Let $R$ be a commutative unital ring and let $M$ be a unital $R$-module. A non-Archimedean ring seminorm on $R$ is a map $|\cdot| \colon R \rightarrow \mathbb{R}\_{\geq 0}$ which satisfies
$$ | 0\_R| = 0, \quad |1\_R| \leq 1, \quad |r - s| \leq \max \{ |r|, |s|\}, \quad |r \cdot s|\leq | r|\cdot |s| $$
for all $r,s \in R$. A non-Archimedean module seminorm on $M$ over $\left( R, |\cdot| \right)$ is a map $\| \cdot \| \colon M \rightarrow \mathbb{R}\_{\geq 0}$ which satisfies
$$ \| 0\_M \| = 0, \quad \| m - n \| \leq \max \{ \|m\|, \|n\| \}, \quad \|r \cdot m\| \leq |r| \cdot \| m \|.$$


Given two non-Archimedean seminormed $\left( R, |\cdot| \right)$-modules $\left( M, \|\cdot\|\_M \right)$ and $\left( N, \|\cdot\|\_N \right)$ one can define a non-Archimedean seminorm on the
tensor product $M \otimes\_R N$ by setting
$$ \| z \|\_{M \otimes N} := \inf \, \left \{ \max\_{i \in I} \| m\_i \|\_M \cdot \| n\_i \|\_N \, \bigg| \, z = \sum\_{i \in I} m\_i \otimes n\_i, \quad I \textrm{ is finite } \right \}. $$


It is written or hinted in various places (for example, in [Introduction To Berkovich Analytic Spaces](http://berkovich-2010.institut.math.jussieu.fr/lecture-notes/Temkin-lecture-notes.pdf) or in [Rigid Analytic Geometry and Its Applications](https://link.springer.com/book/10.1007/978-1-4612-0041-3)) that even if $R$, $M$ and $N$ are normed (or even complete), it is possible for $\| \cdot \|\_{M \otimes N}$ to be only seminorm and not a norm but I couldn't find an example of this.


Can someone describe an example or provide a reference which contains an explicit example? Does the situation change if we assume in addition that the norms are multiplicative (i.e $|x \cdot y| = |x| \cdot |y|$ and $\| r \cdot m \| = |r| \cdot \| m \|$)?



An example where the non-Archimedean tensor product of normed modules is only seminormed?
Let $k$ be a field. Pick some $r \in (0,1)$ and let $R$ be the ring $k[[t]]$ endowed with the absolute value that is trivial on $k$ and such that $\vert t\vert =r$. Let $M$ be $k((t))$ endowed with the absolute value that extends that on $R$. Pick $s \in (0,r)$ and let $N$ be the ring $k[[t]]$ endowed with the absolute value that is trivial on $k$ and such that $\vert t\vert =s$. In this situation, the tensor seminorm on $M\otimes\_R N$ is only a seminorm. Indeed, we can write $1 = t^{-N} \otimes t^N$ for any integer $N$, hence $\| 1\|\_{M\otimes\_R N} \le r^{-N} s^N$, which tends to 0 when $N$ goes to $\infty$.


