# What is wrong with this proof of Wedderburn's little theorem?

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. – martini Sep 18 '12 at 10:15

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@martini: You could turn that comment into an answer so the question can be marked resolved. – joriki Sep 18 '12 at 10:21

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I think the word "domain" is ambiguous. It should be called "not necessarily commutative domain" though it's awkward. – Makoto Kato Sep 18 '12 at 11:20

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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. – lhf Sep 18 '12 at 11:28

## 1 Answer

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. – user25784 Sep 18 '12 at 11:22
@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. – Ragib Zaman Sep 18 '12 at 11:46
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