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Integral closures of Noetherian domains
Accepted Manuscript Integral closures of Noetherian domains
Raymond C. Heitmann
PII: DOI: Reference:
S0022-4049(16)30186-4 http://dx.doi.org/10.1016/j.jpaa.2016.10.014 JPAA 5564
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Journal of Pure and Applied Algebra
Received date: Revised date:
17 July 2016 30 September 2016
Please cite this article in press as: R.C. Heitmann, Integral closures of Noetherian domains, J. Pure Appl. Algebra (2016), http://dx.doi.org/10.1016/j.jpaa.2016.10.014
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INTEGRAL CLOSURES OF NOETHERIAN DOMAINS RAYMOND C HEITMANN I dedicate this article to Tony Geramita, who brought this question to my attention. Abstract. It has long been known that an integral domain A can be Noetherian while its integral closure A is not. However, when Heinzer asked whether A can have an infinitely generated maximal ideal, the question went unanswered. In this article, a local Noetherian domain A is constructed with the property that A has an infinitely generated maximal ideal. Moreover, the completion of A is not Noetherian. It had not previously been shown that this can happen.
1. Introduction From the earliest days of the study of Noetherian commutative rings, there was great interest in the Noetherian properties of the integral closures of Noetherian rings, particularly integral domains. Integral closures are more widely studied in the domain case and the present article is no exception. An understanding of the domain case allows us to easily understand the case where the total quotient ring of R is zero dimensional. Historically, it was quickly observed that whenever r, s ∈ R with r nilpotent and s regular, then r/s is integral over R. Hence, if the nilradical of R is non-trivial and R is not its own total quotient ring, the nilradical of the integral closure of R will be infinitely generated and R cannot be Noetherian. It is interesting to note, especially in regard to the questions we consider here, that any regular ideal of R will contain the entire nilradical and so to study the number of generators of regular ideals, we can work modulo the nilradical and assume R is reduced. Further, the integral closure of a reduced ring is simply the direct sum of the integral closures of R/Pi where Pi ranges over the minimal prime ideals of R and so we have reduced to the domain case. [If the total quotient ring is not zero dimensional, modding out the nilradical changes the total quotient ring and so also the integral closure. That case is clearly more complicated.] Returning to domains, most of the interesting questions about the Noetherian property of integral closures were settled by Krull [3], Akizuki [1], Mori [6], and Nagata [7] long ago. Consider the setting R ⊆ S ⊆ R . Krull and Akizuki observed that while R need not be a finite R-module even in the case of a one dimensional local domain, R and even S must be Noetherian in that case. Mori went on to show that R is Noetherian in the two dimensional case. Nagata then cut short the good news by producing examples of a two dimensional S and a three dimensional R which are not Noetherian. These examples also appear in Nagata’s book [8]. However, there are two intriguing questions as yet unanswered which shall be resolved in this article. The first was posed by Heinzer [4] in 1973. The second is Date: October 21, 2016. 1
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interesting in its own right, but I stumbled upon it as a natural outgrowth of the first. I am surprised that it has not been asked previously. Question (Heinzer). If R is a Noetherian integral domain with integral closure R , must it follow that maximal ideals in R are finitely generated? Question. If R is a local Noetherian domain, is the completion of the integral closure of R Noetherian? To more closely tie the questions together, I will first observe that Heinzer’s question is actually a local one—we can assume that both R and R have a single maximal ideal. To see this, suppose M is an infinitely generated maximal ideal of R . Then M contracts to a maximal ideal M of R and there are only finitely many maximal ideals of R which contract to M [8, 33.10]. So M will completely split in a finite integral extension of R which is of course Noetherian; this allows us to assume that M is the unique maximal ideal which contains M . Now, if T is a local generating set for M , i.e., T RM = M RM , and M = (r1 , r2 , . . . , rk )R, then T ∪ {r1 , r2 , . . . , rk } is a generating set for M . Thus, if M requires infinitely many generators, T must be an infinite set and so we may replace R by RM . Likewise, we can reduce the second question to the case where R has a single maximal ideal. Now, it is always true that ∩(M )n = (0). Hence, R is something which Nagata called a local ring which may not be Noetherian. Such a ring has a completion (R )∗ which contains it and that completion can be Noetherian even when R is not. In fact, a complete local ring which may not be Noetherian will actually be Noetherian if it has a finitely generated maximal ideal [8, 31.7]. Further, as (R )∗ will have a finitely generated maximal ideal whenever R does by [8, 17.5], a positive answer to Heinzer’s question would have guaranteed a positive answer to the second question as well. Below, in Corollary 4, we will see that the converse is also true. If R has an infinitely generated maximal ideal, then so does (R )∗ and so both questions will have negative answers. One reason why these questions have remained unanswered for so long ago is that they are quite difficult to resolve for those of us who owe much of our examplebuilding skills to an understanding of Nagata’s techniques. Nagata actually started with a Noetherian (R )∗ which he understood and built R as a subring of it. In this framework, R will itself be Noetherian if and only if the maximal ideal is finitely generated and finitely generated ideals are closed in the M -adic topology [8, (31.8)]. Nagata found a relatively easy way to put elements into the closure of a finitely generated ideal not primary to the maximal ideal without actually putting them into the ideal. That method cannot resolve the questions here. Section 2 is a brief general treatment of completions of integral closures. In Theorem 2, we see that the M -adic topology is the same as the M R -adic topology on R . This leads to the corollary that our two questions are equivalent. The meat of the article, Section 3, is essentially just the presentation of a single example which answers both questions. The sections can be read independently. Throughout, all rings are commutative with identity. The term local ring will refer to a Noetherian ring with a single maximal ideal. Without the Noetherian assumption, such rings will be called quasi-local. Following Nagata’s lead, we will call quasi-local rings satisfying ∩M n = (0) local rings which may not be Noetherian.
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2. Completions If R is a local ring, the completion of R has been thoroughly studied. However, when R is not Noetherian, the completion will not be a faithfully flat extension and has deservedly attracted much less interest. However, a certain understanding of the properties of (R )∗ is needed to put our example in its proper context. There is a brief treatment of this subject in Nagata’s book [8] and presumably Nagata knew even more than what appears in print. Quite possibly, he was aware of all of the results in Section 2. In truth, none of the Section 2 results are surprising and the proofs are not that difficult. In this section, we will routinely assume that R is a local ring and its integral closure (in the total quotient ring) is quasi-local. As noted in the introduction, the assumption that R has a single maximal ideal is harmless because there are only finitely many maximal ideals of R which contract to the maximal ideal of R and so we may replace R by a localization of a finite extension and produce a setting in which the maximal ideal does not split. We begin with a result familiar to some which is proved here for completeness. Proposition 1. Let R be a local Noetherian domain and suppose that its integral closure R has a unique maximal ideal M . It follows that ∩(M )n = (0), so the local ring R may not be Noetherian. Proof. Since R is Noetherian, it is contained in a discrete valuation ring V which is centered on the maximal ideal. If m is the maximal ideal of V , we clearly have ∩mn = (0) since the intersection can only contain elements of infinite value. As R ⊆ V and M ⊆ m, the result follows. The domain assumption is critical for Proposition 1. If R has nilpotent elements, the intersection will contain the nilradical of R . In this setting, we can still complete R , but here, because the M -adic topology is not T0 , R ⊆ (R )∗ and so the completion contains less information about R and so is of less interest. Next we prove the central result of this section, that the M -adic topology on R is just the M R -adic topology. As the integral domain hypothesis is not useful in the proof, we state the result in greater generality. Theorem 2. Let R be a local ring with maximal ideal M and suppose its integral closure R is a quasi-local ring with maximal ideal M . Then there exists a positive integer m such that (M )m ⊂ M R . Thus the ideals M R and M induce the same topology on R . Proof. Let R∗ be the M -adic completion of R and let N be the nilradical of R∗ . Because R∗ /N is a reduced complete local ring, it follows from [8, 32.1] that its integral closure is a finite module over it and we have a natural sequence of Noetherian rings R ⊆ R∗ −→ R∗ /N ⊆ (R∗ /N ) . We also have R ⊆ (R∗ ) −→ (R∗ /N ) . These last maps allow us to view R∗ [R ] as a subring of (R∗ ) and to define (R∗ /N )◦ as the image of the map R∗ [R ] −→ (R∗ /N ) . Of course, (R∗ /N )◦ is also a finite R∗ /N -module and so is also Noetherian. We now have the following commutative diagram: R R
⊆
⊆
/ R∗ / R∗ [R ]
/ / R∗ /N π
/ / (R∗ /N )◦
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Next we make several somewhat unrelated observations about our setting. First, since (R∗ /N )◦ is a finite R∗ -module, there exists a finite subset Δ ⊂ R such that π(R∗ [Δ]) = (R∗ /N )◦ . Next, let J be the ideal generated by π(M ). J is finitely generated and so there exist δ1 , . . . , δk ∈ M such that J = (π(δ1 ), . . . , π(δk ))(R∗ /N )◦ . If we adjoin δ1 , . . . , δk to Δ, the set is still finite and so, without loss of generality, we may assume that δ1 , . . . , δk ∈ Δ. Finally, as the kernel of (R∗ ) −→ (R∗ /N ) is just the nilradical of (R∗ ) , the kernel of π is just the intersection of R∗ [R ] with the nilradical of (R∗ ) . As R∗ is Noetherian, there exists an integer such that N = (0) and since the numerator of each element in the nilradical of (R∗ ) is in N , we get (Ker π) = (0). Now, (M ) is generated by elements of the form q1 · · · q with each qi ∈ M . We consider an element q1 · · · q . As π(qi ) ∈ J and (R∗ /N )◦ = π(R∗ [Δ]), there exist k elements λij ∈ R∗ [Δ] such that π(qi − j=1 λij δj ) = 0. It follows that k λij δj = 0 , qi − i=1
j=1
from which we see that q1 · · · q ∈ (δ1 , . . . , δk )R∗ [Δ ∪ {q1 , . . . , q }]. As the ring R∗ [Δ] is Noetherian and the ideal M R∗ [Δ] is primary to the maximal ideal, we t ∗ [Δ] ⊆ M R∗ [Δ] for some positive integer t. Let m = t. Then, have m(δ1 , . . . , δk ) R m if i=1 qi ∈ (M ) , we see that m
qi ∈ (δ1 , . . . , δk )t R∗ [Δ ∪ {q1 , . . . , qm }] ⊆ M R∗ [Δ ∪ {q1 , . . . , qm }] .
