. Then b satisfies an equation of the form x n + a n 1 x n a 1 x + a 0 =0 with a i R. But b + Q satisfies the same equation with a i replaced by a i + P for all i, proving integrality of S/Q over R/P. We can now invoke (2.1.10) to prove the following result Proposition Let S be integral over the subring R, and let Q be a prime ideal of S, lying over the prime ideal P = Q R of R. Then P is a maximal ideal of R if and only if Q is a maximal ideal of S. Proof. By (2.1.10), R/P is a field iff S/Q is a field Remarks Some results discussed in (2.1.11) work for arbitrary ideals, not necessarily prime. If R is a subring of S and J is an ideal of S, then I = J R is an ideal of R. As in (2.1.11), R/I can be regarded as a subring of S/J, and if S is integral over R, then S/J is integral over R/I. Similarly, if S is integral over R and T is a multiplicative subset of R, then S T is integral over R T. To prove this, let α/t S T, with α S, t T. Then there is an equation of the form α n + c n 1 α n c 1 α + c 0 = 0, with c i R. Thus with c n j /t j R T. ( α t )n +( c n 1 )( α t t )n 1 + +( c 1 t n 1 )α t + c 0 t n =0

5 2.2. INTEGRALITY AND LOCALIZATION Integrality and Localization Results that hold for maximal ideals can sometimes be extended to prime ideals by the technique of localization. A good illustration follows Proposition Let S be integral over the subring R, and let P 1 and P 2 be prime ideals of S that lie over the prime ideal P of R, that is, P 1 R = P 2 R = P.IfP 1 P 2, then P 1 = P 2. Proof. If P is maximal, then by (2.1.12), so are P 1 and P 2, and the result follows. In the general case, we localize with respect to P. Let T = R\P, a multiplicative subset of R S. The prime ideals P i,i=1, 2, do not meet T, because if x T P i, then x R P i = P, contradicting the definition of T. By the basic correspondence between prime ideals in a ring and prime ideals in its localization, it suffices to show that P 1 S T = P 2 S T. We claim that PR T (P 1 S T ) R T R T. The first inclusion holds because P P 1 and R T S T. The second inclusion is proper, for otherwise R T P 1 S T and therefore 1 P 1 S T, contradicting the fact that P 1 S T is a prime ideal. But PR T is a maximal ideal of R T, so by the above claim, (P 1 S T ) R T = PR T, and similarly (P 2 S T ) R T = PR T. Thus P 1 S T and P 2 S T lie over PR T. By (2.1.13), S T is integral over R T. Asatthe beginning of the proof, P 1 S T and P 2 S T are maximal by (2.1.12), hence P 1 S T = P 2 S T. If S/R is an integral extension, then prime ideals of R can be lifted to prime ideals of S, as the next result demonstrates. Theorem is also a good example of localization technique Lying Over Theorem If S is integral over R and P is a prime ideal of R, there is a prime ideal Q of S such that Q R = P. Proof. First assume that R is a local ring with unique maximal ideal P. If Q is any maximal ideal of S, then Q R is maximal by (2.1.12), so Q R must be P. In general, let T be the multiplicative set R \ P. We have the following commutative diagram. f R S g R T S T The horizontal maps are inclusions, and the vertical maps are canonical (f(r) = r/1 and g(s) =s/1). Recall that S T is integral over R T by (2.1.13). If Q is any maximal ideal of S T, then as at the beginning of the proof, Q R T must be the unique maximal ideal

