BUILDING NOETHERIAN DOMAINS INSIDE

AN IDEAL-ADIC COMPLETION

William Heinzer, Christel Rotthaus and Sylvia Wiegand

December 9, 1997

Abstract. Suppose a is a nonzero nonunit of a Noetherian integral domain R. An interesting construction introduced by Ray Heitmann addresses the question of how ring-theoretically to adjoin a transcendental power series in a to the ring R. We apply this construction, and its natural generalization to finitely many elements, to exhibit Noetherian extension domains of R inside the (a)-adic completion R^{∗} of R. Suppose τ_{1}, …, τ_{s} ∈ aR^{∗} are algebraically independent over K, the field of fractions of R. Starting with U_{0} := R[τ_{1}, …, τ_{s}], there is a natural sequence of nested polynomial rings U_{n} between R and A := K(τ_{1}, …, τ_{s}) ∩ R^{∗}. It is not hard to show that if U := ∪_{n=0}^{∞}U_{n} is Noetherian, then A is a localization of U and R^{∗}[1/a] is flat over U_{0}. We prove, conversely, that if R^{∗}[1/a] is flat over U_{0}, then U is Noetherian and A := K(τ_{1}, …, τ_{s}) ∩ R^{∗} is a localization of U. Thus the flatness of R^{∗}[1/a] over U_{0} implies the intersection domain A is Noetherian.

1. Introduction. Suppose a is a nonzero nonunit of a Noetherian integral domain

R. The (a)-adic completion R^{∗} of R is isomorphic to the ring R[[x]]/(x − a) [N,

(17.5), page 55]. Thus elements of the (a)-adic completion may be regarded as

formal power series in a. Of course if R is already complete in its (a)-adic topol-

ogy, then R = R^{∗}, but often it is the case that there are elements of R^{∗} that are

transcendental over R. An interesting construction first introduced by Ray Heit-

mann in [H, page 126] addresses the question of how ring-theoretically to adjoin a

transcendental (over R) power series in a to the ring R. We have made use of this

construction of Heitmann in [HRW3] in a local or semilocal context. Our purpose

here is to consider this construction in the more general context of an arbitrary

Noetherian integral domain.

The authors would like to thank the National Science Foundation and the University of Nebraska Research Council for support for this research. In addition they are grateful for the hospitality and cooperation of Michigan State, Nebraska and Purdue, where several work sessions on this research were conducted.

Typeset by AMS-TEX 2 WILLIAM HEINZER, CHRISTEL ROTTHAUS AND SYLVIA WIEGAND

There are numerous articles in the literature that have relevance for the building of Noetherian domains inside an ideal-adic completion, for example [BR1], [BR2], [HRS], [H1], [H2], [H3], [L], [N2], [O1], [O2], [R1], [R2], [R3], and [W].

Let R be a Noetherian integral domain with field of fractions K and let a be a nonzero nonunit of R. We are interested in the structure of certain intermediate integral domains between R and R^{∗} := { \atop (R, (a))} = R[[x]]/(x − a), the (a)-adic completion of R. We are particularly interested in domains of the form A := L∩R^{∗}, where L is an intermediate field between K and the total ring of fractions of R^{∗}. It is often difficult to compute this intersection ring A. Thus we seek conditions in order that A be realizable as a localization of a directed union of polynomial ring extensions of R.

This intersection construction inside the completion of R with respect to a principal ideal yields interesting Noetherian rings which are directed unions of localized polynomial rings, as we see below. By contrast, taking the analogous construction inside the completion with respect to a maximal ideal, even of an excellent local domain seems less likely to give Noetherian intersection domains. In [HRW1], it is shown for a countable excellent local domain (R, m) of dimension at least two that there exist infinitely many algebraically independent elements τ_{1}, τ_{2}, … in the m-adic completion R̂ of R such that the corresponding intersection domain is a localized polynomial ring in infinitely many variables over R; that is, R̂ ∩ K(τ_{1}, τ_{2}, …) = R[τ_{1}, τ_{2}, …]_{(m,τ_{1},τ_{2},...)}.

