NOETHERIAN INTEGRALLY CLOSED DUO RINGS ARE KRULL RINGS

Elbert M. Pirtle University of Missouri - Kansas City

Introduction. A ring R with unity is called a duo ring if every right ideal of R is a left ideal and conversely. This is equivalent to the condition that aR = Ra for all a ∈ R . When R is an integral domain, R has a (left and right) division ring D of quotients, and xRx^{−1} = R for all nonzero x ∈ D . In this sense R is an invariant subring of D. In this note we characterize Noetherian, integrally closed duo domains [1] in terms of valuation rings exactly as commutative Krull domains are characterized. We continue in this manner to characterize duo domains which are unique factorization domains [3] as Krull domains in which the height 1 prime ideals are principal. We then show that when R has an Abelian group of divisibility, then R is integrally closed if and only if R is an intersection of valuation rings. Duo domains which are Prüfer domains with Abelian group of divisibility are characterized exactly as commutative Prüfer domains.

In what follows, R denotes a duo ring which is an integral domain, D its division ring of quotients. Prime ideals in duo rings have the same characterization as in the commutative case [10]. Thus P is a prime ideal of R if and only if S = R\P is a multiplicative system in R.

A subset T of R is called invariant if aT = T a for all a ∈ R. This is clearly equivalent to the condition that x^{−1}T x = T for all x ∈ D^{∗} = D\{0}. Invariance is significant when studying localization in duo rings. The following was show in [6].

\underline{T heorem 1}. If S is a multiplicative system in R then R_{S} is a duo ring if and only if S is invariant.

Let U denote the multiplicative group of units of R. It was shown in [3] that U is a normal subgroup of D^{∗} . As in the commutative case, D^{∗}/U is called the group of divisibility of R.

When R is Noetherian and integrally closed, it is shown in [1] that for all a, b ∈ R, aRbR = bRaR. Since aRbR = abR, this is equivalent to the condition that for all a, b ∈ R, ab = bau, for some u ∈ U . We can now state the following.

\underline{T heorem 2}. In a duo domain R, the following are equivalent.

(1) The multiplicative semigroup of ideals of R is Abelian.

(2) The multiplicative semigroup of principal ideals of R is Abelian.

(3) The group of divisibility of R, D^{∗}/U , is Abelian.

(4) Every ideal of R is invariant.

(5) Every principal ideal of R is invariant.

\underline{P roof}. Since every ideal is a sum of principal ideals (1) ⇔ (2), and (4) ⇔ (5). (2) ⇔ (3) follows from the remarks preceding Theorem 2. (5) ⇔ (3) is easy to show.

We say that R satisfies property (A) if it satisfies the equivalent conditions of Theorem 2.

\underline{Corollary 3} [1]. If R satisfies property (A) then every overring T of R (R ⊆ T ⊆ D) is a duo ring. In particular R_{S} is a duo ring for every multiplicative system S in R.

When R is Noetherian and integrally closed, in addition to satisfying property (A), the following was shown in [1]: If {P_{λ}} denotes the collection of height 1 prime ideals of R then

(a) R = ⋂_{λ} R_{P_{λ}}.

(b) P_{λ}R_{P_{λ}} = p_{λ}R_{P_{λ}} for some p_{λ} ∈ P_{λ} .

(c) The only nonzero ideals of p_{λ}R_{P_{λ}} are the ideals of p_{λ}^{n}R_{P_{λ}} .

If we define v_{λ} : R_{P_{λ}} → (Z, +) ∪ {∞} by v_{λ}(x) = n where n is the unique integer such that x ∈ p_{λ}^{n}R_{P_{λ}} − p_{λ}^{n+1}R_{P_{λ}} then v_{λ} satisfies the following: (I) v_{λ}(0) = ∞ , (II) v_{λ}(xy) = v_{λ}(x + y) for all x , y ∈ R_{P_{λ}} , (III) v_{λ}(x + y) ≥ min(v(x) , v(y)) for x , y ∈ R_{P_{λ}} .

Non-commutative valuations and noncommutative valuation rings were studied in [7] and [8].

Let (G, •) be a totally ordered group. Let ∞ be a symbol not in G. Let v : D → G ∪ {∞}. We define x ∗ ∞ = ∞ = ∞ ∗ x for any x ∈ G and x < ∞ for all x ∈ G.

\underline{Definition 4} [8]. v is called a valuation on D if

(1) v(0) = ∞.

(2) v(xy) = v(x)v(y) for any x, y ∈ D.

(3) v(x + y) ≥ min(v(x) , v(y)) . (Here the operation in G is denoted by juxtaposition.)

