# Norm on tensor product of fields

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.

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.