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I am reading about central simple algebras of finite-dimension over a field $F$, and I have a nagging doubt:

Let $A$ be such an algebra. If I let $\overline{F}$ be an algebraic closure of $F$, then I know that $A \otimes_F \overline{F} \cong M_n(\overline{F})$ for some $n$. Choosing an $F$-basis $e_1, \dots, e_m$ for $A$ over $F$, we should have that $e_i \otimes 1$ gives a basis for $M_n(\overline{F})$ over $\overline{F}$ by properties of tensor products. But this tells me that $m = n^2$, while Artin-Wedderburn tells me that $M_\ell(K)$ is a central simple $F$-algebra for any skewfield $K$ with finite dimension over $F$ whose center is $F$. The dimension of $M_\ell(K)$ as an $F$-vector space is $\ell^2 [K : F]$, so can't I contradict that the dimension of $A$ over $F$ must be $n^2$ for some $n$ by taking an skewfield over $F$ whose dimension is not a square?

I feel like I'm missing something, or confusing something in the above arguments, so if anyone can point out where I've gone wrong, it would be greatly appreciated!

stillconfused
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    If $D$ is a central division algebra over $F$, then $D\otimes_F\bar F$ is a central simple algebra over $\bar F$, so its dimension is a square. – Mariano Suárez-Álvarez Feb 19 '23 at 22:58
  • See the answers to this question (which was closed as off topic, in a spat of closure-reason craziness very characteristic of this place…) – Mariano Suárez-Álvarez Feb 19 '23 at 23:01
  • I see, so in fact there is no skew field whose center is $F$ and whose dimension is not a square. I briefly considered this possibility, but then dismissed it as seemingly too strong. I guess it is a good example of how skewfields behave very differently from commutative fields. Thanks for the help! – stillconfused Feb 19 '23 at 23:07
  • All this is very carefully discussed in Pierce's book on associative algebras. – Mariano Suárez-Álvarez Feb 19 '23 at 23:08

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