Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

Hint $\ $ The natural injection of the Laurent series into the fraction field is onto since every fraction can be expanded as a Laurent series by using the expansion for $\rm\:(1-t)^{-1} =\: 1 + t + t^2 +\:\cdots$

$$\rm \frac{f(x)}{a\: x^n (1-x\:g(x))}\ =\ a^{-1} x^{-n} f(x)\:(1 + x\:g(x) + x^2g(x)^2 + \:\cdots\: )$$

Alternatively, if known, you can exploit the universal properties of localizations or fraction fields.

share|improve this answer

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 $\mathbb{F}((X))$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.