Wilson’s Theorem

Prove that x^{p^n-1}-1 = \prod_{\alpha \in \mathbb{F}_{p^n}^\times} (x-\alpha). Conclude that \prod_{\alpha \in \mathbb{F}_{p^n}^\times} \alpha = (-1)^{p^n}. Deduce Wilson’s Theorem: if p is an odd prime, then (p-1)! \equiv -1 mod p.

Recall that x^{p^n}-x = \prod_{\alpha \in \mathbb{F}_{p^n}^\times}(x-\alpha) by definition, and that 0 \in \mathbb{F}_{p^n} is merely the root having minimal polynomial x. So x^{p^n-1}-1 = \prod_{\alpha \in \mathbb{F}_{p^n}^\times} (x-\alpha). Comparing constant coefficients, we have \prod_{\alpha \in \mathbb{F}_{p^n}^\times} (-\alpha) = -1, so that $latex (-1)^{p^n-1} \prod_{\alpha \in \mathbb{F}_{p^n}^\times} \alpha = -1, and hence \prod_{\alpha \in \mathbb{F}_{p^n}^\times} \alpha = (-1)^{p^n}.

Restrict now to the field \mathbb{F}_p \cong \mathbb{Z}/(p) with p odd. Then \prod_{\alpha \in \mathbb{F}_p} \alpha = (p-1)!, and (-1)^p = -1. Thus (p-1)! \equiv -1 mod p.

(I’d like to point out that this is a really roundabout way to prove Wilson’s Theorem. The easy(ier) way is to note that in \mathbb{Z}/(p)^\times, every element but -1 is distinct from its inverse.)

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