Why do Laurentseries with $p$-adic coefficients that converge on some given annulus form a ring

Question

Let $L(r_1, r_2)$ be the set of Laurentseries with entries in $\mathbb{Q}_p$ that are convergent on some annulus $r_1<|x|<r_2$. The question is why $L(r_1, r_2)$ together with formal addition and multiplication forms a ring? I am getting stuck, trying to show that this set is closed under multiplication. So say we have two series $a=\sum_{i\in \mathbb{Z}} a_i x^i$ and $b=\sum_{i\in \mathbb{Z}} b_i x^i$. The product would be $a\cdot b= \sum_{i\in \mathbb{Z}} (\sum_{j\in \mathbb{Z}} a_j b_{i-j})x^i$. But why is $\sum_{j\in \mathbb{Z}} a_j b_{i-j}$ an element of $\mathbb{Q}_p$? This would be the case if $ |a_j b_{i-j}|\to 0$ as $j\to \pm \infty$. But I do not see why this would be true and how to use that the sequences are convergent somewhere.

Answer 1

Here’s my approach:

It makes the typography easier, and it’s essential for any discussion involving Newton polygons, to use the additive valuation $v_p$, say normalized so that $v(p)=1$, instead of the absolute value. You may define $v_p(z)$ to be $-\log_p(|z|)$.

I really like your idea of looking at the set $\mathcal A_\rho=\{z: v(z)=\rho\}$ and the set (we hope it’s a ring) $\mathcal C_\rho$ of series convergent on $\mathcal A_\rho$. Then the ring of series convergent on the set $\{z:\alpha<v(z)<\beta\}$ is just the intersection of all the rings $\mathcal C_\rho$ with $\alpha<\rho<\beta$.

As you recognized in your answer, $\mathcal C_\rho$ is the set of all series $f(x)=\sum_{-\infty}^\infty c_ix^i$ for which $\lim_{i\ge0}v(c_i)+i\rho=\infty$ and $\lim_{i\le0}v(c_i)+i\rho=\infty$. Our first task is to show that the formal product of $\sum_ic_ix^i$ and $\sum_id_ix^i$ is even defined. We get coefficients $C_i=\sum_mc_md_{i-m}$, with $m$ running over all integers, and hope that the infinite series shown here is convergent. We get $$ v(c_md_{i-m})=\bigl[v(c_m)+m\rho\bigr] + \bigl[v(d_{i-m})+(i-m)\rho\bigr]-i\rho\,. $$ Whether we are looking at this as $m\to\infty$ or as $m\to-\infty$, the bracketed items go to $\infty$, that is, approach $0$, so the sum is convergent.

We now have a way of multiplying two elements of $\mathcal C_\rho$, but we need to show that the product is also in $\mathcal C_\rho$, i.e convergent on $\mathcal A_\rho$. I’ll do this for the special but representative case $\rho=0$, i.e. functions on the set of $z$ with $|z|=1$. The general case is the same, mutatis mutandis, but a little messier. For this case, though, you see that to show that $fg$ is convergent, it certainly suffices to show that $(p^rf)(p^sg)$ is convergent. For suitably large $r$, $p^rf=\sum_i(p^rc_i)x^i$ has all coefficients (locally) integral, i.e. $v(p^rc_i)\ge0$. I will assume from now on that our original functions $f=\sum c_ix^i$ and $g=\sum d_ix^i$ have $v(c_i)\ge0$ and $v(d_i)\ge0$.

Still in this special case $\rho=0$, where $fg=\sum C_ix^i$ and $C_i=\sum_mc_md_{i-m}$, where $\lim v(c_i)=\infty$ and $\lim v(d_i)=\infty$, we need to show that $\lim v(C_i)=\infty$. To do this, I’ll show that given $N>0$, there are only finitely many $C_i$ with $v(C_i)<N$. For this, we know that there is $M$ such that for all $m$ with $|m|\ge M$, we have $v(c_m)>N$ and $v(d_m)>N$. Let’s look at $C_k$, where $|k|>2M$. I claim that $v(C_k)>N$.

Indeed, if $k>0$, so that $k>2M$ we’re looking at all the terms of form $c_md_n$ with $m+n=k>2M$. Two indices summing to a number bigger than $2M$, one of them has to be $>M$, giving either $v(c_m)>N$ (and $v(d_n)\ge0$) or $v(d_n)>N$ (and $v(c_m)\ge0$). In either case, $v(c_md_n)>N$. Thus for $k>2M$, $v(C_k)>N$. The proof for $k<0$ runs exactly the same way: two indices summing to a number less than $-2M$, one of them has to be smaller than $-M$. Thus for $|k|>2M$, $v(C_k)>N$, which is what I claimed.

So the set $\mathcal C_0$ has a multiplication, and is closed under this multiplication. I leave the mutandum for general $\mathcal C_\rho$ to you.

Answer 2

I think I can give an answere myself now. Thank you for your help dear commenters. So let $a=\sum_{i\in \mathbb{Z}} a_i x^i$ and $b=\sum_{i\in \mathbb{Z}} b_i x^i$ such that both converge on the annulus $|x|=r<1$ (For r>1 we can find a similar argument). Let $c=a\cdot b= \sum_{i\in \mathbb{Z}} (\sum_{j\in \mathbb{Z}} a_j b_{i-j})x^i$.

So $a$ and $b$ converge on $r$ if and only if $|a_n|r^n$ and $|b_n|r^n$ go to $0$ as $n$ goes to $\pm \infty$. Hence we can define $|a|^{\{r\}}= \max \{ |a_n|r^n\in \mathbb{R}\}$. This implies that $|a_n|<|a|^{\{r\}}r^{-n}$, so that $|a_n|\to 0$ as $n\to -\infty$, since $r<1$.

We want to show that $|a_j b_{i-j}|\to 0$ if $j\to \pm \infty$. It is \begin{align*} |a_j b_{i-j}|r^i = |a_j|r^j |b_{i-j}|r^{i-j}\le |a_j|r^j |b|^{\{r\}} \end{align*} so if we let $j\to -\infty$ the right hand side converges to 0. So as well does the left hand side, which implies $|a_j b_{i-j}|\to 0$ if $j\to -\infty$. Similarly \begin{align*} |a_j b_{i-j}|r^i = |a_j|r^j |b_{i-j}|r^{i-j}\le |a|^{\{r\}} |b_{i-j}|r^{i-j} \end{align*} shows $|a_j b_{i-j}|\to 0$ if $j\to \infty$.

So the coefficients of the series are well defined.