In [HRW2], [HRW3] and the present paper, we study the following element-wise form of the problem. Let τ_{1}, …, τ_{s} ∈ aR^{∗} be elements which are algebraically independent over K. Starting with U_{0} := R[τ_{1}, …, τ_{s}], we define a sequence of nested polynomial rings U_{n} in s variables over R inside A := K(τ_{1}, …, τ_{s}) ∩ R^{∗}. In [HRW3] we consider in the case where R is a semilocal Noetherian integral domain and a is an element of the Jacobson radical of R the condition that the embedding U_{0} → R^{∗}[1/a] is flat. Our goal here is to examine flatness of the embedding U_{0} → R^{∗}[1/a] in a more general context, and to prove the following theorem.^{1}

Theorem 1.1. Suppose R is a Noetherian domain, a ∈ R is a nonzero nonunit, and τ_{1}, …, τ_{s} are elements of the (a)-adic completion R^{∗} of R that are algebraically

1This result generalizes [HRW3, Theorem 2.12].

BUILDING NOETHERIAN DOMAINS 3

independent over R.^{2} Then the following conditions are equivalent:

(1) The ring R^{∗}[1/a] is flat over U_{0} = R[τ_{1}, …, τ_{s}].

(2) The directed union U := ∪_{n=0}^{∞}U_{n} is Noetherian.^{3} Moreover, if these equivalent conditions hold, then the integral domain A := K(τ_{1}, …, τ_{s}) ∩ R^{∗} is a localization of U, and hence A is Noetherian.

Remark 1.2. An example given in [HRW3, (4.4)] shows that it is possible for A to be a localization of U and yet for A, and therefore also U, to fail to be Noetherian. Thus the equivalent conditions of (1.1) are not implied by the property that A is a localization of U.

We present in (2.5) an example that is a modification of [HRW2, Example 2.1] to show that A being Noetherian does not imply that U is Noetherian.

The following diagram displays the situation concerning possible implications between A being a localization of U and A or U being Noetherian:

R^{∗}[1/a] is flat over U_{0} = R[τ_{1}, …, τ_{s}]

U Noetherian

A is a localization of U

A Noetherian

2. The general setting. (2.1) Let R be a Noetherian integral domain of dimension d > 0 with fraction field K. Let a be a nonzero element nonunit of R, let R^{∗} := { \atop (R, (a))} be the (a)-adic completion of R and let R_{a}^{∗} := R^{∗}[1/a]. Suppose τ_{1}, …, τ_{s} ∈ aR^{∗} are regular elements^{4} of R^{∗} that are algebraically independent over K. We consider the polynomial ring

U_{0} := R[τ_{1}, …, τ_{s}].

For every γ ∈ R^{∗} and every n > 0, we define the n^{th}-endpiece γ_{n} with respect to a of γ to be

2We say that elements are algebraically independent over an integral domain if they are algebraically independent over its fraction field.

3Heitmann in [H, page 126] considers the case where there is one transcendental element τ and defines the corresponding extension U to be a simple PS-extension of R for a. Heitmann proves in this case that a certain monomorphism condition on a sequence of maps is equivalent to U being Noetherian [H, Theorem 1.4].

4We say an element of a ring is a regular element if it is not a zero divisor.

(2.1.1)  γ_{n} := ∑_{j=n+1}^{∞} c_{j}a^{j−n}, where γ := ∑_{j=1}^{∞} c_{j}a^{j} with each c_{j} ∈ R.

In particular, we represent each of the τ_{i} by a power series expansion in a; we use these representations to obtain for each positive integer n the n^{th}-endpieces τ_{in} and the corresponding n^{th}-polynomial ring U_{n}: For 1 ≤ i ≤ s, and τ_{i} := ∑_{j=1}^{∞} r_{ij}a^{j}, where the r_{ij} ∈ R, τ_{in} := ∑_{j=n+1}^{∞} r_{ij}a^{j−n}, U_{n} := R[τ_{1n}, …, τ_{sn}], for each n ∈ ℕ. We have a birational inclusion of polynomial rings U_{n} ⊂ U_{n+1}. We define

(2.1.2)  U := ∪_{n=0}^{∞}U_{n} = \underrightarrow{lim} U_{n}  and  A := K(τ_{1}, …, τ_{s}) ∩ R^{∗}.

