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I want to show that every element in the field of fractions of $K[[X]]$ can be written as $\sum_{k=-n}^\infty a_kX^k$ where $a_k\in K, n\in\mathbb{Z}$.

My first attempt was to rewrite

$\displaystyle \frac{\sum_{i=0}^\infty a_iX^i}{\sum_{j=0}^\infty b_jX^j}=\lim_{N\to \infty}\frac{\sum_{i=0}^N a_iX^i}{\sum_{j=0}^N b_jX^j}$

where I am not completely sure, if this is correct. I know that I can write an element of $R[[X]]$ where $R$ is a ring like this. Then I tried polynomial division and some similar things. Also I know that if $b_0$ would be not zero, I can invert the whole polynomial and arrive at the conclusion, even with $n=0$.

However, I was not able to progress from here.

EpsilonDelta
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    $K[[X]]$ is usually the ring of power series. – Randall Oct 11 '18 at 15:27
  • I am pretty sure that general elements don't have infinitely many terms in the denominators. – J. Moeller Oct 11 '18 at 15:30
  • @Randall $R[[X]]$ would be the ring of formal power series. Instead, here, we have the formal power series over a field $K$. – EpsilonDelta Oct 11 '18 at 15:39
  • Right, but they're not "polynomials." – Randall Oct 11 '18 at 15:41
  • @Randall Corrected the title. – EpsilonDelta Oct 11 '18 at 15:46
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    This is not a duplicate of the linked question: there is no requirement for $b_0$ to be a unit and, in fact, the OP has already realised they can solve the question in the case $b_0 \neq 0$ (which is equivalent to $b_0$ being a unit, since $K$ is a field). – Christopher Oct 11 '18 at 16:01
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    @Christopher It is an immediate consequence of the linked dupe, since it implies that the denominator has form $,u x^n$ for a unit $,u\in R[[x]],,$ see the corollary in my answer there. – Bill Dubuque Oct 12 '18 at 22:26
  • I apologize for my possibly confusing voting on this question. The details of my misadventure and a kind of an explanation is given here. – Jyrki Lahtonen Oct 13 '18 at 09:08
  • @JyrkiLahtonen It's ok. I already solved the question myself, so I might leave something for future viewers. HINT: Try to justify why the denominator can be written as $X^{-n}g(X)$ where $g(X)$ is a unit. What has to hold for "denominator polynomials" in the field of quotients of a ring? – EpsilonDelta Oct 13 '18 at 09:29

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