0
$\begingroup$

I'm working on a problem that asks me to show that the ring of formal Laurent series $F((x))$ is actually a field when $F$ is a field. My problem is that the problem doesn't define the product between two elements in $F((x))$. I thought that the product in this ring is somehow similar as the product of formal power series so I tried to define multiplication as follows

Let $\sum _{n\ge N}^{\infty }a_nx^n$ and $\sum _{n\ge M}^{\infty }b_nx^n$ be elements of $F((x))=\{\sum _{n\ge N}^{\infty }c_nx^n |c_n \in F, N\in \mathbb{Z}\}$

$\sum _{n\ge N}^{\infty }a_nx^n*\sum _{n\ge M}^{\infty }b_nx^n=\sum _{n\ge min(N,M)}^{\infty }\left(\sum _{k\ge min(N,M)}^{\infty }a_k b_{n-k}\right)x^n$

But I'm not sure if this makes sense. Can someone tell me if I'm rigth, thanks in advance

$\endgroup$

2 Answers 2

1
$\begingroup$

You suggested that you are able to multiply two power series. Then it is very simple to multiply Laurent series. Say, the initial term of the first one is $x^N$, and the second one $x^M$. Then think about then as power series multiplied by $x^N$ and $x^M$. So multiply the first by $x^{-N}$ and the second by $x^{-M}$, multiply the power series obtained, and then multiply the result by $x^{N+M}$.

$\endgroup$
1
$\begingroup$

You are correct (see, for instance, the Wikipedia article which mentions this explicitly). The idea behind Laurent series is that, when higher-order terms are ignored, they become polynomials in $x$ and $x^{-1}$; they thus must just get multiplied just like polynomials in $x$ and $x^{-1}$, and the formal Laurent series definition preserves this multiplication.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .