p-adic formal series, evaluations and invertible elements

Question

Take an invertible formal series $f\in \mathbb{Z}_p[[T]]$ of inverse $g\in \mathbb{Z}_p[[T]]$ and let $x\in \mathbb{Z}_p$ such that the value of $f$ evaluated at $x$ exists. I have a few questions:

1. Is $f(x)$ a $p$-adic integer ? Since $\mathbb{Z}_p$ is a closed set I thought it was the case, but I have some doubts.
2. Does $g(x)$ also exist ?
3. If $g(x)$ does exist, is it the inverse of $f(x)$ ? I was thinking of evaluating at $x$ in the identity $f(T) g(T) =1$, but still not sure of myself.

I'm new to the p-adic world and to the formal series world, so thanks a lot.

Answer 1

You meant multiplicative inverse. Because the compositional inverse often exists in formal series.

If the series $f(x)$ converges then it does so to an element of $\Bbb{Z}_p$, yes.

Then try $f(T)=1-T$ and $x=1$.

If $g(x)$ converges as well then $f(x)g(x)=1$, yes (consider the truncated series, change the order of summation to make $\sum_n a_n b_{k-n}=0$ appear, show that the remainder is small).

Answer 2

The answer of @reuns was perfect, but telegraphic. If you prefer prolixity, try this:

You’ll find that when $R$ is a unital ring, the set of elements of $R$ that have inverses (i.e. reciprocals) is often denoted $R^\times$. That is for $f\in R$, $f$ is in $R^\times$ if and only if there is a $g\in R$ such that $fg=1_R$ (and if $R$ isn’t commutative, I guess you also need to specify $gf=1$).

Let’s suppose that $R$ is commutative, and look at the ring $R[[T]]$, which I hope you know is complete under the $(T)$-adic topology. You can think of this as saying that if you have an expression $\sum_mf_m(T)$ whose terms are in $R[[T]]$, and if the initial degrees of the separate series $f_m$ have $\lim_m\bigl(\text{init deg}(f_m)\bigr)\to\infty$, then the sum is convergent, with limit a good element of $R[[T]]$.

In particular, if you have a series with constant term equal to $1_R$, i.e. $f(T)=1-\sum_1^\infty a_mT^m$, in other words of the form $1-Tg(T)$ for $g\in R[[T]]$, then the reciprocal of $f$ is easily written as $$ \frac1f=1+\sum_{n=1}^\infty\bigl(Tg(T)\bigr)^n\,, $$ as you know from high-school. I’ll leave it to you now to show that if the constant term of $f(T)$ is in $R^\times$, then $f\in\bigl(R[[x]]\bigr)^\times$.

What about your question about evaluating $f(T)$ somewhere in $R$? I’ll specialize to the case $R=\Bbb Z_p$, as you have done. If you have $f(T)\in\Bbb Z[[T]]$, then you’ll get a $p$-adically convergent series when you plug in for $T$ any element $x\in p\Bbb Z_p$. I’ll let you check that. On the other hand, if $x\notin p\Bbb Z_p$, the convergence of the associated series of elements of $\Bbb Z_p$ is not guaranteed at all. Might converge, might not, mostly depending on whether the coefficients of the original series get $p$-adically small as you look at them farther and farther out.

What you can show, however, is that if $\xi$ is a chosen element of $p\Bbb Z_p$, then the evaluation map $\text{ev}_\xi:\Bbb Z_p[[T]]]\to\Bbb Z_p$ by the rule $\text{ev}_\xi\bigl(f(T)\bigr)=f(\xi)$ is continuous. We’re talking about the $(T)$-adic topology on $\Bbb Z_p[[T]]$ and the $p$-adic topology on $\Bbb Z_p$. This means that when $f$ and $g$ are reciprocals of each other, so that $f(T)g(T)=1$, then also $f(\xi)g(\xi)=1$. And that answers your other question, I hope.