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.
 A: 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 $$s \otimes s(c)=c.$$
Combining these, we have
$$\begin{aligned}|c|_{prod}
&=\inf\left\{\max_{1\le i\le n}\{|c_{1,i}||c_{2,i}| \}\ : \ c=\sum^{n}_{i=1}c_{1,i}\otimes c_{2,i}, \text{where }  c_{1,i}, c_{2,i}\in F\right\}
\\
&\ge \inf\left\{\max_{1\le i\le n}\{|s(c_{1,i})||s(c_{2,i})| \}\ : \ c=\sum^{n}_{i=1}c_{1,i}\otimes c_{2,i}, \text{where }  c_{1,i}, c_{2,i}\in F\right\}
\\
 &=  \inf\left\{\max_{1\le i\le n}\{|s(c_{1,i})||s(c_{2,i})| \}\ : \ s\otimes s (c)=\sum^{n}_{i=1}s(c_{1,i})\otimes s(c_{2,i}), \text{where }  c_{1,i}, c_{2,i}\in F\right\} 
\\
&= \inf\left\{\max_{1\le i\le n}\{|s(c_{1,i})||s(c_{2,i})| \}\ : \  c=\sum^{n}_{i=1}s(c_{1,i})\otimes s(c_{2,i}), \text{where }  c_{1,i}, c_{2,i}\in F\right\} 
\\
&\ge \inf\left\{\max_{1\le i\le n}\{|d_{1,i}||d_{2,i}| \}\ : \ c=\sum^{n}_{i=1}d_{1,i}\otimes d_{2,i}, \text{where } d_{1,i}, d_{2,i}\in {\mathcal{O}_{F}}\right\} 
\\
&= |c|'_{prod}\end{aligned}$$
which together with the trivial inequality proves the identity.
