# Absolute value on tensor product of fields

Suppose that we have the Laurent series fields $$F_1:=\mathbb F_p((X))$$ and $$F_2:=\mathbb F_p((Y))$$.

Equip $$F_1$$ with the $$X$$-adic multiplicative absolute value $$|\cdot|_1$$, i.e. define $$|X|_1=\dfrac{1}{p}$$ and $$|q|_1=1$$ for all $$q\in \mathbb F_p$$. Analogously, equip $$F_2$$ with the $$Y$$-adic multiplicative absolute value $$|\cdot|_2.$$

Define $$|\cdot|_{prod}$$ on the ring $$F:=F_1\otimes _{\mathbb F_p} F_2$$ in the following way. If $$c\in F$$, then

$$|c|_{prod}:=\inf\left(\max_{i}\left(\prod|c_{1,i}|_1|c_{2,i}|_2 \right)\ | \ c=\sum^{n}_{i=1}c_{1,i}\otimes c_{2,i}\right)$$

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 $$F$$?

I am able to show that $$|\cdot|$$ defines semi-norm, which is submultiplicative and is non-archimedean, but I am not able to find whether there does exist $$x\ne 0$$ s.t. $$|x|_{prod}=0$$.

My guess is that such elements don't exist and I am able to show this on the subring $$\mathbb F_p[X]\otimes _{\mathbb F_p} \mathbb F_p[Y]$$. Using this, I thought I could extend the conclusion to $$F$$ using a density argument (using that $$\mathbb F_p[X]\otimes _{\mathbb F_p} \mathbb F_p[Y]$$ is dense in $$\mathbb F_p[[X]]\otimes _{\mathbb F_p} \mathbb F_p[[Y]]$$ ), but I failed.

The product norm is indeed a norm. The key observation is the following: If $$g_1, g_2\in \mathbb{F}_p((Y))$$ have the same norm, then there is $$c\in \mathbb{F}_p$$ such that $$|g_1-cg_2|_2<|g_1|_2$$. Hence, if one has an expression $$f_1\otimes g_1+f_2\otimes g_2$$ with $$|g_1|_2=|g_2|_2$$, one can also express it as $$f_1\otimes g_1+f_2\otimes g_2=f_1\otimes (g_1-cg_2)+f_1\otimes cg_2+f_2\otimes g_2 =f_1\otimes (g_1-cg_2)+(cf_1+f_2) \otimes g_2$$
$$\max\{ |f_1|\cdot |g_1|, |f_2|\cdot |g_2| \} =\max\{ |f_1|, |f_2|\} \cdot |g_2| \geq \max\{ |f_1|\cdot |g_1-cg_2|, |cf_1+f_2|\cdot |g_2| \}$$
This tells us that for computing the infimum, one only has to take into account sums of tensors $$\sum_i f_i\otimes g_i$$ where all $$g_i$$ have different norms.
In the next step, one shows that all such expressions $$c=\sum_{i=1}^n f_i\otimes g_i$$ with $$g_i$$ having different norms, have the same value $$\max_i \{ |f_i|\cdot |g_i|\}$$.
This is done as follows (a bit sketchy here):\ Consider two expressions $$c=\sum_{i=1}^n f_i\otimes g_i$$, $$c=\sum_{j=1}^m f'_j\otimes g'_j$$, with $$|g_1|<\ldots<|g_n|$$ and $$|g'_1|<\ldots<|g'_m|$$ of the same element $$c$$ (all $$f_i$$ and $$f'_j$$ non-zero). Then the set $$\{ g_1,\ldots, g_n, g'_1,\ldots, g'_m\}$$ has to be $$\mathbb{F}_p$$-linearly dependent. In particular, $$|g_n|=|g'_m|$$. Using the same trick as above, we can replace the second expression by one with $$g'_m=g_n$$, and the norm of the new expression does not exceed the old one. Then, $$\sum_{i=1}^{n-1} f_i\otimes g_i- \sum_{j=1}^{m-1} f'_j\otimes g'_j=(f'_m-f_n)\otimes g_n$$, and either we have $$f'_m-f_n=0$$ or $$g_n$$ is linear dependent of the other $$g$$'s. Due to the norms of the $$g$$'s, the latter can't be, and we conclude $$f'_m=f_n$$. Going on inductively in the same way, one can transform one expression into the other, and obtains that $$\max_i \{ |f_i|\cdot |g_i|\} \leq \max_j \{ |f'_j|\cdot |g'_j|\}$$. By switching the roles, we obtain equality.