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Let $k$ be a field. Consider the formal power series ring $k[[x, y]]$. Its fraction field is the formal Laurent series $k((x,y)) = \{\sum_{m, n\in \mathbb{Z}} k_{m, n} x^m y^n \mid k_{m, n}\in k​\}$.

If I want to find the inverse element of $x - y$ in $k((x, y))​$. On one hand I would do \begin{eqnarray} (x - y)^{-1}&=& x^{-1} x (x - y)^{-1}\\ & =& x^{-1} (1 - x^{-1} y)^{-1}\\ &=& x^{-1} [1 + x^{-1} y + x^{-2} y^2 + \cdots+ (x^{-1}y)^n+\cdots]​ \end{eqnarray} On the other hand, I can also do \begin{eqnarray} (x - y)^{-1} &=& (x - y)^{-1} y y^{-1}\\ &=& (xy^{-1} - 1)^{-1} y^{-1}​\\ & =& - [1 + xy^{-1} + x^2 y^{-2} + ⋯ + (xy^{-1})^n + ⋯ ] y^{-1}​ \end{eqnarray} The two results look different! (i.e. the monomials don't match). Where did I do wrong?

Thank you!

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You misdefine $k\left(\left(x,y\right)\right)$. There might be different things called $k\left(\left(x,y\right)\right)$ (the definition of Laurent series in multiple variables is not quite straightforward; see my lecture notes to the relevant part of Kac's vertex-algebras class, §1.2 for one choice that works -- although it is not a field!), but your space $\{\sum_{m, n\in \mathbb{Z}} k_{m, n} x^m y^n \mid k_{m, n}\in k​\}$ is not a ring (how would you multiply $\sum_{n \in \mathbb{Z}} x^n$ with itself?).

This actually answers your question: Your first computation assumes that $\sum_{n\in\mathbb{N}}\left(x^{-1}y\right)^n$ converges, whereas your second computation assumes that $\sum_{n\in\mathbb{N}}\left(xy^{-1}\right)^n$ converges. There is no topological "ring of Laurent series" in which both of these sums converge at the same time.

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  • $\begingroup$ Understood. Thanks! $\endgroup$
    – Steve
    Commented Nov 2, 2015 at 9:23

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