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Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field

Let $\mathbb{F}$ be a finite field and let $A$ be a finite-dimensional associative algebra over $\mathbb{F}$ without zero divisors. Prove $A$ is a field.

Rearmk: Wedderburn's theorem states that every finite division algebra is a field. Is there any way to show $A$ is a finite division algebra?

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user31899
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    Every finite integral domain is a field. Does that help? – Gerry Myerson Oct 22 '12 at 05:31
  • Basically a duplicate of http://math.stackexchange.com/questions/146857/if-a-an-integral-domain-contains-a-field-k-and-a-over-k-is-a-finite-dime and http://math.stackexchange.com/questions/63950/proof-that-an-integral-domain-that-is-a-finite-dimensional-f-vector-space-is-i – rschwieb Oct 22 '12 at 12:24

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If $F$ is a completely arbitrary field (finite or not), then every finite-dimensional associative $F$-algebra $A$ without zero-divisors is a division algebra.

Indeed, given $a\neq0\in A$ the multiplication $\mu_a:A\to A:x\mapsto ax$ is an $F$- linear endomorphism.
It is injective because $a$ is not a zero-divisor (by the assumption on $A$) hence it is surjective by linear algebra.
Surjectivity implies that there exists $b\in A$ with $\mu_a(b)=1=ab $.
Similarly there exists $c\in A$ with $ca=1$.
But then $c(ab)=c1=c=(ca)b=1b=b$ so that $b=c$ is an inverse for $a$.
We have proved that $A$ is a division algebra since each nonzero element $a\in A$ is invertible.

Georges Elencwajg
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