6 6 CHAPTER 2. INTEGRAL EXTENSIONS of R T, namely PR T. By commutativity of the diagram, f 1 (Q R T )=g 1 (Q ) R. (Note that if r R, then f(r) Q R T iff g(r) Q.) If Q = g 1 (Q ), we have f 1 (PR T )=Q R. By the basic localization correspondence [cf.(2.2.1)], f 1 (PR T )=P, and the result follows Going Up Theorem Let S be integral over R, and suppose we have a chain of prime ideals P 1 P n of R, and a chain of prime ideals Q 1 Q m of S, where m < n. If Q i lies over P i for i = 1,...,m, then there are prime ideals Q m+1,...,q n of S such that Q m Q m+1 Q n and Q i lies over P i for every i =1,...,n. Proof. By induction, it suffices to consider the case n =2,m= 1. Thus assume P 1 P 2 and Q 1 R = P 1. By (2.1.11), S/Q 1 is integral over R/P 1. Since P 2 /P 1 is a prime ideal of R/P 1, we may apply the lying over theorem (2.2.2) to produce a prime ideal Q 2 /Q 1 of S/Q 1 such that (Q 2 /Q 1 ) R/P 1 = P 2 /P 1, where Q 2 is a prime ideal of S and Q 1 Q 2. We claim that Q 2 R = P 2, which gives the desired extension of the Q-chain. To verify this, let x 2 Q 2 R. By (2.1.11), we have an embedding of R/P 1 into S/Q 1,sox 2 + P 1 = x 2 + Q 1 (Q 2 /Q 1 ) R/P 1 = P 2 /P 1. Thus x 2 + P 1 = y 2 + P 1 for some y 2 P 2,sox 2 y 2 P 1 P 2. Consequently, x 2 P 2. Conversely, if x 2 P 2 then x 2 + P 1 Q 2 /Q 1, hence x 2 + P 1 = y 2 + Q 1 for some y 2 Q 2. But as above, x 2 + P 1 = x 2 + Q 1,sox 2 y 2 Q 1, and therefore x 2 Q 2. It is a standard result of field theory that an embedding of a field F in an algebraically closed field can be extended to an algebraic extension of F. There is an analogous result for ring extensions Theorem Let S be integral over R, and let f be a ring homomorphism from R into an algebraically closed field C. Then f can be extended to a ring homomorphism g : S C. Proof. Let P be the kernel of f. Since f maps into a field, P is a prime ideal of R. By (2.2.2), there is a prime ideal Q of S such that Q R = P. By the factor theorem, f induces an injective ring homomorphism f : R/P C, which extends in the natural way to the fraction field K of R/P. Let L be the fraction field of S/Q. By (2.1.11), S/Q is integral over R/P, hence L is an algebraic extension of K. Since C is algebraically closed, f extends to a monomorphism g : L C. Ifp : S S/Q is the canonical epimorphism and g = g p, then g is the desired extension of f, because g extends f and f p R = f. In the next section, we will prove the companion result to (2.2.3), the going down theorem. There will be extra hypotheses, including the assumption that R is integrally closed. So it will be useful to get some practice with the idea of integral closure.

7 2.3. GOING DOWN Lemma Let R be a subring of S, and denote by R the integral closure of R in S. If T is a multiplicative subset of R, then (R) T is the integral closure of R T in S T. Proof. Since R is integral over R, it follows from (2.1.13) that (R) T is integral over R T. If α/t S T (α S, t T ) and α/t is integral over R T, we must show that α/t (R) T. There is an equation of the form ( α t )n +( a 1 t 1 )( α t )n a n t n =0 with a i R and t i,t T. Let t 0 = n i=1 t i, and multiply the equation by (tt 0 ) n to conclude that t 0 α is integral over R. Therefore t 0 α R, soα/t = t 0 α/t 0 t (R) T Corollary If T is a multiplicative subset of the integrally closed domain R, then R T is integrally closed. Proof. Apply (2.2.5) with R = R and S = K, the fraction field of R (and of R T ). Then R T is the integral closure of R T in S T. But S T = K, sor T is integrally closed. Additional results on localization and integral closure will be developed in the exercises. The following result will be useful. (The same result was proved in (1.5.1), but a slightly different proof is given here.) Proposition The following conditions are equivalent, for an arbitrary R-module M. (1) M =0; (2) M P = 0 for all prime ideals P of R; (3) M P = 0 for all maximal ideals P of R. Proof. It is immediate that (1) (2) (3). To prove that (3) (1), let m M. IfP is a maximal ideal of R, then m/1 is0inm P, so there exists r P R \ P such that r P m =0 in M. Let I(m) be the ideal generated by the r P. Then I(m) cannot be contained in any maximal ideal M, because r M / Mby construction. Thus I(m) must be R, and in particular, 1 I(m). Thus 1 can be written as a finite sum P a P r P where P is a maximal ideal of R and a P R. Consequently, m =1m = P a P r P m = Going Down We will prove a companion result to the going up theorem (2.2.3), but additional hypotheses will be needed and the analysis is more complicated.