i=1
As Δ ∪ {q1 , . . . , qm } is a finite set, R[Δ ∪ {q1 , . . . , qm }] is Noetherian and so its completion R∗ [Δ ∪ {q1 , . . . , qm }] is a faithfully flat extension. Thus, we have m m i=1 qi ∈ M R[Δ ∪ {q1 , . . . , qm }] ⊆ M R . It follows that (M ) ⊂ M R . Now we use Theorem 2 to show that our two questions are equivalent. Corollary 3. Let R be a local ring with maximal ideal M and suppose its integral closure R is a quasi-local ring. Let I be an ideal of R which is primary to the maximal ideal. Then I(R )∗ ∩ R = I. ui ai with each ui ∈ (R )∗ and each ai ∈ I. Proof. Suppose b ∈ R and b = Adding terms 0 · ak to our sum if necessary, we can presume the ideal (a1 , . . . , an )R is primary to the maximal ideal and so reduce to the case I = (a1 , . . . , an )R . As I and M R are both finitely generated ideals with the same radical, they induce the same adic topology. Hence, by Theorem 2, the I-adic topology on R is the same as the M -adic topology and so (R )∗ is the completion in the I-adic topology. Now each ui is the limit of a Cauchy sequence {uij } in R . It follows that n uij ai b− i=1 is a Cauchy n sequence in R which converges to zero. For sufficiently large j, we have b − i=1 uij ai ∈ I and so b ∈ I as desired.
Corollary 4. Let R be a local ring with maximal ideal M and suppose its integral closure R is a quasi-local ring. Then the maximal ideal M of R is finitely generated if and only if (R )∗ is Noetherian.
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Proof. The forward direction is due to Nagata and has already been noted. For the converse, suppose (R )∗ is Noetherian. Then M (R )∗ is finitely generated and so equals I(R )∗ for some finitely generated ideal I of R . By Corollary 3, M = I and so M is finitely generated. Finally, as it is convenient to do so at this point, we describe (R )∗ . Corollary 5. Let R be a local domain with maximal ideal (a1 , . . . , an )R and suppose its integral closure R is a quasi-local ring. Then (R )∗ = R [[X1 , . . . , Xn ]]/I where I is the closure of (X1 − a1 , . . . , Xn − an ) in the (a1 , . . . , an )-adic topology. Proof. Since the topology on R is just the (a1 , . . . , an )-adic topology, this follows immediately from [8, 17.5]. 3. The Example We begin with a proposition whose proof is due to Nagata. The form we need is slightly more general than that which Nagata considered, but the alterations have no real effect on the actual proof. The original example is bad Noetherian ring example 3 [8, p. 206]. Proposition 6. Let K be a field of characteristic p and assume L is a field such that K p ⊆ L ⊆ K. If R = ((K p [[x1 , . . . , xn ]][L])[[z]])[K], then R is a regular local ring. Proof. Since K[[x1 , . . . , xn , z]] is integral over R, we easily see that R has dimension n + 1. Also clearly, its maximal ideal is generated by x1 , . . . , xn , z. Thus, to prove it is a regular local ring, it suffices to prove that R is Noetherian. To prove this, we first adjust Nagata’s proof to show that K p [[x1 , . . . , xn ]][L] is Noetherian, then observe that (K p [[x1 , . . . , xn ]][L])[[z]] is Noetherian because it is a power series ring over a Noetherian ring, and finally use Nagata’s ideas again to complete the proof. First note that the two rings we must show are Noetherian are quasi-local subrings of their respective completions L[[x1 , . . . , xn ]] and K[[x1 , . . . , xn , z]] and have the induced topologies. If M denotes the maximal ideal of either ring, then we have n ∩∞ i=1 M = (0) and so these rings are local rings which may not be Noetherian. Such rings are Noetherian provided they have a finitely generated maximal ideal (which both do) and have the property that finitely generated ideals are closed [8, (31.8)]. So let a1 , . . . , ad generate a finitely generated ideal of K p [[x1 , . . . , xn ]][L] and suppose b ∈ (a1 , . . . , ad )L[[x1 , . . . , xn ]] ∩ K p [[x1 , . . . , xn ]][L]. Then the coefficients
of a1 , . . . , an , b generate a finite extension K of K p . Now note that L = K B is a direct sum of K - vector spaces. (B is not unique of course, but any choice for this summand will suffice.) Now b ∈ (a1 , . . . , ad )L[[x1 , . . . , xn ]] gives us an equation b = a1 f1 + · · · + ad fd . Using our direct sum, we can project the coefficients that occur in each fi to K and so obtain f¯1 , . . . , f¯d ∈ K [[x1 , . . . , xn ]] such that b = a1 f¯1 + · · · + an f¯n . Thus b ∈ (a1 , . . . , ad )K [[x1 , . . . , xn ]] ⊆ (a1 , . . . , ad )K p [x1 , . . . , xn ]][L] and we see that finitely generated ideals are closed. Showing that b ∈ (a1 , . . . , ad )K[[x1 , . . . , xn , z]] ∩ ((K p [[x1 , . . . , xn ]][L])[[z]])[K] implies b ∈ (a1 , . . . , ad )((K p [[x1 , . . . , xn ]][L])[[z]])[K] is similar though requires several steps. Exactly as before, the coefficients of a1 , . . . , ad , b generate a finite extension L of L and we again see that b ∈ (a1 , . . . , ad )L [[x1 , . . . , xn , z]]. Further,
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as K p [[x1 , . . . , xn ]][L ][[z]] is a Noetherian ring with completion L [[x1 , . . . , xn , z]], this gives b ∈ (a1 , . . . , ad )(K p [[x1 , . . . , xn ]][L ])[[z]]. As K p [[x1 , . . . , xn ]][L ][[z]] = K p [[x1 , . . . , xn ]][L][[z]][L ] ⊆ (K p [[x1 , . . . , xn ]][L])[[z]][K],
the desired result follows.
Throughout, we let K be a fixed field of characteristic p with the property that [K : K p ] is infinite. Let {bi , gijk , | i, j, k ∈ Z≥0 } ⊂ K be a set of p-independent elements over K p . Let L = K p ({gijk , | i, j, k ∈ Z≥0 }). Let R = K p [[x, y]][L][[z]][K]. By Proposition 6, R is a regular local ring. Let {ai | i ∈ Z+ } be a set of distinct elements in K p . Then, setting wi = x + ai z for each i, we have an infinite set of distinct prime elements in R. ∞ Notation 7. Let c = c0 = i=0 bi y i and, for i > 0, let i−1 ∞ ci = c − bj y j /y i = bj y j−i . j=0
Note ci = bi + yci+1 . Let r =
∞ ∞ ∞ i=0
j=0
j=i
k=0 gijk x
i j k
y z .
Two additional bits of notational shorthand will make our presentation much more readable. n Notation 8. Let tn = i=1 wi . So t0 = 1 and tn+1 = tn wn+1 . Let n−1 n σn = wj y i = w2 w3 · · · wn + w3 w4 · · · wn y + · · · + wn y n−2 + y n−1 . i=0
j=i+2
Of course, σ1 = 1, σ2 = w2 + y, and σn+1 = wn+1 σn + y n . Lemma 9. With the notation as above, there exist sets {αi | i ∈ Z≥0 }, {ri | i ∈ Z≥0 } ⊂ L[[x, y, z]] such that if si = ti ci + zαi , then for every i ≥ 0, wi+1 si − ysi+1 = ti+1 bi + zri and r − ri ∈ L[[z]][x] ⊂ R. Moreover, we have wi+1 αi − yαi+1 = ri . Proof. First, we observe that wi+1 si − ysi+1 = wi+1 (ti ci + zαi ) − y(ti+1 ci+1 + zαi+1 ) = ti+1 (ci − yci+1 ) + z(wi+1 αi − yαi+1 ) = ti+1 bi + z(wi+1 αi − yαi+1 ) , so the first part will follow provided we choose αi , ri to satisfy wi+1 αi − yαi+1 = ri . Of course, the quantities on both sides of this equation are power series and so it is natural to verify it by checking that the corresponding coefficients are equal. However, we will do something slightly different. As wi+1 , y are homogeneous in both the y-grading and the (x, z)-grading, we will group our terms using this bigrading. To achieve this, we introduce some slightly unusual notation. For each , rijk of degree k in L[x, z] i, j, k ≥ 0, we intend choose homogeneous forms to ∞αijk ∞ ∞ ∞ and then set αi = j=0 k=0 αijk y j and ri = j=0 k=0 rijk y j . Adopting the convention that rijk = αijk = 0 whenever either j or k is negative, the equation wi+1 αi − yαi+1 = ri is equivalent to wi+1 αi,j,k−1 − αi+1,j−1,k = rijk for all j, k. We shall later refer to this as the crucial equation. It remains to choose our elements rijk , αijk to satisfy this equation and the second conclusion of the lemma. We will
