2
$\begingroup$

For a division ring $D$ with center field $F:=Z(D)$ such that $\dim_F D = n^2$, there is a classical result saying that $D\otimes_{F}\bar{F}\cong M_n(\bar{F})$ as $\bar{F}$-algebras, where $\bar{F}$ is the algebraic closure of $F$. My question is : If $E$ is a subfield of $F$ suth that $F$ is finite and separable over $E$, is $D\otimes_{E}\bar{F}$ still semisimple?

I already know that this is correct when D is a field.

Improve this question
$\endgroup$
2
  • $\begingroup$ You probably assume that $D$ has finite dimension over $F$? $\endgroup$
    – YCor
    Commented Oct 10, 2021 at 7:10
  • $\begingroup$ Oh, yes! I missed this important assumption! Thank you for pointing out it! $\endgroup$
    – GiS
    Commented Oct 10, 2021 at 8:23

1 Answer 1

Reset to default
2
$\begingroup$

I find that the answer of this problem is affirmative and obvious by a trick of tensors: $D\otimes_{E} \bar{F} \cong (D\otimes_{F}F)\otimes_{E} \bar{F}\cong D\otimes_{F}(F\otimes_{E} \bar{F})\cong D\otimes_{F}\bar{F}^m\cong (D\otimes_{F}\bar{F})^m$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .