# How to multiply in the formal Laurent series ring

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

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 $x^{N+M}$.

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.