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.

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

$$|c|_{prod}:=\inf\left(\max_{1\le i\le n}\{|c_{1,i}||c_{2,i}| \}\ : \ c=\sum^{n}_{i=1}c_{1,i}\otimes c_{2,i}\right)$$

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 $F\otimes _{\mathbb F_p} F$?

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 $F\otimes _{\mathbb F_p} F$.


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.

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OK, I found the argument on normed vector spaces in these online notes. I will explain it in the case at hand.

Observe that for $a \in \mathbf{F}_p \subset F$ we have $|a| = 1$ if $a \not = 0$ and $|0| = 0$.

Let $c \in F \otimes_{\mathbf{F}_p} F$. By linear algebra, there are minimal sub $\mathbf{F}_p$-vector spaces $V, W \subset F$ such that $c \in V \otimes_{\mathbf{F}_p} W \subset F \otimes_{\mathbf{F}_p} F$. Then $\dim(V) = \dim(W) < \infty$ and this integer is called the rank of $c$.

Write $c = \sum_{i = 1, \ldots, n} x_i \otimes y_i$ with $x_i \not = 0$ and $y_i \not = 0$ for all $i = 1, \ldots, n$.

If $n$ is minimal, then $n$ is the rank of $c$ and $x_i \in V$ and $y_i \in W$. Since our ground field is $\mathbf{F}_p$ is finite, we have only a finite number of cases here and hence the infimum over these cases is $> 0$.

If $n$ is strictly bigger than the rank of $c$, then $x_1, \ldots, x_n$ must be linearly dependent (otherwise $V$ would be the span of $x_1, \ldots, x_n$ and have bigger dimension). Let $\sum a_i x_i = 0$ be a nontrivial $\mathbf{F}_p$-linear relation. After renumbering we may assume $a_n \not = 0$ and $|x_n| \geq |x_i|$ for all $i$ with $a_i \not = 0$. Thus we may assume $x_n = \sum_{i < n} b_i x_i$ for some $b_i \in \mathbf{F}_p$ not all zero and we may assume $|x_n| \geq |x_i|$ for those $i < n$ with $b_i \not = 0$. It follows that $|x_n| = \max_{i < n} |b_ix_i|$.

Set $y'_i = y_i + b_i y_n$. Then we see that $c = \sum_{i \leq n} x_i \otimes y_i = \sum_{i < n} x_i \otimes y'_i$. Finally, $$ \max_{i \leq n} |x_i|\cdot |y_i| = \max_{i < n} \max(|x_i| \cdot |y_i|, |b_i x_i| \cdot |y_n|) \geq \max_{i < n} |x_i| \cdot |y'_i| $$ So by induction on $n$ we win.

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