It is readily seen that A is a birational extension of U. We say that the τ_{i} have good limit-intersecting behavior if A is a localization of U.

We observe the following properties of (a)-adic completions and an implication of this concerning good limit-intersecting behavior.

Proposition 2.2 (cf. [HRW2],[HRW3, (2.2)]). Assume the notation and setting of (2.1), and let U^{∗} and A^{∗} denote the (a)-adic completions of U and A. Then

(1) a^{k}U = a^{k}A ∩ U = a^{k}R^{∗} ∩ U for each positive integer k.

(2) U^{∗} = A^{∗} = R^{∗}, so R/aR = U/aU = A/aA = R^{∗}/aR^{∗}.

(3) If U is Noetherian, then R^{∗} is flat over U and A is the localization of U at the multiplicative system 1 + aU of U.

Proof. We have R ⊆ U ⊆ A ⊆ R^{∗}. Since R is Noetherian, R^{∗} is flat over R [M1, Theorem 8.8, page 60]. Moreover, a^{k}R is closed in the (a)-adic topology on R, so we have a^{k}R^{∗} ∩ R = a^{k}R for each positive integer k [ZS, Theorem 8, page 261]. Furthermore, A = R^{∗} ∩ K(τ_{1}, …, τ_{s}) implies a^{k}A = a^{k}R^{∗} ∩ A. It is clear that a^{k}U ⊆ a^{k}R^{∗} ∩U, thus for (1) and (2) it suffices to show a^{k}R^{∗} ∩U ⊆ a^{k}U. Moreover, if aR^{∗} ∩U = aU, it follows that a^{k}R^{∗} ∩U = a^{k}R^{∗} ∩aU = a(a^{k−1}R^{∗} ∩U), and by induction we see that a^{k}R^{∗} ∩ U = a^{k}U. Thus we show aR^{∗} ∩ U ⊆ aU.

Let g ∈ aR^{∗}∩U. Then there is a positive integer n with g ∈ U_{n} = R[τ_{1n}, …, τ_{sn}]. Write g = r_{0} + g_{0} where g_{0} ∈ (τ_{1n}, …, τ_{sn})U_{n} and r_{0} ∈ R. From the definition of τ_{in}, we have τ_{in} = aτ_{in+1} + a_{in}a, where a_{in} ∈ R, for each i with 1 ≤ i ≤ s. Thus r_{0} ∈ aR^{∗} ∩ R = aR, τ_{in}U_{n} ⊆ aU_{n+1} and g ∈ aU. This completes the proof of (1)

BUILDING NOETHERIAN DOMAINS 5

and (2). If U is Noetherian, then U^{∗} = R^{∗} is flat over U. Let S be the multiplicative system 1 + aU and let B = S^{−1}U. Then B is Noetherian, the (a)-adic completion of B is R^{∗} and R^{∗} is faithfully flat over B [M1, Theorem 8.14, page 62]. Therefore B = K(τ_{1}, …, τ_{s}) ∩ R^{∗} = A. □

With the notation and setting of (2.1), the representation of the τ_{i} as power series in a with coefficients in R is, in general, not unique. However, as we observe in (2.3), the rings U and U_{n} are uniquely determined by the τ_{i}.

Proposition 2.3 (cf. [HRW3, (2.3)]). Assume the notation and setting of (2.1). Then U and the U_{n} are independent of the representation of the τ_{i} as power series in a with coefficients in R.

Proof. For 1 ≤ i ≤ s, assume that τ_{i} and ω_{i} = τ_{i} have representations

τ_{i} = ∑_{j=1}^{∞} a_{ij}a^{j}  and  ω_{i} = ∑_{j=1}^{∞} b_{ij}a^{j},

where each a_{ij}, b_{ij} ∈ R. We define the n^{th}-endpieces τ_{in} and ω_{in} as in (2.1.1):

τ_{in} = ∑_{j=n+1}^{∞} a_{ij}a^{j−n}  and  ω_{in} = ∑_{j=n+1}^{∞} b_{ij}a^{j−n}.

