Tate algebra

Remarks

The maximum exists because $\abs{a_\nu} \to 0$. $T_n$ is closed under multiplication, and the Gauss norm is multiplicative: $\norm{fg} = \norm{f} \norm{g}$. With this norm $T_n$ is a Banach algebra over $K$, and the polynomial ring $K[X_1, \dots, X_n]$ is dense in it. Let $\bar K$ be an algebraic closure of $K$ with the extended absolute value. A formal power series lies in $T_n$ exactly when it converges at every point $x \in \bar K^n$ with all $\abs{x_i} \le 1$. $T_n$ is Noetherian, regular and a unique factorization domain, and it satisfies a Noether normalization lemma. The residue field of each maximal ideal of $T_n$ is a finite extension of $K$.

Example: over the completion $\Q_p$ of $\Q$ for the $p$-adic absolute value, $f = \sum_{k \ge 0} p^k X^k$ lies in $T_1$, because $\abs{p^k}_p = p^{-k} \to 0$; its Gauss norm is $\abs{1}_p = 1$.

Non-example: $\sum_{k \ge 0} X^k$ has all coefficients of absolute value $1$, which do not tend to $0$, so it is not in $T_1$; it diverges at $X = 1$.