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Let $R$=$\mathbb{F}$$[[x]]$, where $\mathbb{F}$ is a field. Show that $F(R)$(the field of fractions) may be identified with the ring $\mathbb{F}$$((x))$ of formal Laurent series.

A formal Laurent series is a sequence $(a)$=$(a_i)_{i\in\mathbb{Z}}$, with $a_i\in\mathbb{F}$, and for some $k\in\mathbb{Z}$(depending on $a$), $a_i=0$ whenever $i<k$. We formally write $a=\sum_{i\in\mathbb{Z}}a_ix^i=\sum_{i=k}^{\infty}a_ix^i$, and use the addition and multiplication that is suggested by this notion.

How to do this? I have no idea.

Bill Dubuque
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riddikulus
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2 Answers2

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Hint $ $ The natural injection of the Laurent series into the fraction field is onto since every fraction can be expanded as a formal Laurent series by using the expansion for $\,{\large \frac{1}{1-\color{#c00}t}} =\: 1 +\color{#c00} t + \color{#0a0}{t^2} +\:\cdots$

$$\frac{\ f}{a\,x^n+\cdots }\ =\ \frac{f}{a\, x^n (1-\color{#c00}{x\,g)}}\ =\ a^{-1} x^{-n} \:(1 + \color{#c00}{x\:g} + \color{#0a0}{x^2g^2} + \:\cdots\, )\qquad$$

which converges since $\, \lim_{\,j\to \infty}\,{\rm ord}(x^j g^j) = \infty,\, $ see Proposition $1.1.8$.

Alternatively, if known, we can exploit the universal properties of localizations or fraction fields, e.g. see here.

Bill Dubuque
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It is obvious that $\mathbb{F}((X)) \subset F(R)$ (because any Laurent series is a quotient of a formal series by a power of $x$). So it remains to show that if $a = \sum_{n = 0}^{\infty} a_i x^i \in \mathbb{F}[[X]]$ is non zero, then the inverse of $a$ is in $\mathbb{F}((X))$.

If you assume $a_0 \ne 0$, check that $a$ has its inverse in $\mathbb{F}[[X]]$ (write $\left(\sum_{i = 0}^{\infty} a_i x^i \right) \times \left(\sum_{n = 0}^{\infty} b_i x^i\right) = 1$ and recursively compute the coefficients $b_i$).

In the general case, denote $n \ge 0$ the lowest integer such $a_n \ne 0$, then we can write $a = x^n b$ with $b = \sum_{i = 0}^{\infty} a_{i+n} x^i$. We know from the previous paragraph that the inverse of $b$ is in $\mathbb{F}[[X]]$, and the inverse of $x^n$ is in $\mathbb{F}((X))$, so the inverse of $a$ is also in $\mathbb{F}((X))$.

Joel Cohen
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