Then we have

τ_{i} = Σ_{j=1}^{∞}a_{ij}a^{j} = Σ_{j=1}^{n}a_{ij}a^{j} + a^{n}τ_{in} = Σ_{j=1}^{∞}b_{ij}a^{j} = Σ_{j=1}^{n}b_{ij}a^{j} + a^{n}ω_{in} = ω_{i}.

Therefore, for 1 ≤ i ≤ s and each positive integer n,

a^{n}τ_{in} − a^{n}ω_{in} = Σ_{j=1}^{n}b_{ij}a^{j} − Σ_{j=1}^{n}a_{ij}a^{j}, and so τ_{in} − ω_{in} = \frac{Σ_{j=1}^{n}(b_{ij} − a_{ij})a^{j}}{a^{n}} .

Since Σ_{j=1}^{n}(b_{ij} − a_{ij})a^{j} ∈ R is divisible by a^{n} in R^{∗} and since a^{n}R = R ∩ a^{n}R^{∗} because a^{n}R is closed in the (a)-adic topology, it follows that a^{n} divides the sum Σ_{j=1}^{n}(b_{ij} −a_{ij})a^{j} in R. Therefore τ_{in}−ω_{in} ∈ R. It follows that U_{n} and U = ∪_{n=1}^{∞}U_{n} are independent of the representation of the τ_{i}. □

Remark 2.4. With notation as in (2.1), if the embedding U_{0} = R[τ_{1}, …, τ_{s}] → R^{∗}[1/a] is flat, then every nonzero element of U_{0} is a regular element of R^{∗}.

6 WILLIAM HEINZER, CHRISTEL ROTTHAUS AND SYLVIA WIEGAND

Example 2.5. (cf. [HRW2, Example 2.1]) In ℚ[[x, y]], the power series ring in the two variables x and y over the rational numbers, let γ := e^{x}−1 and τ := e^{y} −1; take γ_{n} to be the n^{th}-endpiece of γ with respect to x and take τ_{n} to be the n^{th}-endpiece of τ with respect to y, as described in (2.1). Set R := ∪_{n∈ℕ}ℚ[x, y, γ_{n}]_{(x,y,γ_{n})}. Then R = ℚ[y]_{(y)}[[x]] ∩ ℚ(x, y, γ) is an excellent two-dimensional regular local domain. Now define U in the (y)-adic completion of R using the endpieces τ_{n} as above. Then U ⊇ V := ∪_{n∈ℕ}ℚ[x, y, γ_{n}, τ_{n}]. The ring A := ℚ[[x, y]] ∩ ℚ(x, y, γ, τ) is Noetherian but is different from B := ∪ℚ[x, y, γ_{n}, τ_{n}]_{(x,y,γ_{n},τ_{n})}. The ring B is the localization of U at the multiplicative system 1+yU, and the rings B and U are not Noetherian. It follows that A is not a localization of U.

Proof. Consider the element θ = \frac{γ−τ}{x−y} ∈ A. If θ is an element of B, then

γ − τ ∈ (x − y)B ∩ V = (x − y)V.

Now

V = ∪_{n∈ℕ}ℚ[x, y, γ_{n}, τ_{n}] ⊆ ℚ[x, y, γ, τ][1/x, 1/y] ⊆ ℚ[x, y, γ, τ]_{(x−y)},

and so

γ − τ ∈ (x − y)ℚ[x, y, γ, τ]_{(x−y)} ∩ ℚ[x, y, γ, τ] = (x − y)ℚ[x, y, γ, τ],

but this contradicts the fact that x, y, γ, τ are algebraically independent over ℚ.

