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Can someone please prove the following:

Show that a subalgebra of a finite dimensional division algebra is a (finite dimensional) division algebra.

Thanks! G.

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Let $D$ be a division algebra (not required to be associative) and let $A \subseteq D$ be a subalgebra. Since $1 \neq 0$ in $D$, the same is true in $A$, so that $A \neq 0$. If $a \in A, b \in A \setminus \{0\}$, there is at most one solution for $a=bx$, since this is the case in $D$. For $a=0$ this means that the linear map $A \to A, x \mapsto bx$ is injective. Since $A$ is finite-dimensional, the map is surjective, i.e. there are solutions of $a=bx$. Similarily one proves that $a=xb$ has a unique solution.

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thanks! isn't it enough to take a=1 and say that there is a solution for ax=1? –  cruvadom Jan 13 '13 at 2:28
    
Which definition of division algebra do you choose? –  Martin Brandenburg Jan 13 '13 at 2:57
    
a ring A with a module structure over a commutative ring, which is embedded in the ring's center, usual definition I think –  cruvadom Jan 13 '13 at 12:12

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