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

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    Please always make the body of your questions contain the actual question. Have you ever seen a book whose first sentence starts in the title?2012-03-24

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

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    Great! thank you, dear Georges Elencwajg.2012-03-24
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    I have learned a lot from you :)2012-03-24
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    Dear @ehsanno, thanks a lot: no words could be more appreciated by a teacher.2012-03-24
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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?

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    Dear Mariano Suárez-Alvarez: yes, thanks. I thought about that. In fact, $\text{dim}k[[x,y]]=2.$ So based on your words and my knowledge, $\text{dim}k[[x,y]]_{xy} \leq 2.$ Am I right? what is it exactly, then? and is it obvious that $k[[x,y]]_{xy} \cong k[[x,y]][x^{-1},y^{-1}]?$2012-03-24
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    Dear @ehsanno, since $k[[x,y]]_{xy}$ contains $x,y$ and $\frac {1}{xy}$, it also contains $x\cdot \frac {1}{xy}=\frac {1}{y}$ and similarly it contains $\frac {1}{x}$. The other inclusion follows from $\frac {1}{xy}=\frac {1}{x}\cdot \frac {1}{y}$2012-03-24
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    Dear @Georges Elencwajg: Oops! yes, that's right!2012-03-24