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.
Best Answer
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.