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.



Absolute value on tensor product of fields
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.


