I have a question regarding the proof of quaternion algebras (denoted $A$) being either a division algebra or isomorphic to $M_2(F)$.

I can understand that $A$ is central simple. And since $A$ is finite dimensional, minimal right ideal of $A$ exists, so $A$ is semisimple.

This is the part where I don't understand: Most books then say that by Wedderburn's Structure Theorem, $A$ is either a division algebra or $A\cong M_2(F)$.

Why is that so? I suppose $M_1(D)$ is a division algebra, so I can still accept the division algebra part.

But why not $A\cong M_2(D)$, or even $M_3(D)$ for that matter? Why must the matrix be 2 by 2 and over a field?

Thanks! This has been puzzling me for some time.


1 Answer 1


If $A$ is a central simple $F$-algebra of dimension $4$ over $F$, it is certainly semisimple, hence $A\cong M_n(D)$, where $D$ is a (finite dimensional) division $F$-algebra.

Since $\dim_F(M_n(D))=n^2\dim_FD=4$, there aren't many choices. If $n=1$, then we need $\dim_F(D)=4$, so $D=A$. Otherwise $n=2$ and $\dim_F(D)=1$, so $D=F$.


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