We let $D$ be an open or closed disc in $\mathbb{C}_p$ centered at $0$. I have to prove that

$\text{(a)}$ $\{f \in 1 + x\mathbb{C}_p[[x]] : f \text{ converges in }D\}$ is closed with respect to multiplication but does not form a group; and

$\text{(b)}$ $\{f = 1 + \sum_{i=1}^{\infty} a_i x^i : \text{ord}_p a_i − λi > 0\ \forall i \text{, and tends to }\infty\text{ as }i \rightarrow \infty\}$ forms a group with respect to multiplication, where $\lambda \in \mathbb{R}$ is fixed.

While showing that these sets are closed with respect to multiplication is straightforward, I don't know how to show that $\text{(a)}$ is not a group. Clearly it has identity, but how do I show that it's candidate for inverse (which will exist since the constant term is $1$) doesn't converge in the disk?

Coming to $\text{(b)}$, I'm unable to show how a candidate for inverse's coefficients will satisfy the given constrain, may be mathematical induction will help. But not getting anywhere.

I know asking multiple questions in a single question is frowned upon, but all these questions come from a single question, which is an exercise problem in Koblitz's book on $p$ - adic analysis, but here's another question I have which I can't make heads or tails of:

Let $f_j ∈ 1+x\mathbb{Z}_p[[x]]$ converge in the closed unit disc $D(1)$ for any $j$. Does $f := \prod_{j \geq 1}f_j (x^j)$ converge in $D(1)$?

  • 3
    $\begingroup$ With $B = \{ \|x \| < r \}$ don't we have $\{f \in 1 + x\mathbb{C}_p[[x]] : f \text{ converges and has no zeros in } B\}$ is a group. If $f$ has a zero in $B$ its inverse doesn't converge in $B$. And $ord_pa_i \ge \lambda i$ in $(b)$ would be the condition for $f$ having no zeros while $ord_pa_i > \lambda i$ would be the extension to the closed disk. $\endgroup$ – reuns Nov 9 '18 at 0:55

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