Let $F$ be an algebraically closed field of characteristic $p$ equipped with an absolute value $|\cdot|:F \rightarrow \mathbb{R}\_{\ge 0}$ with respect to which $F$ is complete.



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> Define $|\cdot|\_{prod}$ on the ring $F\otimes \_{\mathbb F\_p} F$ in the following way. If $c\in F\otimes \_{\mathbb F\_p} F$, then
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> $$|c|\_{prod}:=\inf\left(\max\_{1\le i\le n}\{|c\_{1,i}||c\_{2,i}| \}\ : \ c=\sum^{n}\_{i=1}c\_{1,i}\otimes c\_{2,i}\right)$$
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> where the infimum is taken over all the possible ways to write $c$ as a sum of pure tensors. Does $|\cdot|\_{prod}$ define a norm on $F\otimes \_{\mathbb F\_p} F$?
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I am able to show that $|\cdot|\_{prod}$ defines a semi-norm, which is submultiplicative and non-archimedean, but I am not able to find whether there does exist some $x\ne 0$ s.t. $|x|\_{prod}=0$.


My guess is that such elements don't exist and I am also able to show that no non-zero pure tensor has absolute value zero, but I still can't show this for a general element of $F\otimes \_{\mathbb F\_p} F$.



Norm on tensor product of fields
You can probably find this in most books on non-Archimedean functional analysis, see for instance Proposition 17.4 in Schneider's book.


The rough idea is to reduce to a tensor product of finite-dimensional spaces and then to norms associated to bases. You can now compute directly.


By the way, you probably want to assume your norms to be non-Archimedean.


