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.



Completeness of Localizations of Completions of Commutative Rings
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.