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actually satisfy the second conclusion by choosing rijk so that r − ri is a polynomial in L[[z]][x] of x-degree at most i. The plan of the proof is to define the elements as needed. We will choose our elements in the following order. We shall first define rijk for j > 0. Next we define αijk for i > 0. Then we define ri0k and finally α0jk . ∞ ∞ If we set hjk = d+e=k gdje xd z e , we may write r = j=0 k=0 hjk y j . Now, for j > 0, set rijk = hjk . This ensures that r −ri ∈ L[[x, z]]. For the second conclusion, it only remains to bound the degree of x. Next we develop a recursive definition for the alphas which allows us to find each αijk with i > 0 in terms of known quantities and the as yet unspecified α0jk . Set αi+1,j−1,k = wi+1 αi,j,k−1 − rijk whenever i, k ≥ 0 and j > 0. This equation is also trivially true when k < 0, a fact which will simplify our calculations later on. This guarantees that the crucial equation will hold whenever j > 0. In the j = 0 case, the desired equation reduces to wi+1 αi,0,k−1 = ri0k and we may define ri0k to force this equation to hold as well. Like {αijk | i > 0}, the set {ri0k } will depend upon our choice of {α 0jk }. ∞ It only remains to bound the degree of r − ri . Since r − ri = k=0 h0k − ri0k , to bound the x-degree of r − ri , we need to choose {α0jk } so that the x-degree of h0k − ri0k is at most i. Now we use our definitions to recursively calculate αi,0,k−1 = −(ri−1,1,k−1 + wi ri−2,2,k−2 + wi wi−1 ri−3,3,k−3 + · · · + wi · · · w2 r0,i,k−i ) +wi · · · w1 α0,i,k−i−1 for i > 0 and so note ri0k = wi+1 αi,0,k−1 = −(wi+1 ri−1,1,k−1 + wi+1 wi ri−2,2,k−2 + wi+1 wi wi−1 ri−3,3,k−3 + · · · +wi+1 · · · w2 r0,i,k−i ) + wi+1 · · · w1 α0,i,k−i−1 . We can write this in the simplified form ri0k = τik + ti+1 α0,i,k−i−1 where τik is a specific homogeneous polynomial in L[x, z] of degree k. Thus, we have h0k − ri0k = h0k − τik − ti+1 α0,i,k−i−1 for i > 0. We note that this equation is also true for i = 0 since r00k = w1 α0,0,k−1 (here τik = 0). There are two cases, k ≤ i or k > i. In the first case, α0,i,k−i−1 = 0 by definition, but clearly h0k − ri0k is a form of degree k ≤ i and so necessarily has x-degree ≤ i. In the second case, we define α0,i,k−i−1 to achieve our goal. Note that each α0im occurs in precisely one expression and so we will not define the same element twice. View h0k − τik and ti+1 as polynomials in x, and note that the latter is monic of degree i + 1. Perform long division to obtain h0k − τik = δik ti+1 + fik where fik is a polynomial in x of degree at most i. Then set α0,i,k−i−1 = δik and we get h0k − ri0k = fik , completing the proof. Theorem 10. With the notation as above and s = s0 , let A = R[r, s]. Then A , the integral closure of A, is a non-Noetherian integral domain of dimension three which has an infinitely generated maximal ideal. Moreover, the completion of A also has an infinitely generated maximal ideal and so is also not Noetherian. The rest of this article will be devoted to proving the first conclusion of this theorem, which will follow immediately from Theorem 17. The moreover statement then is a consequence of Corollary 4. We need to show that the maximal ideal of A
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is infinitely generated. A logical way to proceed would be to compute A and then we would expect the final steps to be easy. Unfortunately, except for the p = 2 case, the computation is messy and not very useful. As we want to show that the example is valid for arbitrary p, an alternate approach is needed. If M denotes the maximal ideal of A , then clearly the pth power of any element in M is in (xp , y p , z p )R and it follows that A /(x, y, z)A is a zero dimensional ring with a single prime ideal. Since a non-Noetherian ring necessarily has an infinitely generated prime ideal [2], it will suffice to prove that A /(x, y, z)A is not Noetherian. Our goal then will be to understand A well enough to find an infinite set of elements in A that, together with x, y, z, generate an ideal that is not finitely generated. Remark 11. As K[[x, y, z]] is integral over R, if S is any ring betwen R and K[[x, y, z]], notice that the integral closure S will just be the intersection of the quotient field with K[[x, y, z]]. Moreover, since rp , sp ∈ R, any element in the p−1 p−1 integral closure of A necessarily has the form ( i=0 j=0 uij ri sj )/v with v ∈ R and each uij ∈ R. Lemma 12. Let K be a field of characteristic p such that [K : K p ] = ∞. Suppose J is a field, K p ⊆ J ⊂ K, and e1 , e2 , . . . ∈ K are such that the infinite set {ei } is a minimal generating set for K over J. Let S be a Noetherian integrally closed ∞ domain, K p [[X1 , . . . , Xn ]][K][⊆ S ⊆ J[[X1 , . . . , Xn ]][K], and let τ = i=1 ei X1i +h with h ∈ J[[X1 , . . . , Xn ]][K]. Let T = S[τ ]. If γ ∈ T , X1k γ ∈ T for some k. Proof. If the lemma is false, we can choose a counterexample γ ∈ / T such that the conductor of γ is a prime ideal which does not contain X1 . Thus we may assume there is an element u ∈ S such that uγ ∈ T and X1 , u is a regular sequence. As noted in Remark 11, γ ∈ K[[X1 , . . . , Xn ]] and of course uγ ∈ T implies that we may write uγ = vd τ d + · · · + v0 with each vi ∈ S and 0 < d < p. For any i, we may adjoin all ej ’s but one to J and produce a field Ki such that K = Ki [ei ]. This makes K a p-dimensional vector space over Ki and of course K[[X1 , . . . , Xn ]] is a rank p free module over Ki [[X1 , . . . , Xn ]]. The elements u, vd , . . . , v0 , h are in J[[X1 , . . . , Xn ]][K] and so can only involve finitely many ej . Thus we may choose i so that u, vd , . . . , v0 , h ∈ Ki [[X1 , . . . , Xn ]]. Now we consider the equation vd τ d + · · · + v0 · 1 = uγ in our rank p free module, noting the coefficients are all in the ground ring. The projection of the left hand side on the edi coordinate is just vd X1id , while the projection of the right hand side is uγd where γd is the projection of γ on the edi coordinate. However this forces u to divide vd X1id in Ki [[X1 , . . . , Xn ]] and so also in the integrally closed S. As X1 , u is a regular sequence, we see that vd ∈ uS and so we may reduce to the case vd = 0. By inducting on d, we see that the lemma holds. Lemma 13. (1) R[r] is integrally closed. (2) If v ∈ A , then y n v ∈ A for some n. (3) if (t, y)A is a height two ideal, v ∈ A, and tv ∈ y n A , then v ∈ y n A . Proof. To prove will apply Lemma 12 twice to S = K p [[x, y]][K] ∼ = R/zR ∞(1),we ∞ with τ = r¯ = i=0 j=0 gij0 xi y j . In the first application, we let X1 = x, X2 = y, ei = gi00 , and J be an intermediate field which contains gij0 for j > 0. We then conclude that γ ∈ T implies xk γ ∈ T . In the second application, we let X1 = y, X2 = x, ei = g0i0 , and J be an intermediate field which contains gij0 for i > 0.
INTEGRAL CLOSURES OF NOETHERIAN DOMAINS
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We then conclude that γ ∈ T implies y k γ ∈ T . Since S is Cohen-Macaulay and so the free S-module T is also Cohen-Macaulay, either γ ∈ T or the conductor of γ is contained in a height one prime ideal. As no height one prime ideal contains both x and y, T is integrally closed. Next, we note that R/zR ⊂ (R/zR)[¯ r] is an integral extension of degree p and so zR[r] is a prime ideal. Moreover, R[r]/zR[r] is isomorphic to T , which is integrally closed. Since normality lifts (see for example [5, p. 439]), R[r] is integrally closed. Statement (2) follows from Lemma 12, using S = R[r], X1 = y, X2 = x, X3 = z, τ = s, and ei = bi . For statement (3), let tv = y n γ with γ ∈ A . Then, we have γ ∈ K[[x, y, z]] and we see that in that unique factorization domain, v = y n δ. But then δ ∈ A by Remark 11 and the statement is proved. n In light of this lemma, A = ∪∞ n=1 An where An = (A :A y ). Fix n. Since tn is a monic polynomial of degree n in x, there is a unique polynomial μ ∈ L[[z]][x, y] with degx μ < n, degy μ < n such that r − μ ∈ (y n , tn )L[[x, y, z]]. Let rˆ = r − μ and n−1 sˆ = y n cn + zα0 . Since sˆ − s = − j=0 bj y j ∈ K[y] ⊂ R and μ ∈ L[[z]][x, y] ⊂ R, A = R[ˆ r, sˆ]. Strictly speaking, we should perhaps use the notation rˆn and sˆn as these elements depend on n. However, we shall only these terms in a setting where n is fixed and no confusion should result. Accordingly, we will avoid the more cumbersome notation.
Lemma 14. There exist polynomials p1 , p2 ∈ L[[z]][x, y] such that tn sˆ = y n sn + zˆ rσn + tn zp1 + y n zp2 . Proof. By the definitions, tn sˆ = tn (y n cn + zα0 ) = y n (sn − zαn ) + tn zα0 = y n sn + z(tn α0 − y n αn ) and so it only remains to show that we can solve tn α0 − y n αn − rˆσn = tn p1 + y n p2 , i.e., show tn α0 − y n αn − rˆσn ∈ (tn , y n )L[[z]][x, y]. Now this element is clearly in (tn , y n )L[[x, y, z]] and, as (tn , y n )L[[x, y, z]] ∩ L[[z]][x, y] = (tn , y n )L[[z]][x, y], it is enough to show tn α0 − y n αn − rˆσn ∈ L[[z]][x, y]. We verify this by proving tk α0 − y k αk − rˆσk ∈ L[[z]][x, y] for 1 ≤ k ≤ n by induction on k. For k = 1, t1 α0 − yα1 − rˆσ1 = r0 − rˆ by Lemma 9, and since r0 − rˆ = (r0 − r) + μ ∈ L[[z]][x, y], the k = 1 case is proved. For the induction step, tk+1 α0 − y k+1 αk+1 − rˆσk+1 = wk+1 tk α0 − y k+1 αk+1 − rˆ(wk+1 σk + y k ) = wk+1 (tk α0 − y k αk − rˆσk ) + y k (wk+1 αk − yαk+1 ) − rˆy k = wk+1 (tk α0 − y k αk − rˆσk ) + y k (rk − rˆ) ∈ L[[z]][x, y] .
