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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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closed as off-topic by T. Bongers, Claude Leibovici, John B, Sahiba Arora, MathOverview Jan 21 '18 at 20:19

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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 $\dim_K(D)=\dim_K(D\otimes_K\overline K)=n^2$.

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  • $\begingroup$ For the fact $D\otimes_K\overline K\cong M_n(\overline K)$, is it by using Artin Wedderburn? $\endgroup$ – Confused on math Jan 20 '18 at 17:51
  • $\begingroup$ @Confusedonmath Yes. $\endgroup$ – Lord Shark the Unknown Jan 20 '18 at 17:53
  • $\begingroup$ Thank you! I will think about it. $\endgroup$ – 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

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    $\begingroup$ A matrix ring generally has zero-divisors. $\endgroup$ – Lord Shark the Unknown Jan 20 '18 at 17:44

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