\underline{T heorem 5} [12]. Let v : R → G ∪ \{∞\} satisfy (1), (2), (3) above. There is a unique extension v′ of v to a valuation on D.

\underline{P roof}. For x = b^{−1}a ∈ D^{∗}, define v′(x) = v(b)^{−1}v(a). Clearly (1) is satisfied. To see (2), let x = b^{−1}a, y = d^{−1}c. If x = 0 or y = 0,

(2) is clear. Suppose x , y ≠ 0 . Then v′(xy) = v′(b^{−1}ad^{−1}c) . Now ad = d_{1}a for some d_{1} ∈ R, so ad^{−1} = d_{1}^{−1}a, so

v′(xy) = v′(b^{−1}d_{1}^{−1}ac) = v′((d_{1}b)^{−1}(ac)) = v(d_{1}b)^{−1}v(ac)
= v(b)^{−1}v(d_{1})^{−1}v(a)v(c) = [v(b)^{−1}v(a)][v(d)^{−1}v(c)]
= v′(x)v′(y) ,

and (2) is proved.

To see (3), let x = b^{−1}a, y = d^{−1}c. If x+y = 0 there is nothing to show. So suppose x + y ≠ 0 . Then x + y = (b_{1}d)^{−1}(d_{1}a + b_{1}c) , where d_{1}b = b_{1}d.

v′(x + y) = v(b_{1}d)^{−1}v(d_{1}a + b_{1}c)
≥ v(b_{1}d)^{−1} min(v(d_{1}a) , v(b_{1}c))
= min(v(b_{1}d)^{−1}v(d_{1}a) , v(b_{1}d)^{−1}v(b_{1}d)) .

Now d_{1}b = b_{1}d, so v(d_{1}b) = v(b_{1}d) , and hence

v(b_{1}d)^{−1} = v(b)^{−1}v(d_{1})^{−1} = v(d)^{−1}v(b_{1})^{−1} .

So v′(x + y) ≥ min(v′(b^{−1}a) , v′(d^{−1}c)) , that is,

v′(x + y) ≥ min(v′(x) , v′(y)) .

Uniqueness is clear, since extension w of v satisfies w(b^{−1}a) = w(b^{−1})w(a) = w(b^{−1})v(a)v(b)^{−1}v(a) . For w(b^{−1}b) = w(1) = e , the identity of G, and w(b^{−1}) = w(b)^{−1} = v(b)^{−1} .

We observe that the proof of Theorem 5 did not require that R satisfy property (A).

It follows from Theorem 5 that each v_{λ} can be extended to a valuation v_{λ} on D, and that each valuation ring R_{v_{λ}} = R_{P_{λ}} and is rank one, discrete.

So when R is Noetherian and integrally closed, the family {P_{λ}} of height 1 prime ideals induces a family F of valuations on D which has the following properties.

(i) R = ⋂{R_{v} : v ∈ F }. (ii) Each v ∈ F has rank one and is discrete. (iii) F is of finite character, i.e., for a ∈ R , v(a) ≠ 0 for only

finitely many v ∈ F . (iv) R_{v} = R_{P (v)} , where P (v) = {x ∈ R : v(x) > 0}.

So a Noetherian, integrally closed duo domain satisfies the classical definition of a commutative Krull domain [12].

Accordingly, we shall call a duo domain R a Krull domain if there is a family F of valuations on the division ring D of quotients of R satisfying (i), (ii), (iii), (iv) above. Clearly, if R is a Krull domain then R satisfies property (A) since each R_{v} has an Abelian group of divisibility. Thus every multiplicative system S in a Krull domain is invariant.

We now state the following. The proof is like the commutative case.

\underline{T heorem 6}. Let R be a Krull domain with family F of essential valuations. Let S be a multiplicative system in R. Let

H = {v ∈ F : P (v) ∩ S = ∅} .

Then R_{S} is a Krull domain with H as family of essential valuations.

Duo rings which are unique factorization domains (UFD) were first studied in [3, pg. 85]. We state definitions from [3].

\underline{Definition 7}. Let R be a duo domain.

(1) x ∈ R is a prime if xR = Rx is a prime ideal of R.

(2) x, y ∈ R are associates if Rx = Ry . (Then xR = yR.)

(3) For a, b ∈ R, a divides b (a|b) if bR ⊆ aR.

(4) x ∈ R is irreducible if x = yz implies either y or z is a unit.

(5) R is called a UFD if the following are satisfied.

\underline{UF 1}. Every nonzero element is a product of a finite number of irreducible elements.

\underline{UF 2}. If a_{1}a_{2} ⋯ a_{s} = b_{1}b_{2} ⋯ b_{t} where a_{1} , a_{2} , …, a_{s} , b_{1} , b_{2} , …, b_{t} are nontrivial irreducibles, then s = t and for a suitable ordering of the subscripts, a_{i} is an associate of b_{i} .