If U were Noetherian, then B would be Noetherian. But the maximal ideal of B is (x, y)B, so if B were Noetherian, then it would be a regular local domain with completion ℚ[[x, y]]. Since the completion of a local Noetherian ring is a faithfully flat extension of it, and since the fraction field of B is ℚ(x, y, γ, τ), then B would equal A.

That A is Noetherian follows from [V, Proposition 3]. If A were a localization of U, then A would be a localization of B. But each of A and B has a unique maximal ideal and the maximal ideal of A contains the maximal ideal of B. Therefore B ⊊ A implies that A is not a localization of B. □

3. The proof of the main theorem.

Proof of Theorem 1.1. Assume that U is Noetherian. By (2.2), the (a)-adic completion U^{∗} of U is equal to R^{∗}. Since U is Noetherian, U^{∗} = R^{∗} is flat over

BUILDING NOETHERIAN DOMAINS 7

U [M1, Theorem 8.8]. Therefore the localization R^{∗}[1/a] is flat over U. Since U[1/a] = U_{0}[1/a], the localization R^{∗}[1/a] is also flat over U_{0}.

To prove the converse we use results of Heitmann in [H1, Theorem 1.4].

First we show in (3.1) that the flatness condition for R^{∗}[1/a] over U_{0} behaves well under certain residue class formations. Proposition 3.1. Let R be a Noetherian domain, let a be a nonzero nonunit of R, let R^{∗} be the y-adic completion of R and let τ_{1}, …, τ_{s} ∈ aR^{∗} be algebraically independent over R. Suppose that R_{a}^{∗} := R^{∗}[1/a] is flat over U_{0}, using the notation of (2.1) and that Q is a prime ideal of R with a ∉ Q. Assume that Q is the contraction of a prime ideal of R^{∗}. Let ¯ denote image in R_{a}^{∗}/QR_{a}^{∗} and let (R/Q)^{∗} denote the (ā)-adic completion of R/Q. Then (R/Q)∗ ; ā := (R/Q)^{∗}[¯1/ā] is flat over (R/Q)[τ̄_{1}, …, τ̄_{s}]. Proof. The (ā)-adic completion (R/Q)^{∗} of R/Q is canonically isomorphic to R^{∗}/QR^{∗}. Therefore τ̄_{1}, …, τ̄_{s} are regular elements of (R/Q)^{∗}. We show τ̄_{1}, …, τ̄_{s} are algebraically independent over R/Q. Since R[τ_{1}, …, τ_{s}] ⟶ R_{a}^{∗} is flat, a ∉ Q, and Q is the contraction of a prime ideal of R^{∗}, we have QR[τ_{1}, …, τ_{s}] = QR_{a}^{∗}∩R[τ_{1}, …, τ_{s}]. Thus

R[τ_{1}, …, τ_{s}]/(QR_{a}^{∗} ∩ R[τ_{1}, …, τ_{s}]) ≅ (R/Q)[τ̄_{1}, …, τ̄_{s}]

is a polynomial ring in s variables τ̄_{1}, …, τ̄_{s} over R/Q. Therefore τ̄_{1}, …, τ̄_{s} are algebraically independent over R/Q.

We show flatness of the map:

ϕ̄: (R/Q)[τ̄_{1}, …, τ̄_{s}] ⟶ R_{a}^{∗}/QR_{a}^{∗} = (R/Q)∗ ; ā.

Let P̄ be a prime ideal of R^{∗}/QR^{∗} with ā ∉ P̄. The ideal P̄ lifts to a prime ideal P of R^{∗} with a ∉ P and QR^{∗} ⊆ P. By assumption the map

ϕ_{P} : R[τ_{1}, …, τ_{s}] ⟶ R_{P}^{∗}

is flat. The map on the residue class rings:

ϕ̄P̄ : (R/Q)[τ̄_{1}, …, τ̄_{s}] ⟶ (R^{∗}/QR^{∗})P̄

is obtained from ϕ_{P} by tensoring with (R/Q)[τ_{1}, …, τ_{s}] over the ring R[τ_{1}, …, τ_{s}]. Hence ϕ̄ is flat. □

8 WILLIAM HEINZER, CHRISTEL ROTTHAUS AND SYLVIA WIEGAND

Theorem 3.2. Assume the notation and setting of (2.1). Also assume that s = 1, τ := τ_{1} and that the localization R^{∗}[1/a] is flat over U_{0} = R[τ]. Then U is Noetherian and A = R^{∗} ∩ K(τ) is a localization of U.

