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Wedderburn's little theorem $\quad$ every finite domain $A$ is a field.

Proof $\quad$ Let $x$ be a nonzero element of $A$. Because $A$ is finite, there exist positive integers $n$, $k$ such that $x^n = x^{n + k}$. It is easy to see by induction that the set $E = \left \{x^i : i \in \mathbf{N}^*\right\}$ does not contain $0$; it follows therefore from $x^n\left(1 - x^k\right) = 0$ that $x^k = 1$. Thus, $x^{k - 1}$ is the inverse of $x$ (when $k = 1$, $x$ has inverse $1$).

All the proofs I have seen of this result are much more sophisticated than mine. Hence, I am doubting its correctness and could use a second opinion.

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    Your proof only shows that every *commutative* domain is a field. Wedderburn's theorem states moreover that your $A$ is commutative.2012-09-18
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    @martini: You could turn that comment into an answer so the question can be marked resolved.2012-09-18
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    I think the word "domain" is ambiguous. It should be called "not necessarily commutative domain" though it's awkward.2012-09-18
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    The existence of inverses can also be proved by considering the map $y \mapsto xy$ and arguing that it is injective and hence surjective.2012-09-18

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Wedderburn's little theorem states as you wrote above, that a finite domain is a field. A field is a commutative domain $A$, such that every nonzero $x \in A$ has a multiplicative inverse. Your proof only shows that any finite domain is a skew field. You must also prove that $A$ is commutative, which needs more sophisticated arguments.

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    Well, this is embarrassing. I keep thinking that "field" is the translation for "corps" (French), which are not assumed to be commutative. Thank you.2012-09-18
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    @user25784 Indeed I believe Fields were originally not assumed to be commutative, especially by french mathematicians. Andre Weil's number theory text follows that convention.2012-09-18