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Let $F$ be a field. The ring of Laurent series is defines as: $$F((t))=\left\{\sum\limits_{n=N}^{\infty}a_nt^n\ :\ N\in\mathbb Z, a_n\in F\right\}$$ I am trying to show that it is actually a field.

So if $a=\sum\limits_{n=N}^{\infty}a_nt^n\in F((t))$ we can split $a$ into two parts: $$a=\sum\limits_{n=N}^{-1}a_nt^n+\sum\limits_{n=0}^{\infty}a_nt^n$$ (if $N\geq0$ then there is no problem as $a_0+a_1t+...+a_nt^n$ is a unit in $F[[t]]=\left\{\sum\limits_{n=0}^{\infty}a_nt^n\right\}$ since $a_0$ is a unit in $F$ we can find the inverse of $a$: $$a^{-1}=a_0^{-1}+b_1t+b_2t^2+\ \ldots$$ where $ b_n := -a_0^{-1} \sum_{k=1}^n a_k b_{n-k} $

Starting from scratch, can I write $F((t))=F[t,{1\over t}]$ ? And then since $t$ and $1/t$ are units in $F[t,{1\over t}]$ $F((t))$ is a field?

John Cataldo
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2 Answers2

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Concerning the title question, one can review the proof that $F((t))$ is the fraction field of the ring $F[[t]]$. Hence it is a field.

Reference: Prove the field of fractions of $F[[x]]$ is the ring $F((x))$ of formal Laurent series.

Dietrich Burde
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Note that $F[[t]]$ is a DVR, and $t$ is a uniformizing parameter: every powerseries can be written uniquely as $ut^N$ where $u$ is a unit and $N\geqslant 0$. This means that inverting $t$ suffices to obtain $F((t))$, and indeed this is more or less the definition of this field.

Pedro
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