Proposition 15. (1) Each element of An has the form q/y n where q = B0 +B1 sˆ+· · ·+Bp−1 sˆp−1 with each Bm ∈ R[ˆ r]. r] (2) B0 ∈ (z, y n )R[ˆ r] = ((x, y)n , rˆ, z)R[ˆ r] (3) B1 ∈ ({y n−i ti | 0 ≤ i ≤ n}, rˆ, z)R[ˆ
10
RAYMOND C HEITMANN
Proof. The first statement has already been observed. By Lemma 13(3), an element n of this form is in An if and only if tp−1 n q/y is integral. We compute: p−1 p−2 tp−1 ˆ) + · · · + Bp−1 (tn sˆ)p−1 n q = tn B0 + tn B1 (tn s p−2 n = tp−1 rσn + tn zp1 + y n zp2 ) + · · · n B0 + tn B1 (y sn + zˆ
+Bp−1 (y n sn + zˆ rσn + tn zp1 + y n zp2 )p−1 We want to know for which values of Bm this quantity is divisible by y n in A . We can ignore any terms which are divisible by y n and, since sn ∈ A , this means we can ignore all terms involving either y n sn or y n p2 . Thus it suffices to consider the p−2 rσn + tn p1 ) + · · · + Bp−1 z p−1 (ˆ rσn + tn p1 )p−1 ∈ R[ˆ r] and element tp−1 n B0 + tn B1 z(ˆ now we can take advantage of the fact that for such elements, divisibility by y n in r] are all equivalent. It is clear that this element can only be A , K[[x, y, z]], and R[ˆ n r] when tp−1 r], and since tn , z, y n is a regular sequence, we in y n R[ˆ n B0 ∈ (z, y )R[ˆ get the desired B0 ∈ (z, y n )R[ˆ r] and so part (2) is proved. Expanding the powers of rˆσn +tn p1 and regrouping, our element can be rewritten as the sum p−1 p−1 tnp−1−k rˆk σnk B z p−k 1 k k=0
=k
Noting that the k = 0 term is divisible by tp−1 and that each term for k ≥ 2 is n divisible by rˆ2 , the element can be divisible by y n only if we have tp−2 ˆσn n r
p−1
B z p−1 ∈ (y n , tp−1 ˆ2 )R[ˆ r] . n ,r 1
=1
p−1 Next, for every , we write B = j=0 Bj rˆj with each Bj ∈ R. Then we see p−1 ˆσn =1 B0 z p−1 ∈ (y n , tp−1 ˆ2 )R[ˆ r]. As R[ˆ r] is a free R-module and tp−2 n r n ,r 1 p np p rˆ ∈ (y , tn )R, this reduces to a condition in R, namely tp−2 n σn
p−1
B0 z p−1 ∈ (y n , tp−1 n )R . 1
=1
p−1 Further, as y, tn is a regular sequence, this implies σn =1 B0 z p−1 ∈ (y n , tn )R. 1 , z is a regular sequence, we may divide the element by z. Now set Similarly, as y, t n p−1 v = =1 B0 (zp1 )−1 . Then our condition becomes simply σn v ∈ (y n , tn )R. We now claim that v ∈ ({y n−i ti | 0 ≤ i ≤ n})R. We prove the claim by using induction to show that for 1 ≤ d ≤ n, σd v ∈ (y d , td )R implies v ∈ ({y d−i ti | 0 ≤ i ≤ d})R. The d = 1 case is obvious since σ1 = 1. Certainly (y d , td )R ⊂ (y k , wk )R for each k ≤ d, so we have σd v ∈ (y k , wk )R. d−1 d d Recall σd = i=0 ( j=i+2 wj )y i . Now, we see that ( j=i+2 wj )y i ∈ (y k , wk )R if d k ≥ i + 2 or i ≥ k, i.e., if i = k − 1. Thus ( j=k+1 wj )y k−1 v ∈ (y k , wk )R. As y, wk , wj is regular for j = k, this gives v ∈ (wk , y)R. Using the regularity of the sequence, y, wi , j=i wj , this leads to v ∈ (td , y)R. We write v = td h1 + yh2 . As td h1 ∈ {y d−i ti | 0 ≤ i ≤ d}R and also satisfies σd td h1 ∈ (y d , td )R, we may reduce to the case v = yh2 . As σd = wd σd−1 + y d−1 , (wd σd−1 + y d−1 )yh2 ∈ (y d , td )R gives σd−1 h2 ∈ (y d−1 , td−1 )R. The induction assumption implies that h2 ∈ ({y d−1−i ti | 0 ≤ i ≤ d − 1})R , so the claim follows.
INTEGRAL CLOSURES OF NOETHERIAN DOMAINS
Recalling that v =
p−1 =1
11
B0 (zp1 )−1 , we see that
B10 ∈ ({y n−i ti | 0 ≤ i ≤ n}, z)R r] as desired. Finally, as wi − x ∈ zR and and so B1 ∈ ({y n−i ti | 0 ≤ i ≤ n}, z, rˆ)R[ˆ so ti − xi ∈ zR, the last equation in conclusion (3) holds. Remark 16. Proposition 15 givesus all of the information we need to compute p−1 A in the case p = 2. Here v = =1 B0 (zp1 )−1 just means v = B10 and so n−i B1 ∈ ({y ti | 0 ≤ i ≤ n}, rˆ)R[ˆ r]. So An is generated over A by xs1 , x2 s2 , . . . , xn sn plus one additional element (βz + rˆsˆ)/y n . Hence An /zAn will be generated by xc1 , x2 c2 , . . . , xn cn , xn cn ρ for some ρ ∈ K[[x, y]]. Thus A /zA bears a strong resemblance to the example due to Kaplansky which appears at the end of [4]. On the other hand, B2 is not necessarily in the ideal ((x, y)n , rˆ, z)R[ˆ r] and so A will be larger for p > 2. I suspect that A /zA contains cki for all i and all k ≥ 2 but I have not actually verifed this as it is not needed for the proof. Theorem 17. The elements x, y, z, s − b0 , s1 , s2 , s3 , . . . generate an infinitely generated ideal of A . Proof. We know that ri , s0 ∈ A. Thus the relation wi+1 si − ysi+1 = ti+1 bi + zri from Lemma 9 tells us that each si is in the quotient field of A and so in A . Hence the listed elements are in A and it only remains to show that the ideal is infinitely generated. Suppose not. Then for all sufficiently large k, we have sk ∈ (x, y, z, s0 − b0 , s1 , . . . , sk−1 )A . As A = ∪An , for sufficiently large n, we have sk ∈ (x, y, z, s0 − b0 , s1 , . . . , sk−1 )An . We now will use Proposition 15 and the notation which accompanies it. n−1 n−1 Recall sˆ = y n cn + zα0 = s0 − i=0 bi y i . It follows that s0 − b0 = sˆ + i=1 bi y i and so (y, s0 −b0 )An = (y, sˆ)An . Thus we may replace s0 −b0 by sˆ in our generating n−1 set. Now assume i > 0. Combining ci = j=i bj y j−i + y n−i cn , si = ti ci + zαi , n−1 and sˆ = y n cn + zα0 , we see that ti sˆ − y i si = z(ti α0 − y i αi ) − ti j=i bj y j . By a r]. simple induction argument using Lemma 9, we see that δi = −(ti α0 − y i αi ) ∈ R[ˆ r]. As ti ∈ (x, z)A , we Consequently si = (zδi + ti sˆ)/y i + ti ρi for some δi , ρi ∈ R[ˆ may discard the second term and replace si by the first. More conveniently, using a common denominator, we use s˜i = (zβi + sˆy k−i ti )/y k where βi = y k−i δi . The finitely generated assumption then gives us an equation with each element written in the format of Proposition 15. (zβk + sˆtk )/y k = x(E0 + E1 sˆ + · · · + Ep−1 sˆp−1 )/y n + y(F0 + F1 sˆ + · · · + Fp−1 sˆp−1 )/y n +z(G0 + G1 sˆ + · · · + Gp−1 sˆp−1 )/y n + sˆ(H0 + H1 sˆ + · · · + Hp−1 sˆp−1 )/y n +
k−1
((γi0 + γi1 sˆ + · · · + γi,p−1 sˆp−1 )/y n )((zβi + sˆy k−i ti )/y k )
i=1
As R[ˆ r, sˆ] is free over R[ˆ r], we can regard the coefficients of 1, sˆ, sˆ2 , . . . , sˆp−1 separately and obtain p distinct equations in R[ˆ r]. We actually only need the equation related to the coefficient of sˆ to derive our contradiction. That equation k−1 is tk /y k = xE1 /y n + yF1 /y n + zG1 /y n + H0 /y n + i=1 (γi0 y k−i ti + γi1 zβi )/y n+k . Multiplying through by y n+k , we get y n tk = xy k E1 + y k+1 F1 + zy k G1 + y k H0 + k−1 k−i ti + γi1 zβi ). i=1 (γi0 y
12
RAYMOND C HEITMANN
We may use this equation in R[ˆ r] to induce a congruence modulo r] + (z, rˆ)R[ˆ r] . I = (x, y)n+k+1 R[ˆ k−1 n k k k+1 k F1 + y H0 + i=1 γi0 y k−i xi . Finally, As wj ≡ x, we have y x ≡ xy E1 + y r] and H0 , γi0 ∈ (z, y n )R[ˆ r]. Thus by Proposition 15, E1 , F1 ∈ (z, rˆ, (x, y)n )R[ˆ k k+1 n n F1 ≡ 0 and H0 ≡ y H0 , γi0 ≡ y γi0 . This gives xy E1 ≡ y y n xk ≡ y n+k H0 +
k−1
k+n−i i γi0 y x .
i=1
As k > 0, this gives y n xk ∈ (xk+1 , y n+1 , z, rˆ)R[ˆ r] ∩ R = (xk+1 , y n+1 , z, rˆp )R and p np np p since rˆ ∈ (x , y , z )R, we have our desired contradiction. Remark 18. We note that (A )∗ contains the element xs1 + x2 s2 + x3 s3 + · · · and so it is not contained in the integral closure of A∗ . References [1] Y. Akizuki Einige bemerkunge u ¨ber prim¨ are integrit¨ atsbereiche mit teilerkettensatz, Proc. Phys.-Math. Soc. Japan, 3rd Series 17 (1935), 327–336. [2] I. S. Cohen Commutative rings with restricted minimum condition, Duke Math. J. 17 (1950), 27–42. [3] W. Krull Ein satz u ¨ber prim¨ are integrit¨ atsberereiche, Math. Ann. 103 (1930), 450–465. [4] W. Heinzer, Minimal primes of ideals and integral ring extensions, Proc. AMS 40 (1973), 370–372. [5] R. Heitmann, Lifting Seminormality, Michigan Math. J. 57 (2008), 439–445. [6] Y. Mori On the integral closure of an integral domain, II, Bull. Kyoto Gakugei Univ. B7 (1955), 19–30. [7] M. Nagata, Note on integral closures of Noetherian domains, Mem. Coll. Sci., Univ. Kyoto 28 (1953-54), 121–124. [8] M. Nagata, Local Rings, Krieger, Huntington, NY, 1975.