It follows from [3] that if R is UFD then irreducible elements are prime.

\underline{P roposition 8}. If R is UFD and if x , y ∈ R are nonassociate prime elements then (xR)(yR) = (yR)(xR) , i.e., xyR = yxR.

\underline{P roof}. Since R is duo, xy = yx′ for some x′ ∈ R . Then y^{−1}xy = x′ , and y^{−1}xyR = y^{−1}(xR)y = x′R . Then x′R is the conjugate of a prime ideal and so is a prime ideal. So x′ ∈ R is prime and so by UF2, x′ is an associate of x, say x′ = xt where t is a unit. Then xyR = yxtR = yxR.

The above proposition shows that if R is UFD then R satisfies property (A), by UF1.

Let {x_{λ}} denote a collection of primes of R such that

(a) x_{α} is not an associate of x_{β} for α ≠ β .

(b) If r ∈ R is not a unit, then x_{λ}|r for some λ.

Let P_{λ} = x_{λ}R . Then as in [12], each x_{λ}R_{P_{λ}} is a rank one discrete valuation ring, and it is like the commutative case to show that each P_{λ} induces a family F of valuations satisfying (i), (ii), (iii), (iv).

\underline{T heorem 9}. Let R be a duo domain with D as a division ring of quotients. D is UFD if and only if

(1) R is a Krull domain.

(2) Every height 1 prime ideal is principal.

The proof is like the commutative case [12].

As in the commutative case, an element x ∈ D is integral over R if there exist a_{0} , …, a_{n−1} ∈ R such that x^{n} + a_{n−1}x^{n−1} + ⋯ + a_{1}x + a_{0} = 0 . This definition is symmetric:

\begin{aligned}x^{n} & + a_{n−1}x^{n−1} + ⋯ + a_{1}x + a_{0} \\ & = x^{n} + x^{n−1}(x^{n−1})^{−1}a_{n−1}x^{n−1} + ⋯ + x(x^{−1}a_{1}x) + a_{0} , \end{aligned}

and (x^{j})^{−1}a_{j}x^{j} ∈ R, since R is duo, for j = 1 , …, n − 1 . It was shown in [7] that (noncommutative) valuation rings are integrally closed. As was shown earlier, when R is Noetherian and integrally closed, then R is an intersection of valuation rings. It is well known [12] that an integrally closed commutative integral domain is an intersection of valuation rings. We have been able to prove the following:

\underline{T heorem 10}. Let R be a duo ring satisfying property (A). Then R is integrally closed if and only if R is an intersection of valuation rings.

The proof of Theorem 10 will follow from Lemmas 11 and 12.

\underline{Lemma 11} [12]. Let I be an ideal in R. Then for any x ∈ D , at least one of R[x]I , R[x^{−1}]I is a proper ideal of R[x] , R[x^{−1}] respectively.

\underline{P roof}. The proof is like the commutative case, using property (A) in R and the invariance of R. See [12, pg. 11].

\underline{Lemma 12} [12]. Let P be a prime ideal of R . There exists a valuation overring V of R (R ⊆ V ⊆ D) with maximal ideal P such that P ∩ R = P .

\underline{P roof}. We may assume R = R_{P} , so that P is the unique maximal ideal of R . Let S = {T : T is the overring of R , P T ≠ T } , S ≠ ∅. A Zorn’s lemma argument shows S has a maximal element, say T . The maximality of P in R_{P} gives that P T is maximal. If not, then P T < M for some maximal ideal M of T and then T < T_{M} , T_{M} ∈ S , a contradiction. Clearly P T ∩ R = P .

Applying Lemma 11 to R = T, I = P T, as in the commutative case [12, pg. 12] we get that for x ∈ D, either x ∈ T or x^{−1} ∈ T , and T is a valuation ring, since T is duo.

So if b ∈ D\R , then consider R[b^{−1}] (see [5, pg. 38]). If b^{−1} (∑_{i=0}^{n} a_{i}(b^{−1})^{i}) = 1 , a_{0} , …, a_{n} ∈ R, then b is integral over R, which is not the case if R is integrally closed. So b^{−1} is not a unit in R[b^{−1}] . Then b^{−1} belongs to a maximal, hence prime ideal of R[b^{−1}] . Applying Lemma 13, we get a valuation ring V , R ⊆ R[b^{−1}] ⊆ V ⊆ D with b^{−1} a nonunit in V . Then b ∉ V . So if R is integrally closed it is the intersection of valuation rings.

Since (noncommutative) valuation rings are integrally closed [7], this completes the proof of Theorem 10.

