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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}$$|\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.

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.

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.

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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 $c\in \mathcal{O}_{F}\otimes _{\mathbb F_p} \mathcal{O}_{F}$$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|$$|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.

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 $c\in \mathcal{O}_{F}\otimes _{\mathbb F_p} \mathcal{O}_{F}$, 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. 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.

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.

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