Raymond C. Heitmann
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INTEGRAL CLOSURES OF NOETHERIAN DOMAINS RAYMOND C HEITMANN I dedicate this article to Tony Geramita, who brought this question to my attention. Abstract. It has long been known that an integral domain A can be Noetherian while its integral closure A is not. However, when Heinzer asked whether A can have an infinitely generated maximal ideal, the question went unanswered. In this article, a local Noetherian domain A is constructed with the property that A has an infinitely generated maximal ideal. Moreover, the completion of A is not Noetherian. It had not previously been shown that this can happen.
1. Introduction From the earliest days of the study of Noetherian commutative rings, there was great interest in the Noetherian properties of the integral closures of Noetherian rings, particularly integral domains. Integral closures are more widely studied in the domain case and the present article is no exception. An understanding of the domain case allows us to easily understand the case where the total quotient ring of R is zero dimensional. Historically, it was quickly observed that whenever r, s ∈ R with r nilpotent and s regular, then r/s is integral over R. Hence, if the nilradical of R is non-trivial and R is not its own total quotient ring, the nilradical of the integral closure of R will be infinitely generated and R cannot be Noetherian. It is interesting to note, especially in regard to the questions we consider here, that any regular ideal of R will contain the entire nilradical and so to study the number of generators of regular ideals, we can work modulo the nilradical and assume R is reduced. Further, the integral closure of a reduced ring is simply the direct sum of the integral closures of R/Pi where Pi ranges over the minimal prime ideals of R and so we have reduced to the domain case. [If the total quotient ring is not zero dimensional, modding out the nilradical changes the total quotient ring and so also the integral closure. That case is clearly more complicated.] Returning to domains, most of the interesting questions about the Noetherian property of integral closures were settled by Krull [3], Akizuki [1], Mori [6], and Nagata [7] long ago. Consider the setting R ⊆ S ⊆ R . Krull and Akizuki observed that while R need not be a finite R-module even in the case of a one dimensional local domain, R and even S must be Noetherian in that case. Mori went on to show that R is Noetherian in the two dimensional case. Nagata then cut short the good news by producing examples of a two dimensional S and a three dimensional R which are not Noetherian. These examples also appear in Nagata’s book [8]. However, there are two intriguing questions as yet unanswered which shall be resolved in this article. The first was posed by Heinzer [4] in 1973. The second is Date: October 21, 2016. 1
2
RAYMOND C HEITMANN
interesting in its own right, but I stumbled upon it as a natural outgrowth of the first. I am surprised that it has not been asked previously. Question (Heinzer). If R is a Noetherian integral domain with integral closure R , must it follow that maximal ideals in R are finitely generated? Question. If R is a local Noetherian domain, is the completion of the integral closure of R Noetherian? To more closely tie the questions together, I will first observe that Heinzer’s question is actually a local one—we can assume that both R and R have a single maximal ideal. To see this, suppose M is an infinitely generated maximal ideal of R . Then M contracts to a maximal ideal M of R and there are only finitely many maximal ideals of R which contract to M [8, 33.10]. So M will completely split in a finite integral extension of R which is of course Noetherian; this allows us to assume that M is the unique maximal ideal which contains M . Now, if T is a local generating set for M , i.e., T RM = M RM , and M = (r1 , r2 , . . . , rk )R, then T ∪ {r1 , r2 , . . . , rk } is a generating set for M . Thus, if M requires infinitely many generators, T must be an infinite set and so we may replace R by RM . Likewise, we can reduce the second question to the case where R has a single maximal ideal. Now, it is always true that ∩(M )n = (0). Hence, R is something which Nagata called a local ring which may not be Noetherian. Such a ring has a completion (R )∗ which contains it and that completion can be Noetherian even when R is not. In fact, a complete local ring which may not be Noetherian will actually be Noetherian if it has a finitely generated maximal ideal [8, 31.7]. Further, as (R )∗ will have a finitely generated maximal ideal whenever R does by [8, 17.5], a positive answer to Heinzer’s question would have guaranteed a positive answer to the second question as well. Below, in Corollary 4, we will see that the converse is also true. If R has an infinitely generated maximal ideal, then so does (R )∗ and so both questions will have negative answers. One reason why these questions have remained unanswered for so long ago is that they are quite difficult to resolve for those of us who owe much of our examplebuilding skills to an understanding of Nagata’s techniques. Nagata actually started with a Noetherian (R )∗ which he understood and built R as a subring of it. In this framework, R will itself be Noetherian if and only if the maximal ideal is finitely generated and finitely generated ideals are closed in the M -adic topology [8, (31.8)]. Nagata found a relatively easy way to put elements into the closure of a finitely generated ideal not primary to the maximal ideal without actually putting them into the ideal. That method cannot resolve the questions here. Section 2 is a brief general treatment of completions of integral closures. In Theorem 2, we see that the M -adic topology is the same as the M R -adic topology on R . This leads to the corollary that our two questions are equivalent. The meat of the article, Section 3, is essentially just the presentation of a single example which answers both questions. The sections can be read independently. Throughout, all rings are commutative with identity. The term local ring will refer to a Noetherian ring with a single maximal ideal. Without the Noetherian assumption, such rings will be called quasi-local. Following Nagata’s lead, we will call quasi-local rings satisfying ∩M n = (0) local rings which may not be Noetherian.
INTEGRAL CLOSURES OF NOETHERIAN DOMAINS
3
2. Completions If R is a local ring, the completion of R has been thoroughly studied. However, when R is not Noetherian, the completion will not be a faithfully flat extension and has deservedly attracted much less interest. However, a certain understanding of the properties of (R )∗ is needed to put our example in its proper context. There is a brief treatment of this subject in Nagata’s book [8] and presumably Nagata knew even more than what appears in print. Quite possibly, he was aware of all of the results in Section 2. In truth, none of the Section 2 results are surprising and the proofs are not that difficult. In this section, we will routinely assume that R is a local ring and its integral closure (in the total quotient ring) is quasi-local. As noted in the introduction, the assumption that R has a single maximal ideal is harmless because there are only finitely many maximal ideals of R which contract to the maximal ideal of R and so we may replace R by a localization of a finite extension and produce a setting in which the maximal ideal does not split. We begin with a result familiar to some which is proved here for completeness. Proposition 1. Let R be a local Noetherian domain and suppose that its integral closure R has a unique maximal ideal M . It follows that ∩(M )n = (0), so the local ring R may not be Noetherian. Proof. Since R is Noetherian, it is contained in a discrete valuation ring V which is centered on the maximal ideal. If m is the maximal ideal of V , we clearly have ∩mn = (0) since the intersection can only contain elements of infinite value. As R ⊆ V and M ⊆ m, the result follows. The domain assumption is critical for Proposition 1. If R has nilpotent elements, the intersection will contain the nilradical of R . In this setting, we can still complete R , but here, because the M -adic topology is not T0 , R ⊆ (R )∗ and so the completion contains less information about R and so is of less interest. Next we prove the central result of this section, that the M -adic topology on R is just the M R -adic topology. As the integral domain hypothesis is not useful in the proof, we state the result in greater generality. Theorem 2. Let R be a local ring with maximal ideal M and suppose its integral closure R is a quasi-local ring with maximal ideal M . Then there exists a positive integer m such that (M )m ⊂ M R . Thus the ideals M R and M induce the same topology on R . Proof. Let R∗ be the M -adic completion of R and let N be the nilradical of R∗ . Because R∗ /N is a reduced complete local ring, it follows from [8, 32.1] that its integral closure is a finite module over it and we have a natural sequence of Noetherian rings R ⊆ R∗ −→ R∗ /N ⊆ (R∗ /N ) . We also have R ⊆ (R∗ ) −→ (R∗ /N ) . These last maps allow us to view R∗ [R ] as a subring of (R∗ ) and to define (R∗ /N )◦ as the image of the map R∗ [R ] −→ (R∗ /N ) . Of course, (R∗ /N )◦ is also a finite R∗ /N -module and so is also Noetherian. We now have the following commutative diagram: R R
⊆
⊆
/ R∗ / R∗ [R ]
/ / R∗ /N π
/ / (R∗ /N )◦
4
RAYMOND C HEITMANN
Next we make several somewhat unrelated observations about our setting. First, since (R∗ /N )◦ is a finite R∗ -module, there exists a finite subset Δ ⊂ R such that π(R∗ [Δ]) = (R∗ /N )◦ . Next, let J be the ideal generated by π(M ). J is finitely generated and so there exist δ1 , . . . , δk ∈ M such that J = (π(δ1 ), . . . , π(δk ))(R∗ /N )◦ . If we adjoin δ1 , . . . , δk to Δ, the set is still finite and so, without loss of generality, we may assume that δ1 , . . . , δk ∈ Δ. Finally, as the kernel of (R∗ ) −→ (R∗ /N ) is just the nilradical of (R∗ ) , the kernel of π is just the intersection of R∗ [R ] with the nilradical of (R∗ ) . As R∗ is Noetherian, there exists an integer such that N = (0) and since the numerator of each element in the nilradical of (R∗ ) is in N , we get (Ker π) = (0). Now, (M ) is generated by elements of the form q1 · · · q with each qi ∈ M . We consider an element q1 · · · q . As π(qi ) ∈ J and (R∗ /N )◦ = π(R∗ [Δ]), there exist k elements λij ∈ R∗ [Δ] such that π(qi − j=1 λij δj ) = 0. It follows that k λij δj = 0 , qi − i=1
j=1
from which we see that q1 · · · q ∈ (δ1 , . . . , δk )R∗ [Δ ∪ {q1 , . . . , q }]. As the ring R∗ [Δ] is Noetherian and the ideal M R∗ [Δ] is primary to the maximal ideal, we t ∗ [Δ] ⊆ M R∗ [Δ] for some positive integer t. Let m = t. Then, have m(δ1 , . . . , δk ) R m if i=1 qi ∈ (M ) , we see that m
qi ∈ (δ1 , . . . , δk )t R∗ [Δ ∪ {q1 , . . . , qm }] ⊆ M R∗ [Δ ∪ {q1 , . . . , qm }] .
i=1
As Δ ∪ {q1 , . . . , qm } is a finite set, R[Δ ∪ {q1 , . . . , qm }] is Noetherian and so its completion R∗ [Δ ∪ {q1 , . . . , qm }] is a faithfully flat extension. Thus, we have m m i=1 qi ∈ M R[Δ ∪ {q1 , . . . , qm }] ⊆ M R . It follows that (M ) ⊂ M R . Now we use Theorem 2 to show that our two questions are equivalent. Corollary 3. Let R be a local ring with maximal ideal M and suppose its integral closure R is a quasi-local ring. Let I be an ideal of R which is primary to the maximal ideal. Then I(R )∗ ∩ R = I. ui ai with each ui ∈ (R )∗ and each ai ∈ I. Proof. Suppose b ∈ R and b = Adding terms 0 · ak to our sum if necessary, we can presume the ideal (a1 , . . . , an )R is primary to the maximal ideal and so reduce to the case I = (a1 , . . . , an )R . As I and M R are both finitely generated ideals with the same radical, they induce the same adic topology. Hence, by Theorem 2, the I-adic topology on R is the same as the M -adic topology and so (R )∗ is the completion in the I-adic topology. Now each ui is the limit of a Cauchy sequence {uij } in R . It follows that n uij ai b− i=1 is a Cauchy n sequence in R which converges to zero. For sufficiently large j, we have b − i=1 uij ai ∈ I and so b ∈ I as desired.
