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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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marked as duplicate by rschwieb, Qiaochu Yuan Oct 22 '12 at 12:40

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Every finite integral domain is a field. Does that help? –  Gerry Myerson Oct 22 '12 at 5:31
    
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up vote 3 down vote accepted

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.

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Why is $A$ unital? Does a finite-dim algebra always have an indentity? –  user31899 Oct 22 '12 at 12:03
    
For me an algebra (or a ring) always has an identity by definition. –  Georges Elencwajg Oct 22 '12 at 13:39
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