# Equivalence of norms in finite dimension over complete fields is true, but false for finite rank modules over complete rings

We know that if $k$ is complete valued field and $V$ a finite dimensional vector space then all norms on $V$ are equivalent. (The field is not necessarily of characteristic $0$ and its absolute value is not necessarily non-trivial.)

Now, if $k$ is a complete valued ring and if $V$ is $k$-module of finite rank, can we hope to have an analogue result ?

Endow $\mathbf{Z}$ with its usual absolute value, for which it is complete, for instance because it is closed in $\mathbf{R}$. Consider the sub-$\mathbf{Z}$-module $$V = \{a+b\sqrt{2}\in\mathbf{R}\;|\;(a,b)\in\mathbf{Z}^2\}$$ of rank $2$ of $\mathbf{R}$. Defined two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on $V$ by setting $\|a+b\sqrt{2}\|_1 = |a+b\sqrt{2}|$ and $\|a+b\sqrt{2}\|_2 = |a|+|b|$ for all $(a,b)\in\mathbf{Z}^2$.
Now consider the two sequences $(x_n)_n$ and $(y_n)_n$ in $V$ defined by $x_n = (1+\sqrt{2})^n$ and $y_n = (1-\sqrt{2})^n$ for all $n\in\mathbf{N}$. Write $x_n = a_n - b_n\sqrt{2}$ with $a_n,b_n \in\mathbf{Z}$ for all $n\in\mathbf{N}$ so that $y_n = a_n + b_n \sqrt{2}$ for all $n\in\mathbf{N}$. Note that $x_n \to 0$ and $y_n \to +\infty$ for the usual topology of $\mathbf{R}$ as $n\to +\infty$, so that $a_n = \frac{x_n + y_n}{2} \to +\infty$ as $n\to +\infty$. Now $\|x_n\|_1 \to 0$ and $\|x_n\|_2 \to +\infty$ so that the sequence of term $\frac{\|x_n\|_2}{\|x_n\|_1}$ is not bounded, showing that the two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are not equivalent.