Corollary 4. Let R be a local ring with maximal ideal M and suppose its integral closure R is a quasi-local ring. Then the maximal ideal M of R is finitely generated if and only if (R )∗ is Noetherian.
INTEGRAL CLOSURES OF NOETHERIAN DOMAINS
5
Proof. The forward direction is due to Nagata and has already been noted. For the converse, suppose (R )∗ is Noetherian. Then M (R )∗ is finitely generated and so equals I(R )∗ for some finitely generated ideal I of R . By Corollary 3, M = I and so M is finitely generated. Finally, as it is convenient to do so at this point, we describe (R )∗ . Corollary 5. Let R be a local domain with maximal ideal (a1 , . . . , an )R and suppose its integral closure R is a quasi-local ring. Then (R )∗ = R [[X1 , . . . , Xn ]]/I where I is the closure of (X1 − a1 , . . . , Xn − an ) in the (a1 , . . . , an )-adic topology. Proof. Since the topology on R is just the (a1 , . . . , an )-adic topology, this follows immediately from [8, 17.5]. 3. The Example We begin with a proposition whose proof is due to Nagata. The form we need is slightly more general than that which Nagata considered, but the alterations have no real effect on the actual proof. The original example is bad Noetherian ring example 3 [8, p. 206]. Proposition 6. Let K be a field of characteristic p and assume L is a field such that K p ⊆ L ⊆ K. If R = ((K p [[x1 , . . . , xn ]][L])[[z]])[K], then R is a regular local ring. Proof. Since K[[x1 , . . . , xn , z]] is integral over R, we easily see that R has dimension n + 1. Also clearly, its maximal ideal is generated by x1 , . . . , xn , z. Thus, to prove it is a regular local ring, it suffices to prove that R is Noetherian. To prove this, we first adjust Nagata’s proof to show that K p [[x1 , . . . , xn ]][L] is Noetherian, then observe that (K p [[x1 , . . . , xn ]][L])[[z]] is Noetherian because it is a power series ring over a Noetherian ring, and finally use Nagata’s ideas again to complete the proof. First note that the two rings we must show are Noetherian are quasi-local subrings of their respective completions L[[x1 , . . . , xn ]] and K[[x1 , . . . , xn , z]] and have the induced topologies. If M denotes the maximal ideal of either ring, then we have n ∩∞ i=1 M = (0) and so these rings are local rings which may not be Noetherian. Such rings are Noetherian provided they have a finitely generated maximal ideal (which both do) and have the property that finitely generated ideals are closed [8, (31.8)]. So let a1 , . . . , ad generate a finitely generated ideal of K p [[x1 , . . . , xn ]][L] and suppose b ∈ (a1 , . . . , ad )L[[x1 , . . . , xn ]] ∩ K p [[x1 , . . . , xn ]][L]. Then the coefficients
of a1 , . . . , an , b generate a finite extension K of K p . Now note that L = K B is a direct sum of K - vector spaces. (B is not unique of course, but any choice for this summand will suffice.) Now b ∈ (a1 , . . . , ad )L[[x1 , . . . , xn ]] gives us an equation b = a1 f1 + · · · + ad fd . Using our direct sum, we can project the coefficients that occur in each fi to K and so obtain f¯1 , . . . , f¯d ∈ K [[x1 , . . . , xn ]] such that b = a1 f¯1 + · · · + an f¯n . Thus b ∈ (a1 , . . . , ad )K [[x1 , . . . , xn ]] ⊆ (a1 , . . . , ad )K p [x1 , . . . , xn ]][L] and we see that finitely generated ideals are closed. Showing that b ∈ (a1 , . . . , ad )K[[x1 , . . . , xn , z]] ∩ ((K p [[x1 , . . . , xn ]][L])[[z]])[K] implies b ∈ (a1 , . . . , ad )((K p [[x1 , . . . , xn ]][L])[[z]])[K] is similar though requires several steps. Exactly as before, the coefficients of a1 , . . . , ad , b generate a finite extension L of L and we again see that b ∈ (a1 , . . . , ad )L [[x1 , . . . , xn , z]]. Further,
6
RAYMOND C HEITMANN
as K p [[x1 , . . . , xn ]][L ][[z]] is a Noetherian ring with completion L [[x1 , . . . , xn , z]], this gives b ∈ (a1 , . . . , ad )(K p [[x1 , . . . , xn ]][L ])[[z]]. As K p [[x1 , . . . , xn ]][L ][[z]] = K p [[x1 , . . . , xn ]][L][[z]][L ] ⊆ (K p [[x1 , . . . , xn ]][L])[[z]][K],
the desired result follows.
Throughout, we let K be a fixed field of characteristic p with the property that [K : K p ] is infinite. Let {bi , gijk , | i, j, k ∈ Z≥0 } ⊂ K be a set of p-independent elements over K p . Let L = K p ({gijk , | i, j, k ∈ Z≥0 }). Let R = K p [[x, y]][L][[z]][K]. By Proposition 6, R is a regular local ring. Let {ai | i ∈ Z+ } be a set of distinct elements in K p . Then, setting wi = x + ai z for each i, we have an infinite set of distinct prime elements in R. ∞ Notation 7. Let c = c0 = i=0 bi y i and, for i > 0, let i−1 ∞ ci = c − bj y j /y i = bj y j−i . j=0
Note ci = bi + yci+1 . Let r =
∞ ∞ ∞ i=0
j=0
j=i
k=0 gijk x
i j k
y z .
Two additional bits of notational shorthand will make our presentation much more readable. n Notation 8. Let tn = i=1 wi . So t0 = 1 and tn+1 = tn wn+1 . Let n−1 n σn = wj y i = w2 w3 · · · wn + w3 w4 · · · wn y + · · · + wn y n−2 + y n−1 . i=0
j=i+2
Of course, σ1 = 1, σ2 = w2 + y, and σn+1 = wn+1 σn + y n . Lemma 9. With the notation as above, there exist sets {αi | i ∈ Z≥0 }, {ri | i ∈ Z≥0 } ⊂ L[[x, y, z]] such that if si = ti ci + zαi , then for every i ≥ 0, wi+1 si − ysi+1 = ti+1 bi + zri and r − ri ∈ L[[z]][x] ⊂ R. Moreover, we have wi+1 αi − yαi+1 = ri . Proof. First, we observe that wi+1 si − ysi+1 = wi+1 (ti ci + zαi ) − y(ti+1 ci+1 + zαi+1 ) = ti+1 (ci − yci+1 ) + z(wi+1 αi − yαi+1 ) = ti+1 bi + z(wi+1 αi − yαi+1 ) , so the first part will follow provided we choose αi , ri to satisfy wi+1 αi − yαi+1 = ri . Of course, the quantities on both sides of this equation are power series and so it is natural to verify it by checking that the corresponding coefficients are equal. However, we will do something slightly different. As wi+1 , y are homogeneous in both the y-grading and the (x, z)-grading, we will group our terms using this bigrading. To achieve this, we introduce some slightly unusual notation. For each , rijk of degree k in L[x, z] i, j, k ≥ 0, we intend choose homogeneous forms to ∞αijk ∞ ∞ ∞ and then set αi = j=0 k=0 αijk y j and ri = j=0 k=0 rijk y j . Adopting the convention that rijk = αijk = 0 whenever either j or k is negative, the equation wi+1 αi − yαi+1 = ri is equivalent to wi+1 αi,j,k−1 − αi+1,j−1,k = rijk for all j, k. We shall later refer to this as the crucial equation. It remains to choose our elements rijk , αijk to satisfy this equation and the second conclusion of the lemma. We will
INTEGRAL CLOSURES OF NOETHERIAN DOMAINS
7
actually satisfy the second conclusion by choosing rijk so that r − ri is a polynomial in L[[z]][x] of x-degree at most i. The plan of the proof is to define the elements as needed. We will choose our elements in the following order. We shall first define rijk for j > 0. Next we define αijk for i > 0. Then we define ri0k and finally α0jk . ∞ ∞ If we set hjk = d+e=k gdje xd z e , we may write r = j=0 k=0 hjk y j . Now, for j > 0, set rijk = hjk . This ensures that r −ri ∈ L[[x, z]]. For the second conclusion, it only remains to bound the degree of x. Next we develop a recursive definition for the alphas which allows us to find each αijk with i > 0 in terms of known quantities and the as yet unspecified α0jk . Set αi+1,j−1,k = wi+1 αi,j,k−1 − rijk whenever i, k ≥ 0 and j > 0. This equation is also trivially true when k < 0, a fact which will simplify our calculations later on. This guarantees that the crucial equation will hold whenever j > 0. In the j = 0 case, the desired equation reduces to wi+1 αi,0,k−1 = ri0k and we may define ri0k to force this equation to hold as well. Like {αijk | i > 0}, the set {ri0k } will depend upon our choice of {α 0jk }. ∞ It only remains to bound the degree of r − ri . Since r − ri = k=0 h0k − ri0k , to bound the x-degree of r − ri , we need to choose {α0jk } so that the x-degree of h0k − ri0k is at most i. Now we use our definitions to recursively