We use the same proof as in [H1, Theorem 1.4] and prove first the following lemma.

Lemma 3.3. With notation as in Theorem 3.2, if P is a nonzero prime ideal of U such that P ∩ R = (0), then there exists f ∈ P, r ∈ R and a positive integer N such that P = (fU :_{U} ra^{N}).

Proof. The localization D := (R − {0})^{−1}U of U at the nonzero elements of R is also a localization of the polynomial ring U_{0} := R[τ]. Hence P D is a a principal maximal ideal of D and there exists a polynomial f ∈ R[τ] such that P D = fD.

We use the fact that U is the directed union of the polynomial rings U_{n} := R[τ_{n}], U = ∪_{n=0}^{∞}U_{n}. Let P_{n} = P ∩ U_{n}. Since D_{P D} = (U_{0})_{P_{0}} and U_{0} is Noetherian, there exists r ∈ R such that P_{0} = (fU_{0} :_{U_{0}} r). Also for g ∈ U there exists a positive integer b(g), depending on g, such that a^{b(g)}g ∈ U_{0}. Hence for g ∈ P we have ra^{b(g)}g ∈ fU_{0}.

The Artin-Rees Lemma [N1, (3.7)] applied to the ideals aR^{∗} and fR^{∗} of the Noetherian ring R^{∗} implies the existence of a positive integer N such that for m ≥ N we have

fR^{∗} ∩ (aR^{∗})^{m} = (aR^{∗})^{m−N}((fR^{∗} ∩ (aR^{∗})^{N}) = (a^{m−N})R^{∗}(fR^{∗} ∩ a^{N}R^{∗}).

We may assume that b(g) ≥ N.

Suppose g ∈ P. Then ra^{b(g)}g ∈ fU_{0} ⊆ fU, so

ra^{b(g)}g ∈ fR^{∗} ∩ a^{b(g)}R^{∗} = a^{b(g)−N}R^{∗}(fR^{∗} ∩ a^{N}R^{∗}).

Since a is not a zero-divisor in R^{∗}, it follows that ra^{N}g ∈ fR^{∗} ∩ a^{N}R^{∗}. Thus ra^{N}g = ft, where t ∈ R^{∗}. Since we also have ra^{b(g)}g ∈ fU, it follows that a^{b(g)−N}ft ∈ fU, and therefore a^{b(g)−N}t ∈ U, as f is not a zero-divisor in R^{∗}. Therefore a^{b(g)−N}t ∈ a^{b(g)−N}R^{∗} ∩ U = a^{b(g)−N}U by (2.2.1) and so t ∈ U. Hence for every g ∈ P we have g ∈ (fU :_{U} ra^{N}). It follows that P = (fU :_{U} ra^{N}). □

As in [H1, Lemma 1.5], we have:

BUILDING NOETHERIAN DOMAINS 9

Lemma 3.4. With notation as in Theorem 3.2, if each prime ideal P of U such that P ∩ R ≠ (0) is finitely generated, then U is Noetherian.

Proof. By a Theorem of Cohen [N1, (3.4)], it suffices to show each P ∈ Spec(U) such that P ∩ R = (0) is finitely generated. Let P be a nonzero prime ideal of U such that P ∩ R = (0). Since the localization of U at the nonzero elements of R is also a localization of the polynomial ring U_{0} := R[τ], every prime ideal of U properly containing P has a nonzero intersection with R. Therefore the hypothesis implies that U/P is Noetherian. By (3.3), there exist r ∈ R and f ∈ P such that P = (fU :_{U} ra^{N}). Since ra^{N} is a nonzero element of R, every prime ideal of U containing ra^{N} is finitely generated, so U/ra^{N}U is Noetherian. Therefore U/(P ∩ ra^{N}U) is Noetherian [N1, (3.16)]. Since ra^{N} ∉ P and P is prime, we have ra^{N}U ∩ P = ra^{N}P. Therefore U/ra^{N}P is Noetherian. We have ra^{N}P ⊆ fU ⊆ P. Hence U/fU, as a homomorphic image of U/ra^{N}P, is Noetherian, and P/fU is finitely generated. It follows that P is finitely generated. □

