Ring of Formal Laurent Series which are Dedekind domains

Question

Let $R$ be an integral domain and $R((x))$ be the ring of formal Laurent series over $R$. (The answer to this question has a good explanation for our ring.)

Is it true that $R$ is a Dedekind domain if and only if $R((x))$ is a Dedekind domain? If not, is there any better necessary and sufficient condition for $R((x))$ to be a Dedekind domain?

Note that in case of $R[x]$, $R[x]$ is a Dedekind domain if and only if $R$ is a field. (We can use a similar proof to this one.) The same goes for $R[[x]]$ with a similar proof, and for $R(x)$ since $R(x)=S^{-1}R[x]$ where $S=\{1,x,x^2,\ldots\}$. Unfortunately, the same thing doesn't happen in $R((x))$, since if $R$ is a field, then $R((x))$ is a field also. (The proof is also available in that earlier question whose link I've put above.) If we weaken the condition of $R$ to be a Euclidean domain, then $R((x))$ will also be a Euclidean domain. (This has been discussed here.) Hence, of course $R$ being a field is not a necessary condition to make $R((x))$ a Dedekind domain. Thus, a question sparks in my head:

What is a necessary and sufficient condition for $R$ to make $R((x))$ a Dedekind domain?

Thank you very much in advance.

Answer 1

If I am not mistaken the following sequence of arguments eventually shows that if $R$ is a principal ideal domain, then $R((x))$ is a principal ideal domain too. I thought that from this fact through an appropriate localization argument one could show that $R((x))$ is Dedekind if $R$ is. But this turned out to be more complicated than expected - if it is true at all.

The case of a field $R$ is well-known.

In the sequel rings are always commutative .

(1) For an arbitrary ring $R$ the map $\phi: R[[x]]\rightarrow R,\;\sum\limits_{k=0}^\infty a_kx^k\mapsto a_0$ is a surjective ring homomorphism.

(2) For a noetherian ring $R$ the rings $R[[x]]$ and $R((x))$ are noetherian.

The first statement is a well-known extension of Hilbert's Basis Theorem. The second statement follows from $R((x))=\{1,x,x^2,\ldots\}^{-1}R[[x]]$.

(3) Let $R$ be a principal ideal ring and let $I\subseteq R[[x]]$ be a prime ideal with the property $x\not\in I$, then $I$ is a principal ideal.

Proof: let $s$ be a generator of $\phi(I)$ and $g\in I$ an element with $\phi(g)=s$. For an arbitrary $f=\sum\limits_{k=0}^\infty a_kx^k\in I$ one then has $a_0=b_0s$, hence $f-b_0g=c_1x+c_2x^2+\ldots\in I$. Since $x\not\in I$ this shows that $f$ has the form

$f=b_0g+xh_0,\; b_0\in R,h_0\in I.$

Recursively this gives a sequence $b_0,b_1,b_2\ldots\in R$ such that

$f=b_0g+b_1xg+b_2x^2g+\ldots =(b_0+b_1x+b_2x^2+\ldots )g$,

hence $g$ generates $I$.

(4) If $R$ is a principal ideal domain, then $R((x))$ is a principal ideal domain.

Proof: the prime ideals of $R((x))$ by localization correspond bijectively to the prime ideals of $R[[x]]$ not containing $x$. By (3) these primes are principal. We conclude that $R((x))$ is a noetherian domain in which every prime ideal is principal. In particular $\dim(R((x)))=1$ by Krull's Principal Ideal Theorem. Now by (Matsumura, Commutative Ring Theory, Thm . 20.1) one concludes that $R((x))$ is factorial and hence a principal ideal domain.

Note: the factoriality of $R((x))$ by (Matsumura, Commutative Ring Theory, Thm . 20.2) implies the factoriality of $R[[x]]$. This is interesting, because in general it is not true that $R[[x]]$ is factorial if $R$ is.

Answer 2

Following a path different from the one started with in my previous post it seems that one can indeed prove that $R((x))$ is a Dedekind domain if $R$ is a Dedekind domain: In the sequel I am always assuming that $R$ is not a field. Dedekind domains can then be characterized as 1-dimensional, noetherian, integrally closed domains. By (2) of my previous post it therefore suffices to prove $\dim(R((x)))=1$ and that $R((x))$ is integrally closed.

Integral closedness: In general it is not true that $R((x))$ is integrally closed if $R$ is. By (R.Gilmer, Multiplicative Ideal Theory, Theorem 13.09) however $R((x))$ is integrally closed if $R$ is completely integrally closed and by (R.Fossum, Divisor Class Group of Krull Domains, Theorem 3.6) and (Matsumura, Commutative Ring Theory, Theorem 12.4) every noetherian, integrally closed domain is completely integrally closed. Hence for a Dedekind domain $R$ the ring $R((x))$ is integrally closed.

Dimension: Let $I$ be a proper ideal of $R[[x]]$ with the property $x\not\in I$. Then the ideal generated by $I$ and $x$ is a proper ideal: assume that $(I,x)=R[[x]]$. Then there exists $f\in R[[x]]$ and $g\in I$ such that $g+xf=1$. Therefore $g=1-xf$, which shows that $g$ is a unit, a contradiction.

The property just proven shows that prime ideals $I$ of $R[[x]]$ with $x\not\in I$ are not maximal. Now if $R$ has dimension $1$ one knows (R.Gilmer, Multiplicative Ideal Theory, Theorem 30.06) that $\dim(R[[x]])=\dim(R)+1=2$, hence such ideals must have height $1$ or $0$. Passing to $R((x))$ by localization at $\{1,x,x^2,\ldots\}$ yields $\dim(R((x)))=1$.

Note: as for the conclusion $R((x))$ Dedekind implies $R$ Dedekind (R. Gilmer, Multiplicative Ideal Theory, Theorem 13.10) at least shows that $R$ is integrally closed.