calculate αi,0,k−1 = −(ri−1,1,k−1 + wi ri−2,2,k−2 + wi wi−1 ri−3,3,k−3 + · · · + wi · · · w2 r0,i,k−i ) +wi · · · w1 α0,i,k−i−1 for i > 0 and so note ri0k = wi+1 αi,0,k−1 = −(wi+1 ri−1,1,k−1 + wi+1 wi ri−2,2,k−2 + wi+1 wi wi−1 ri−3,3,k−3 + · · · +wi+1 · · · w2 r0,i,k−i ) + wi+1 · · · w1 α0,i,k−i−1 . We can write this in the simplified form ri0k = τik + ti+1 α0,i,k−i−1 where τik is a specific homogeneous polynomial in L[x, z] of degree k. Thus, we have h0k − ri0k = h0k − τik − ti+1 α0,i,k−i−1 for i > 0. We note that this equation is also true for i = 0 since r00k = w1 α0,0,k−1 (here τik = 0). There are two cases, k ≤ i or k > i. In the first case, α0,i,k−i−1 = 0 by definition, but clearly h0k − ri0k is a form of degree k ≤ i and so necessarily has x-degree ≤ i. In the second case, we define α0,i,k−i−1 to achieve our goal. Note that each α0im occurs in precisely one expression and so we will not define the same element twice. View h0k − τik and ti+1 as polynomials in x, and note that the latter is monic of degree i + 1. Perform long division to obtain h0k − τik = δik ti+1 + fik where fik is a polynomial in x of degree at most i. Then set α0,i,k−i−1 = δik and we get h0k − ri0k = fik , completing the proof. Theorem 10. With the notation as above and s = s0 , let A = R[r, s]. Then A , the integral closure of A, is a non-Noetherian integral domain of dimension three which has an infinitely generated maximal ideal. Moreover, the completion of A also has an infinitely generated maximal ideal and so is also not Noetherian. The rest of this article will be devoted to proving the first conclusion of this theorem, which will follow immediately from Theorem 17. The moreover statement then is a consequence of Corollary 4. We need to show that the maximal ideal of A
8
RAYMOND C HEITMANN
is infinitely generated. A logical way to proceed would be to compute A and then we would expect the final steps to be easy. Unfortunately, except for the p = 2 case, the computation is messy and not very useful. As we want to show that the example is valid for arbitrary p, an alternate approach is needed. If M denotes the maximal ideal of A , then clearly the pth power of any element in M is in (xp , y p , z p )R and it follows that A /(x, y, z)A is a zero dimensional ring with a single prime ideal. Since a non-Noetherian ring necessarily has an infinitely generated prime ideal [2], it will suffice to prove that A /(x, y, z)A is not Noetherian. Our goal then will be to understand A well enough to find an infinite set of elements in A that, together with x, y, z, generate an ideal that is not finitely generated. Remark 11. As K[[x, y, z]] is integral over R, if S is any ring betwen R and K[[x, y, z]], notice that the integral closure S will just be the intersection of the quotient field with K[[x, y, z]]. Moreover, since rp , sp ∈ R, any element in the p−1 p−1 integral closure of A necessarily has the form ( i=0 j=0 uij ri sj )/v with v ∈ R and each uij ∈ R. Lemma 12. Let K be a field of characteristic p such that [K : K p ] = ∞. Suppose J is a field, K p ⊆ J ⊂ K, and e1 , e2 , . . . ∈ K are such that the infinite set {ei } is a minimal generating set for K over J. Let S be a Noetherian integrally closed ∞ domain, K p [[X1 , . . . , Xn ]][K][⊆ S ⊆ J[[X1 , . . . , Xn ]][K], and let τ = i=1 ei X1i +h with h ∈ J[[X1 , . . . , Xn ]][K]. Let T = S[τ ]. If γ ∈ T , X1k γ ∈ T for some k. Proof. If the lemma is false, we can choose a counterexample γ ∈ / T such that the conductor of γ is a prime ideal which does not contain X1 . Thus we may assume there is an element u ∈ S such that uγ ∈ T and X1 , u is a regular sequence. As noted in Remark 11, γ ∈ K[[X1 , . . . , Xn ]] and of course uγ ∈ T implies that we may write uγ = vd τ d + · · · + v0 with each vi ∈ S and 0 < d < p. For any i, we may adjoin all ej ’s but one to J and produce a field Ki such that K = Ki [ei ]. This makes K a p-dimensional vector space over Ki and of course K[[X1 , . . . , Xn ]] is a rank p free module over Ki [[X1 , . . . , Xn ]]. The elements u, vd , . . . , v0 , h are in J[[X1 , . . . , Xn ]][K] and so can only involve finitely many ej . Thus we may choose i so that u, vd , . . . , v0 , h ∈ Ki [[X1 , . . . , Xn ]]. Now we consider the equation vd τ d + · · · + v0 · 1 = uγ in our rank p free module, noting the coefficients are all in the ground ring. The projection of the left hand side on the edi coordinate is just vd X1id , while the projection of the right hand side is uγd where γd is the projection of γ on the edi coordinate. However this forces u to divide vd X1id in Ki [[X1 , . . . , Xn ]] and so also in the integrally closed S. As X1 , u is a regular sequence, we see that vd ∈ uS and so we may reduce to the case vd = 0. By inducting on d, we see that the lemma holds. Lemma 13. (1) R[r] is integrally closed. (2) If v ∈ A , then y n v ∈ A for some n. (3) if (t, y)A is a height two ideal, v ∈ A, and tv ∈ y n A , then v ∈ y n A . Proof. To prove will apply Lemma 12 twice to S = K p [[x, y]][K] ∼ = R/zR ∞(1),we ∞ with τ = r¯ = i=0 j=0 gij0 xi y j . In the first application, we let X1 = x, X2 = y, ei = gi00 , and J be an intermediate field which contains gij0 for j > 0. We then conclude that γ ∈ T implies xk γ ∈ T . In the second application, we let X1 = y, X2 = x, ei = g0i0 , and J be an intermediate field which contains gij0 for i > 0.
INTEGRAL CLOSURES OF NOETHERIAN DOMAINS
9
We then conclude that γ ∈ T implies y k γ ∈ T . Since S is Cohen-Macaulay and so the free S-module T is also Cohen-Macaulay, either γ ∈ T or the conductor of γ is contained in a height one prime ideal. As no height one prime ideal contains both x and y, T is integrally closed. Next, we note that R/zR ⊂ (R/zR)[¯ r] is an integral extension of degree p and so zR[r] is a prime ideal. Moreover, R[r]/zR[r] is isomorphic to T , which is integrally closed. Since normality lifts (see for example [5, p. 439]), R[r] is integrally closed. Statement (2) follows from Lemma 12, using S = R[r], X1 = y, X2 = x, X3 = z, τ = s, and ei = bi . For statement (3), let tv = y n γ with γ ∈ A . Then, we have γ ∈ K[[x, y, z]] and we see that in that unique factorization domain, v = y n δ. But then δ ∈ A by Remark 11 and the statement is proved. n In light of this lemma, A = ∪∞ n=1 An where An = (A :A y ). Fix n. Since tn is a monic polynomial of degree n in x, there is a unique polynomial μ ∈ L[[z]][x, y] with degx μ < n, degy μ < n such that r − μ ∈ (y n , tn )L[[x, y, z]]. Let rˆ = r − μ and n−1 sˆ = y n cn + zα0 . Since sˆ − s = − j=0 bj y j ∈ K[y] ⊂ R and μ ∈ L[[z]][x, y] ⊂ R, A = R[ˆ r, sˆ]. Strictly speaking, we should perhaps use the notation rˆn and sˆn as these elements depend on n. However, we shall only these terms in a setting where n is fixed and no confusion should result. Accordingly, we will avoid the more cumbersome notation.
Lemma 14. There exist polynomials p1 , p2 ∈ L[[z]][x, y] such that tn sˆ = y n sn + zˆ rσn + tn zp1 + y n zp2 . Proof. By the definitions, tn sˆ = tn (y n cn + zα0 ) = y n (sn − zαn ) + tn zα0 = y n sn + z(tn α0 − y n αn ) and so it only remains to show that we can solve tn α0 − y n αn − rˆσn = tn p1 + y n p2 , i.e., show tn α0 − y n αn − rˆσn ∈ (tn , y n )L[[z]][x, y]. Now this element is clearly in (tn , y n )L[[x, y, z]] and, as (tn , y n )L[[x, y, z]] ∩ L[[z]][x, y] = (tn , y n )L[[z]][x, y], it is enough to show tn α0 − y n αn − rˆσn ∈ L[[z]][x, y]. We verify this by proving tk α0 − y k αk − rˆσk ∈ L[[z]][x, y] for 1 ≤ k ≤ n by induction on k. For k = 1, t1 α0 − yα1 − rˆσ1 = r0 − rˆ by Lemma 9, and since r0 − rˆ = (r0 − r) + μ ∈ L[[z]][x, y], the k = 1 case is proved. For the induction step, tk+1 α0 − y k+1 αk+1 − rˆσk+1 = wk+1 tk α0 − y k+1 αk+1 − rˆ(wk+1 σk + y k ) = wk+1 (tk α0 − y k αk − rˆσk ) + y k (wk+1 αk − yαk+1 ) − rˆy k = wk+1 (tk α0 − y k αk − rˆσk ) + y k (rk − rˆ) ∈ L[[z]][x, y] .