Proof of Theorem 3.2. Suppose U is not Noetherian and let Q ∈ Spec(R) be maximal with respect to being the contraction to R of a non-finitely generated prime ideal of U. Since R/aR = U/aU = R^{∗}/aR^{∗} by (2.2), we have a ∉ Q. Since U = ∪_{n=0}^{∞}U_{n} and QU_{n} is prime, we have QU is prime in U. We claim that Q is the contraction of a prime ideal of R^{∗}, for otherwise we have (Q, a)R = R. But this means that the image of a in U/QU is a unit which implies that U/QU = U_{0}/QU_{0} is Noetherian, and this implies that P is finitely generated. Therefore Q is the contraction of a prime of R^{∗}, and (3.1) implies that, passing to the image τ̄ of τ in U/QU, the localization (R/Q)∗ ; ā is flat over (R/Q)[τ̄]. But Lemma 3.4 then implies that U/QU is Noetherian. This contradicts the existence of a non-finitely generated prime ideal of U lying over Q in R. We conclude that U is Noetherian. Therefore U^{∗} = R^{∗} is flat over U and if S is the multiplicative system 1 + aU, then S^{−1}U = R^{∗} ∩ K(τ). □

Remark 3.5. The proof of Theorem 3.2 is essentially due to Ray Heitmann. In his paper [H1] Heitmann defines simple PS-extensions. For a regular element x in a ring R and a formal power series in x transcendental over R, a simple PSextension of R for x is an infinite direct union of simple transcendental extensions of R. If R is Noetherian and T is a simple PS-extension of R, Heitmann proves in [H1, Theorem 1.4] that a certain monomorphism condition is equivalent to T being

Noetherian. Heitmann’s monomorphism condition insures that the element f in

the proof of Lemma 3.3 is a regular element in R^{∗}. In our situation our flatness

condition on the embedding U_{0} ⟶ R_{a}^{∗}, and hence on U ⟶ R_{a}^{∗}, implies the

regularity of f in R^{∗}. Thus Proposition 3.1 yields that if s = 1 and the embedding

U_{0} ⟶ R_{a}^{∗} is flat, then the ring U = \underrightarrow{lim} R[τ_{n}] is a simple PS-extension satisfying

the monomorphism condition of Heitmann. In view of Theorem 1.1, Heitmann’s

monomorphism condition on the PS-extension determined by τ is equivalent to τ

yielding a flat extension. The flat extension concept however extends to more than

one element τ.

Completion of Proof of Theorem 1.1. If U is Noetherian, we have already shown

that R^{∗}[1/a] is flat over U_{0}. Assume, conversely, that R^{∗}[1/a] is flat over U_{0} =

R[τ_{1}, …, τ_{s}]. It follows that R^{∗}[1/a] is flat over R[τ_{1}]. By Theorem 3.2, U(1),

the directed union ring constructed with respect to τ_{1} in (2.1) is Noetherian and

R^{∗} ∩ K(τ_{1}) is a localization of U(1). It also follows that U(1)^{∗}[1/a] = R^{∗}[1/a] is

flat over U(1)[τ_{2}, …, τ_{s}] (cf. [HRW2, Proposition 5.10]). Hence a simple induction

argument implies that U is Noetherian. Hence U^{∗} = R^{∗} is flat over U and A is a

localization of U. □

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Department of Mathematics, Purdue University, West Lafayette, IN 47907-1395

Department of Mathematics, Michigan State University, East Lansing, MI 48824-1027

Department of Mathematics and Statistics, University of Nebraska, Lincoln, NE 68588-0323