I learned that the tensor product of two division algebras may not be a division algebra. Thus, I am curious if there is some case in which this is true.
To be precise, given a division algebra $A$ over $\mathbb R$, are $A\otimes_{\mathbb R}\mathbb R[x]$ and $A\otimes_{\mathbb R}\mathbb R(x)$ division algebras?
I searched on the internet but find no result. Any help would be appreciated.
If $A$ is finite dimensional with basis $a_1\ldots a_J$ then so is $B=A\otimes_\Bbb{R} \Bbb{R}(x)$ with basis $a_1\otimes 1\ldots a_J\otimes 1$,
For $b\in B-0$, by the characteristic polynomial stuff, if the determinant of the left-multiplication by $b$ (as a $\Bbb{R}(x)$-linear map) is non-zero then $b$ has an inverse in $B$.
So let's assume that $\det(b)=0$. Write $b=\sum_{j=1}^J a_j \otimes f_j(x)$.
If $r\in \Bbb{R}$ is not a pole of any $f_j(x)$ and not a zero of some $f_j(x)$ then $\sum_{j=1}^J a_j f_j(r)$ is a non-zero element of $A$, but its determinant is $0$, contradicting that $A$ is a division algebra.
Whence $B$ is itself a division algebra.