Completeness of Localizations of Completions of Commutative Rings

Question

Let $R$ be an integral domain. Let $x,y\in R\setminus\{0\}$ be distinct. Let $\hat R$ be the $x$-adic completion of $R$ (the ring of all sequences $(r_n+Rx^n)_{n\ge0}$ where for $n\ge0$, $r_n\in R$ and $r_n+Rx^{n}=r_{n+1}+Rx^n$). Denote the image of $r\in R$ under the canonical ring-homomorphism by $\hat r$. Localize $\hat R$ using the multiplicative subset $\{1,\hat y,\hat y^2,\hat y^3,\dots\}$ to get the ring $\hat R_{\hat y}$. Is $\hat R_{\hat y}$ $\frac{\hat x}{1_{\hat R}}$-complete? By this we mean the following: Consider a sequence $\alpha_1,\alpha_2,\alpha_3,\dots$ in $\hat R_{\hat y}$ and let $\beta\in\hat R_{\hat y}$. The sequence is ${\it Cauchy}$ if for all $n\ge1$, there exists $N\ge1$ such that for all $h,k\ge N$, $\frac{\hat x}{1_{\hat R}}^n$ divides $\alpha_h-\alpha_k$. The sequence has ${\it limit}$ $\beta$ if for all $n\ge1$, there exists $N\ge1$ such that for all $k\ge N$, $\frac{\hat x}{1_{\hat R}}^n$ divides $\beta-\alpha_k$. We want every Cauchy sequence to have a limit.

If need be, you can assume $R$ is a UFD and $x$ and $y$ are distinct primes and you can assume the canonical homomorphism is 1-1.

Answer 1

No, $\hat{R}_y$ need not be $\hat{x}$-adically complete. The polynomial ring $R=k[x,y]$ is a counterexample. The $x$-adic completion of $R$ is identified with $k[y][[x]]$, the ring of power series in $x$ whose coefficients are polynomials in $y$. If we invert $y$, we get a ring whose elements are power series in $x$ with coefficients in $k[y,y^{-1}]$ having bounded powers of $y$ in the denominators. The $x$-adic completion of $\hat{R}_{y}$ should be $k[y,y^{-1}][[x]]$. Explicitly, the sequence $$ \alpha_n:=\sum_{i=0}^n \frac{x^i}{y^i}\in\hat{R}_y $$ is Cauchy ($x^n$ divides $\alpha_{n}-\alpha_{n-1}$), but this sequence does not converge.