8 8 CHAPTER 2. INTEGRAL EXTENSIONS Lemma Let S be integral over the subring R, with I an ideal of R. Then IS is the set of all s S satisfying an equation of integral dependence s m + r m 1 s m r 1 s + r 0 =0 with the r i I. Proof. If s satisfies such an equation, then s m IS,sos IS. Conversely, let s n IS, n 1, so that s n = k i=1 r is i for some r i I and s i S. Then S 1 = R[s 1,...,s k ] is a subring of S, and is also a finitely generated R-module by (2.1.5). Now s n S 1 = k r i s i S 1 i=1 k r i S 1 IS 1. Moreover, S 1 is a faithful R[s n ]-module, because an element that annihilates S 1 annihilates 1 and is therefore 0. By (2.1.2), s n, hence s, satisfies an equation of integral dependence with coefficients in I Lemma Let R be an integral domain with fraction field K, and assume that R is integrally closed. Let f and g be monic polynomials in K[x]. If fg R[x], then both f and g are in R[x]. Proof. In a splitting field containing K, we have f(x) = i (x a i) and g(x) = j (x b j). Since the a i and b j are roots of the monic polynomial fg R[x], they are integral over R. The coefficients of f and g are in K and are symmetric polynomials in the roots, hence are integral over R as well. But R is integrally closed, and the result follows Proposition Let S be integral over the subring R, where R is an integrally closed domain. Assume that no nonzero element of R is a zero-divisor of S. (This is automatic if S itself is an integral domain.) If s S, define a homomorphism h s : R[x] S by h s (f) =f(s); thus h s is just evaluation at s. Then the kernel I of h s is a principal ideal generated by a monic polynomial. Proof. IfK is the fraction field of R, then IK[x] is an ideal of the PID K[x], and IK[x] 0 because s is integral over R. (If this is unclear, see the argument in Step 1 below.) Thus IK[x] is generated by a monic polynomial f. Step 1 : f R[x]. By hypothesis, s is integral over R, so there is a monic polynomial h R[x] such that h(s) = 0. Then h I IK[x], hence h is a multiple of f, sayh = fg, with g monic in K[x]. Since R is integrally closed, we may invoke (2.3.2) to conclude that f and g belong to R[x]. Step 2 : f I. Since f IK[x], we may clear denominators to produce a nonzero element r R such that rf IR[x] =I. By definition of I we have rf(s) = 0, and by hypothesis, r is not a zero-divisor of S. Therefore f(s) =0,sof I. Step 3 : f generates I. Let q I IK[x]. Since f generates IK[x], we can take a common denominator and i=1

9 2.3. GOING DOWN 9 write q = q 1 f/r 1 with 0 r 1 R and q 1 R[x]. Thus r 1 q = q 1 f, and if we pass to residue classes in the polynomial ring (R/Rr 1 )[x], we have q 1 f = 0. Since f is monic, the leading coefficient of q 1 must be 0, which means that q 1 itself must be 0. Consequently, r 1 divides every coefficient of q 1,soq 1 /r 1 R[x]. Thus f divides q in R[x] Going Down Theorem Let the integral domain S be integral over the integrally closed domain R. Suppose we have a chain of prime ideals P 1 P n of R and a chain of prime ideals Q m Q n of S, with 1 <m n. If Q i lies over P i for i = m,..., n, then there are prime ideals Q 1,...,Q m 1 such that Q 1 Q m and Q i lies over P i for every i =1,...,n. Proof. By induction, it suffices to consider n = m = 2. Let T be the subset of S consisting of all products rt, r R \ P 1, t S \ Q 2. In checking that T is a multiplicative set, we must make sure that it does not contain 0. If rt = 0 for some r/ P 1 (hence r 0) and t/ Q 2, then the hypothesis that r is not a zero-divisor of S gives t = 0, which is a contradiction (because 0 Q 2 ). Note that R \ P 1 T (take t = 1), and S \ Q 2 T (take r = 1). First we prove the theorem under the assumption that T P 1 S =. Now P 1 S T is a proper ideal of S T, else 1 would belong to T P 1 S. Therefore P 1 S T is contained in a maximal ideal M. By basic localization theory, M corresponds to a prime ideal Q 1 of S that is disjoint from T. Explicitly, s Q 1 iff s/1 M. We refer to Q 1 as the contraction of M to S; it is the preimage of M under the canonical map s s/1. With the aid of the note at the end of the last paragraph, we have (R \ P 1 ) Q 1 =(S \ Q 2 ) Q 1 =. Thus Q 1 R P 1 and Q 1 = Q 1 S Q 2. We must show that P 1 Q 1 R. We do this by taking the contraction of both sides of the inclusion P 1 S T M. Since the contraction of P 1 S T to S is P 1 S, we have P 1 S Q 1,soP 1 (P 1 S) R Q 1 R, as desired. Finally, we show that T P 1 S is empty. If not, then by definition of T, T P 1 S contains an element rt with r R \ P 1 and t S \ Q 2. We apply (2.3.1), with I = P 1 and s replaced by rt, to produce a monic polynomial f(x) =x m + r m 1 x m r 1 x + r 0 with coefficients in P 1 such that f(rt) = 0. Define v(x) =r m x m + r m 1 r m 1 x m r 1 rx + r 0. Then v(x) R[x] and v(t) = 0. By (2.3.3), there is a monic polynomial g R[x] that generates the kernel of the evaluation map h t : R[x] S. Therefore v = ug for some u R[x]. Passing to residue classes in the polynomial ring (R/P 1 )[x], we have v = u g. Since r i P 1 for all i =0,...,m 1, we have v = r m x m. Since R/P 1 is an integral domain and g, hence g, is monic, we must have g = x j for some j with 0 j m. (Note that r/ P 1,sov is not the zero polynomial.) Consequently, g(x) =x j + a j 1 x j a 1 x + a 0 with a i P 1,i =0,...,j 1. But g ker h t,sog(t) = 0. By (2.3.1), t belongs to the radical of P 1 S, so for some positive integer l, we have t l P 1 S P 2 S Q 2 S = Q 2,so t Q 2. This contradicts our choice of t (recall that t S \ Q 2 ).

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