Proposition 15. (1) Each element of An has the form q/y n where q = B0 +B1 sˆ+· · ·+Bp−1 sˆp−1 with each Bm ∈ R[ˆ r]. r] (2) B0 ∈ (z, y n )R[ˆ r] = ((x, y)n , rˆ, z)R[ˆ r] (3) B1 ∈ ({y n−i ti | 0 ≤ i ≤ n}, rˆ, z)R[ˆ
10
RAYMOND C HEITMANN
Proof. The first statement has already been observed. By Lemma 13(3), an element n of this form is in An if and only if tp−1 n q/y is integral. We compute: p−1 p−2 tp−1 ˆ) + · · · + Bp−1 (tn sˆ)p−1 n q = tn B0 + tn B1 (tn s p−2 n = tp−1 rσn + tn zp1 + y n zp2 ) + · · · n B0 + tn B1 (y sn + zˆ
+Bp−1 (y n sn + zˆ rσn + tn zp1 + y n zp2 )p−1 We want to know for which values of Bm this quantity is divisible by y n in A . We can ignore any terms which are divisible by y n and, since sn ∈ A , this means we can ignore all terms involving either y n sn or y n p2 . Thus it suffices to consider the p−2 rσn + tn p1 ) + · · · + Bp−1 z p−1 (ˆ rσn + tn p1 )p−1 ∈ R[ˆ r] and element tp−1 n B0 + tn B1 z(ˆ now we can take advantage of the fact that for such elements, divisibility by y n in r] are all equivalent. It is clear that this element can only be A , K[[x, y, z]], and R[ˆ n r] when tp−1 r], and since tn , z, y n is a regular sequence, we in y n R[ˆ n B0 ∈ (z, y )R[ˆ get the desired B0 ∈ (z, y n )R[ˆ r] and so part (2) is proved. Expanding the powers of rˆσn +tn p1 and regrouping, our element can be rewritten as the sum p−1 p−1 tnp−1−k rˆk σnk B z p−k 1 k k=0
=k
Noting that the k = 0 term is divisible by tp−1 and that each term for k ≥ 2 is n divisible by rˆ2 , the element can be divisible by y n only if we have tp−2 ˆσn n r
p−1
B z p−1 ∈ (y n , tp−1 ˆ2 )R[ˆ r] . n ,r 1
=1
p−1 Next, for every , we write B = j=0 Bj rˆj with each Bj ∈ R. Then we see p−1 ˆσn =1 B0 z p−1 ∈ (y n , tp−1 ˆ2 )R[ˆ r]. As R[ˆ r] is a free R-module and tp−2 n r n ,r 1 p np p rˆ ∈ (y , tn )R, this reduces to a condition in R, namely tp−2 n σn
p−1
B0 z p−1 ∈ (y n , tp−1 n )R . 1
=1
p−1 Further, as y, tn is a regular sequence, this implies σn =1 B0 z p−1 ∈ (y n , tn )R. 1 , z is a regular sequence, we may divide the element by z. Now set Similarly, as y, t n p−1 v = =1 B0 (zp1 )−1 . Then our condition becomes simply σn v ∈ (y n , tn )R. We now claim that v ∈ ({y n−i ti | 0 ≤ i ≤ n})R. We prove the claim by using induction to show that for 1 ≤ d ≤ n, σd v ∈ (y d , td )R implies v ∈ ({y d−i ti | 0 ≤ i ≤ d})R. The d = 1 case is obvious since σ1 = 1. Certainly (y d , td )R ⊂ (y k , wk )R for each k ≤ d, so we have σd v ∈ (y k , wk )R. d−1 d d Recall σd = i=0 ( j=i+2 wj )y i . Now, we see that ( j=i+2 wj )y i ∈ (y k , wk )R if d k ≥ i + 2 or i ≥ k, i.e., if i = k − 1. Thus ( j=k+1 wj )y k−1 v ∈ (y k , wk )R. As y, wk , wj is regular for j = k, this gives v ∈ (wk , y)R. Using the regularity of the sequence, y, wi , j=i wj , this leads to v ∈ (td , y)R. We write v = td h1 + yh2 . As td h1 ∈ {y d−i ti | 0 ≤ i ≤ d}R and also satisfies σd td h1 ∈ (y d , td )R, we may reduce to the case v = yh2 . As σd = wd σd−1 + y d−1 , (wd σd−1 + y d−1 )yh2 ∈ (y d , td )R gives σd−1 h2 ∈ (y d−1 , td−1 )R. The induction assumption implies that h2 ∈ ({y d−1−i ti | 0 ≤ i ≤ d − 1})R , so the claim follows.
INTEGRAL CLOSURES OF NOETHERIAN DOMAINS
Recalling that v =
p−1 =1
11
B0 (zp1 )−1 , we see that
B10 ∈ ({y n−i ti | 0 ≤ i ≤ n}, z)R r] as desired. Finally, as wi − x ∈ zR and and so B1 ∈ ({y n−i ti | 0 ≤ i ≤ n}, z, rˆ)R[ˆ so ti − xi ∈ zR, the last equation in conclusion (3) holds. Remark 16. Proposition 15 givesus all of the information we need to compute p−1 A in the case p = 2. Here v = =1 B0 (zp1 )−1 just means v = B10 and so n−i B1 ∈ ({y ti | 0 ≤ i ≤ n}, rˆ)R[ˆ r]. So An is generated over A by xs1 , x2 s2 , . . . , xn sn plus one additional element (βz + rˆsˆ)/y n . Hence An /zAn will be generated by xc1 , x2 c2 , . . . , xn cn , xn cn ρ for some ρ ∈ K[[x, y]]. Thus A /zA bears a strong resemblance to the example due to Kaplansky which appears at the end of [4]. On the other hand, B2 is not necessarily in the ideal ((x, y)n , rˆ, z)R[ˆ r] and so A will be larger for p > 2. I suspect that A /zA contains cki for all i and all k ≥ 2 but I have not actually verifed this as it is not needed for the proof. Theorem 17. The elements x, y, z, s − b0 , s1 , s2 , s3 , . . . generate an infinitely generated ideal of A . Proof. We know that ri , s0 ∈ A. Thus the relation wi+1 si − ysi+1 = ti+1 bi + zri from Lemma 9 tells us that each si is in the quotient field of A and so in A . Hence the listed elements are in A and it only remains to show that the ideal is infinitely generated. Suppose not. Then for all sufficiently large k, we have sk ∈ (x, y, z, s0 − b0 , s1 , . . . , sk−1 )A . As A = ∪An , for sufficiently large n, we have sk ∈ (x, y, z, s0 − b0 , s1 , . . . , sk−1 )An . We now will use Proposition 15 and the notation which accompanies it. n−1 n−1 Recall sˆ = y n cn + zα0 = s0 − i=0 bi y i . It follows that s0 − b0 = sˆ + i=1 bi y i and so (y, s0 −b0 )An = (y, sˆ)An . Thus we may replace s0 −b0 by sˆ in our generating n−1 set. Now assume i > 0. Combining ci = j=i bj y j−i + y n−i cn , si = ti ci + zαi , n−1 and sˆ = y n cn + zα0 , we see that ti sˆ − y i si = z(ti α0 − y i αi ) − ti j=i bj y j . By a r]. simple induction argument using Lemma 9, we see that δi = −(ti α0 − y i αi ) ∈ R[ˆ r]. As ti ∈ (x, z)A , we Consequently si = (zδi + ti sˆ)/y i + ti ρi for some δi , ρi ∈ R[ˆ may discard the second term and replace si by the first. More conveniently, using a common denominator, we use s˜i = (zβi + sˆy k−i ti )/y k where βi = y k−i δi . The finitely generated assumption then gives us an equation with each element written in the format of Proposition 15. (zβk + sˆtk )/y k = x(E0 + E1 sˆ + · · · + Ep−1 sˆp−1 )/y n + y(F0 + F1 sˆ + · · · + Fp−1 sˆp−1 )/y n +z(G0 + G1 sˆ + · · · + Gp−1 sˆp−1 )/y n + sˆ(H0 + H1 sˆ + · · · + Hp−1 sˆp−1 )/y n +
k−1
((γi0 + γi1 sˆ + · · · + γi,p−1 sˆp−1 )/y n )((zβi + sˆy k−i ti )/y k )
i=1
As R[ˆ r, sˆ] is free over R[ˆ r], we can regard the coefficients of 1, sˆ, sˆ2 , . . . , sˆp−1 separately and obtain p distinct equations in R[ˆ r]. We actually only need the equation related to the coefficient of sˆ to derive our contradiction. That equation k−1 is tk /y k = xE1 /y n + yF1 /y n + zG1 /y n + H0 /y n + i=1 (γi0 y k−i ti + γi1 zβi )/y n+k . Multiplying through by y n+k , we get y n tk = xy k E1 + y k+1 F1 + zy k G1 + y k H0 + k−1 k−i ti + γi1 zβi ). i=1 (γi0 y
12
RAYMOND C HEITMANN
We may use this equation in R[ˆ r] to induce a congruence modulo r] + (z, rˆ)R[ˆ r] . I = (x, y)n+k+1 R[ˆ k−1 n k k k+1 k F1 + y H0 + i=1 γi0 y k−i xi . Finally, As wj ≡ x, we have y x ≡ xy E1 + y r] and H0 , γi0 ∈ (z, y n )R[ˆ r]. Thus by Proposition 15, E1 , F1 ∈ (z, rˆ, (x, y)n )R[ˆ k k+1 n n F1 ≡ 0 and H0 ≡ y H0 , γi0 ≡ y γi0 . This gives xy E1 ≡ y y n xk ≡ y n+k H0 +
k−1
k+n−i i γi0 y x .
i=1
As k > 0, this gives y n xk ∈ (xk+1 , y n+1 , z, rˆ)R[ˆ r] ∩ R = (xk+1 , y n+1 , z, rˆp )R and p np np p since rˆ ∈ (x , y , z )R, we have our desired contradiction. Remark 18. We note that (A )∗ contains the element xs1 + x2 s2 + x3 s3 + · · · and so it is not contained in the integral closure of A∗ . References [1] Y. Akizuki Einige bemerkunge u ¨ber prim¨ are integrit¨ atsbereiche mit teilerkettensatz, Proc. Phys.-Math. Soc. Japan, 3rd Series 17 (1935), 327–336. [2] I. S. Cohen Commutative rings with restricted minimum condition, Duke Math. J. 17 (1950), 27–42. [3] W. Krull Ein satz u ¨ber prim¨ are integrit¨ atsberereiche, Math. Ann. 103 (1930), 450–465. [4] W. Heinzer, Minimal primes of ideals and integral ring extensions, Proc. AMS 40 (1973), 370–372. [5] R. Heitmann, Lifting Seminormality, Michigan Math. J. 57 (2008), 439–445. [6] Y. Mori On the integral closure of an integral domain, II, Bull. Kyoto Gakugei Univ. B7 (1955), 19–30. [7] M. Nagata, Note on integral closures of Noetherian domains, Mem. Coll. Sci., Univ. Kyoto 28 (1953-54), 121–124. [8] M. Nagata, Local Rings, Krieger, Huntington, NY, 1975.