Absolute value on tensor product of fields

Question

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.

Answer 1

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 $f_1\otimes g_1+f_2\otimes g_2=f_1\otimes (g_1-cg_2)+f_1\otimes cg_2+f_2\otimes g_2 =f_1\otimes (g_1-cg_2)+(cf_1+f_2) \otimes g_2$

and

$\max\{ |f_1|\cdot |g_1|, |f_2|\cdot |g_2| \} =\max\{ |f_1|, |f_2|\} \cdot |g_2| \geq \max\{ |f_1|\cdot |g_1-cg_2|, |cf_1+f_2|\cdot |g_2| \}$

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.