Counter-example for the Multiplication of Classical Laurent Series

Question

It can be shown that the set of formal Laurent series over a ring $R$ themselves form a ring $R[[X]]$.

Let $f(x) = \sum_{i=n}^\infty a_i z^i$ and $g(x) = \sum_{j=m}^\infty b_j z^j$ where $a_i, b_j \in R$ and $m, n \in \mathbb{Z}$. Define multiplication by
\begin{align*} \times : R[[X]] \times R[[X]] &\to R[[X]]\\ f(x)g(x) &\mapsto \sum_{k = n+m}^\infty \left( \sum_{i+j=k} a_ib_j \right)z^k \end{align*}

I wish to better understand why this multiplication fails for classical Laurent series (i.e., $n=-\infty$). Could anyone please provide a concrete counter-example? Thanks!

Answer 1

The problem is since a Laurent series has negative powers of $z$, there are infinitely many terms $a_iz^i$ and $b_jz^j$ where $i + j = k$, thus the coefficient of $z^k$ is an infinite series and hence not necessarily an element of the ring $R$. As an example, if you tried to compute

$$\left(\sum_{k=-\infty}^{\infty}z^{k}\right)^2 = (\cdots + z^{-2} + z^{-1} + 1 + z + z^2 + \cdots)^2$$ with this multiplication rule, then every coefficient would be $1 + 1 + 1 + \cdots$, which is a divergent series, hence not in $R$.

Answer 2

The problem is that the sum over $i$ and $j$ with sum $k$ is now an infinite sum, and there is no reason for it to converge. For example, how would you do $f(x)^2$ where $f(x) = \sum_{k=-\infty}^\infty x^k$?