Let $D$ be a division ring and let $K$ be the center of $D$.
Assume $\dim_K(D)<\infty$.
Why is $\dim_K(D)$ a square?
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Let $\overline K$ be an algebraic closure of $K$. Then $D\otimes_K\overline K$ is a finite-dimensional simple algebra over $\overline K$. As $\overline K$ is algebraically closed, $D\otimes_K\overline K\cong M_n(\overline K)$. Thus $\dim_K(D)=\dim_K(D\otimes_K\overline K)=n^2$.
Angina Seng
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For the fact $D\otimes_K\overline K\cong M_n(\overline K)$, is it by using Artin Wedderburn? – Confused on math Jan 20 '18 at 17:51
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@Confusedonmath Yes. – Angina Seng Jan 20 '18 at 17:53
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Thank you! I will think about it. – Confused on math Jan 20 '18 at 17:54
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This is a consequence of Artin Wedderburn by extending the scalar over the algebraic closure, it is simple ring so it is isonorphic to a matrix ring.
https://en.m.wikipedia.org/wiki/Artin%E2%80%93Wedderburn_theorem
Tsemo Aristide
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