# Product absolute value in rings of integers

Let $$F$$ be an algebraically closed field of characteristic $$p$$ equipped with a nonarchimedean dense absolute value $$|\cdot|:F \rightarrow \mathbb{R}_{\ge 0}$$ with respect to which $$F$$ is complete. Let $$\mathcal{O}_{F}$$ denote the ring of integers of $$F$$.

First define the product norm $$|\cdot|_{prod}$$ on $$F\otimes _{\mathbb F_p} F$$ in the following way. If $$c\in F\otimes _{\mathbb F_p} F$$, then

$$|c|_{prod}:=\inf\left\{\max_{1\le i\le n}\{|c_{1,i}||c_{2,i}| \}\ : \ c=\sum^{n}_{i=1}c_{1,i}\otimes c_{2,i}, \text{where } c_{1,i}, c_{2,i}\in F\right\}$$ On the other hand, define $$|\cdot|'_{prod}$$ on the subring $$\mathcal{O}_{F}\otimes _{\mathbb F_p} \mathcal{O}_{F}$$ in the following way. If $$d\in \mathcal{O}_{F}\otimes _{\mathbb F_p} \mathcal{O}_{F}$$, then

$$|d|'_{prod}:=\inf\left\{\max_{1\le i\le n}\{|d_{1,i}||d_{2,i}| \}\ : \ d=\sum^{n}_{i=1}d_{1,i}\otimes d_{2,i}, \text{where } d_{1,i}, d_{2,i}\in \color{red}{\mathcal{O}_{F}}\right\}$$

In both definitions the infimum is taken over all the possible ways to write the element as a sum of pure tensors in the respective rings. Is it true that $$|\cdot|_{prod}$$ and $$|\cdot|'_{prod}$$ coincide when restricting them to elements of $$\mathcal{O}_{F}\otimes _{\mathbb F_p} \mathcal{O}_{F}$$?

My first guess is that the answer is true. It is clear that $$|c|_{prod}\le |c|'_{prod}$$ for $$c\in \mathcal{O}_{F}\otimes_{\mathbb{F}_{p}}\mathcal{O}_{F}$$, since there are more ways to write $$c$$ as a sum of pure tensors in $$F$$.

I am able to show that they do coincide on pure tensors in $$\mathcal{O}_{F}\otimes _{\mathbb F_p} \mathcal{O}_{F}$$, more precisely I showed that $$|x\otimes y|_{prod}=|x\otimes y|'_{prod}=|x|\cdot|y|$$ for $$x,y\in \mathcal{O}_{F}$$, but I am not able to prove that in the general case.

The natural map $$i \colon \mathcal O_F \to F$$ is an injective map of $$\mathbb F_p$$-vector spaces, hence we can choose a splitting, i.e. an $$\mathbb F_p$$-linear map $$s \colon F \to \mathcal O_F$$ such that $$s \circ i$$ is the identity.
We have $$|s(x)| \leq |x|$$ since if $$x \in \mathcal O_F$$ then $$|s(x) | = |x|$$ and if $$x \notin \mathcal O_F$$ then $$|s(x)| \leq 1 \leq |x|$$.
By $$\mathbb F_p$$-linearity, if $$c = \sum_{i=1}^n c_{1,i} \otimes c_{2,i}$$ then $$s \otimes s(c) = \sum_{i=1}^n s(c_{1,i}) \otimes s(c_{2,i})$$ and if $$c \in \mathcal O_F \otimes \mathcal O_F$$ then by definition $$s \otimes s(c)=c.$$
\begin{aligned}|c|_{prod} &=\inf\left\{\max_{1\le i\le n}\{|c_{1,i}||c_{2,i}| \}\ : \ c=\sum^{n}_{i=1}c_{1,i}\otimes c_{2,i}, \text{where } c_{1,i}, c_{2,i}\in F\right\} \\ &\ge \inf\left\{\max_{1\le i\le n}\{|s(c_{1,i})||s(c_{2,i})| \}\ : \ c=\sum^{n}_{i=1}c_{1,i}\otimes c_{2,i}, \text{where } c_{1,i}, c_{2,i}\in F\right\} \\ &= \inf\left\{\max_{1\le i\le n}\{|s(c_{1,i})||s(c_{2,i})| \}\ : \ s\otimes s (c)=\sum^{n}_{i=1}s(c_{1,i})\otimes s(c_{2,i}), \text{where } c_{1,i}, c_{2,i}\in F\right\} \\ &= \inf\left\{\max_{1\le i\le n}\{|s(c_{1,i})||s(c_{2,i})| \}\ : \ c=\sum^{n}_{i=1}s(c_{1,i})\otimes s(c_{2,i}), \text{where } c_{1,i}, c_{2,i}\in F\right\} \\ &\ge \inf\left\{\max_{1\le i\le n}\{|d_{1,i}||d_{2,i}| \}\ : \ c=\sum^{n}_{i=1}d_{1,i}\otimes d_{2,i}, \text{where } d_{1,i}, d_{2,i}\in {\mathcal{O}_{F}}\right\} \\ &= |c|'_{prod}\end{aligned}