How to find Krull dimension of $k[[x,y]][x^{-1},y^{-1}]$ where $k$ is a field?

Question

I don't know how to find Krull dimension of $k[[x,y]][x^{-1},y^{-1}],$ where $k$ is a field?

Answer 1

Since you know that $dim(A)=dim(k[[x,y]])=2$, here is how to compute the dimension of $B=k[[x,y]][x^{-1},y^{-1}]=k[[x,y]][\frac {1}{xy}]=A[\frac {1}{f}]$, where we put $f=x\cdot y$

We are lucky to be in the happy situation where $A$ is a local ring with maximal ideal $\mathfrak m=(x,y)$, so that all maximal chains of primes of $A$ are of the form $0\subsetneq \mathfrak p\subsetneq \mathfrak m$.
Since the primes of $B$ correspond to the primes of $A$ not containing $f$, the maximal chains in $B$ will thus correspond to chains in $A$ of the form $0\subsetneq \mathfrak p$ with $f\notin \mathfrak p$.
Such $\mathfrak p$ do exist: for example $\mathfrak p=(x+y)$.

Conclusion $$ \dim ( k[[x,y]][x^{-1},y^{-1}])=1 $$

Answer 2

Your ring is the localization at the multiplicative set generated by $xy$ of the completion of $k[x,y]$ at the maximal ideal $(x,y)$.

Do you know how the procedures of localization and completion interact with Krull dimension?