Strong use of property (A) was made in the proof. The author believes the following general case to be true but has been unable to prove it: For any duo domain R, R is integrally closed if and only if R is an intersection of valuation rings.

As in the commutative case, R is called a Dedekind domain if R is a Krull domain in which each height 1 prime ideal is maximal, i.e., P (v) is maximal for each v ∈ F. Since R satisfies property (A), all the equivalent conditions for a commutative integral domain to be a Dedekind domain are satisfied in R. So R is Dedekind if and only if every nonzero ideal of R is invertible [5], [4].

The following is found in [9]: A (not necessarily commutative) ring T is called a left AM-ring (respectively right AM-ring) is for all ideal A < B of T (here < denotes proper containment) there is an ideal C such that A = BC (respectively A = CB ). T is an AM-ring if it is both a left and right AM-ring. (In the commutative case, T is sometimes called a multiplication ring [4, pg. 209].)

Theorem 3.1 of [9] shows that if R is an AM-ring then R satisfies property (A). As in the commutative case, we have the following [4, pg. 210].

\underline{T heorem 13}. If R is an AM-ring then R is a Dedekind domain.

\underline{P roof}. Let A be a nonzero ideal of R . If A = Ra then A is invertible. If A is not principal, then there is nonzero a ∈ A such that Ra < A. Then Ra = AC for some ideal C , and A is invertible, with A^{−1} = CRa^{−1} . Thus R is Dedekind.

We call R a Prüfer domain if R_{M} is a (noncommutative) valuation ring for each maximal ideal M of R. Noncommutative valuation rings [7] are examples of duo rings which are Prüfer domains. Ten equivalent conditions for a commutative integral domain to be a Prüfer domain are given in Theorem 6.6 of [4]. Theorem 6.6 of [4] also holds when R is a duo domain which satisfies property (A). The proof is the same as the commutative case because of property (A).

An example of a noncommutative valuation ring in which R_{P} is not a valuation ring for every prime P was given in [7]. Thus there exist noncommutative valuation rings which do not satisfy the ten equivalent conditions of Theorem 6.6 of [4].

When R is commutative, it is well known that if R is a Krull domain, then so is the polynomial ring R[X] . Commutativity appears to be necessary, for even when R is duo and satisfies property (A), R[X] is in general not a duo ring. For example, if D is a noncommutative division ring, then D satisfies property (A) but D[X] is not a duo ring.

An example of a duo domain which is a Krull domain is the power series ring D[[X]] where D is any division ring and X is a central indeterminate. D[[X]] is a noncommutative rank one discrete valuation ring with XD[[X]] as the unique maximal ideal.

So D[[X]] is a noncommutative duo UFD, and a Dedekind domain.

Other examples of integrally closed duo rings which satisfy property (A) may be constructed from results in [2]. If D is a 2finite division algebra over its center k and satisfies the equivalent conditions of Theorem 1 of [2], then any real valued valuation v on k may be extended to a valuation w on D . So the valuation w has Abelian value group, and R_{w}, the valuation ring of w, satisfies property (A) and is integrally closed. If {R_{w_{λ}}} is any collection of such valuation rings, then R = ∩_{λ}R_{w_{λ}} is a duo ring which satisfies property (A) and is integrally closed.

\underline{References}

1. H. H. Brungs, “Three Questions on Duo Rings,” Pacific J. Math., 58(1975) 345–349.

2. P. M. Cohn, “On Extending Valuations in Division Algebras,” Studia Sci. Math. Hung., 16(1981) 65–70.

3. E. H. Feller, “Properties of Primary Non-Commutative Rings,” Trans. Amer. Math. Soc., 89(1958) 79–91.

4. M. D. Larsen and P. J. McCarthy, Multiplicative Theory of Ideals, Academic Press, New York, 1971.

5. M. Nagata, Local Rings, Interscience, New York, 1962.

6. E. M. Pirtle, “Localization in Duo Rings,” Pub. Math. Debrecen, 31 (1984) 47–52.

7. E. M. Pirtle, “Noncommutative Valuation Rings,” Pub. de l’Institut Mathematique, 39 (1986) 83–87.

8. E. F. G. Schilling, “Noncommutative Valuations,” Bull. Amer. Math. Soc., 51 (1945) 297–304.

9. P. F. Smith, “On Non-Commutative AM-Rings,” Houston Journal of Math., 11 (1985) 405–422.

10. G. Thierrin, “On Duo Rings,” Can. Math. Bull., 3 (1960) 167– 11. O. Zariski and P. Samuel, Commutative Algebra, New York,

Vol. I, 1958.

12. O. Zariski and P. Samuel, Commutative Algebra, New York, Vol. II, 1962.