Wedderburn's little theorem

In mathematics, Wedderburn's little theorem states that every finite domain is a field. In other words, for finite rings, there is no distinction between domains, skew-fields and fields.

The Artin–Zorn theorem generalizes the theorem to alternative rings.

1 History
2 Sketch of proof
3 Notes
4 References
5 External links

History

The original proof was given by Joseph Wedderburn in 1905,^[1] who went on to prove it two other ways. Another proof was given by Leonard Eugene Dickson shortly after Wedderburn's original proof, and Dickson acknowledged Wedderburn's priority. However, as noted in (Parshall 1983), Wedderburn's first proof was incorrect – it had a gap – and his subsequent proofs came after he had read Dickson's correct proof. On this basis, Parshall argues that Dickson should be credited with the first correct proof.

A simplified version of the proof was later given by Ernst Witt.^[1] Witt's proof is sketched below. Alternatively, the theorem is a consequence of the Skolem–Noether theorem.

Sketch of proof

Let $A$ be a finite domain. For each nonzero $x \in A$ , the map

$a \mapsto ax: A \to A$

is injective; thus, surjective. Hence, $x$ has a left inverse. By the same argument, $x$ has a right inverse. A is thus a skew-field. Since the center $Z(A)$ of $A$ is a field, $A$ is a vector space over $Z(A)$ with finite dimension n. Our objective is then to show $n = 1$ . If $q$ is the order of $Z(A)$ , then A has order $q^n$ . For each $x \in A$ that is not in the center, the centralizer $Z_x$ of x has order $q^d$ where d divides n. Viewing $Z(A)^*$ , ${Z_x}^*$ and $A^*$ as groups under multiplication, we can write the class equation

$q^n - 1 = q - 1 %2B \sum {q^n - 1 \over q^d - 1}$

where the sum is taken over all representatives $x$ that is not in $Z(A)$ and d are the numbers discussed above. $q^n-1$ and $q^d-1$ both admit factorization in terms of cyclotomic polynomials $\Phi_n$ . After cancellation, we see that $\Phi_n(q)$ divides ${q^n - 1 \over q^d - 1}$ and $q^n - 1$ , so it must divide $q - 1$ . So we reach contradiction unless $n = 1$ .

Notes

^ ^a ^b Lam (2001), p. 204

References

Parshall, K. H. (1983), In pursuit of the finite division algebra theorem and beyond: Joseph H M Wedderburn, Leonard Dickson, and Oswald Veblen, Archives of International History of Science, 33, pp. 274–99
Lam, Tsit-Yuen (2001). A first course in noncommutative rings. Graduate texts in mathematics. 131 (2 ed.). Springer. ISBN 0387951830.

External links

Proof of Wedderburn's Theorem at Planet Math