I’ve been reading some algebra from Larry C. Grove’s Algebra from Dover. It’s a nice book with lots of exercises, and while doing some of them I found out a nice way to prove Wilson’s theorem in any finite field \(F\). This post does require the reader to know what a group is and what a field is. Wilson’s theorem (or maybe a generalization of it) states that
$$\prod_{a \in F^*} a = -1$$
First, we have need a basic result in group theory, Lagrange’s theorem. It states that for a group \(G\) and a subgroup \(H \) we have the equality
$$ |G| = |H| [G : H] $$
where \([G:H]\) is the number of different cosets, that is, sets \(gH = \{ gh \mid h \in H \}\) for any \(g \in G\). This can be seen from several observations. We have \(|H| = |gH|\) because the group operation \(h \mapsto gh\) is a bijection since it has an inverse \(h \mapsto g^{-1}h\). Also, the cosets must be disjoint, for if \(a \in g_1 H \cap g_2 H\) or explicitly written \(a = g_1 h_1 = g_2 h_2\), we have $$ g_1 H = g_1 (h_1 H) = a H = g_2 (h_2 H) = g_2 H$$
From these facts we can see that \(G\) partitions into subsets of the size \(|H|\) and we define that \([G : H]\) is the number of these sets.
If we have a finite group \(G\), and take any element \(x \in G\), we must have $$x^a = x^b,\; a \neq b,\; a, b \in \mathbb{N}$$
By division we get that \(x^n = 1\) for some \(n\) and we can assume that \(n\) is minimal.
We have the subgroup \(H\) of \(G\) generated by \(x\) that can be defined to be the smallest subgroup of \(G\) containing \(x\).
We can define a group \(H_2 = \{ x^a \mid a \in \mathbb{Z}\}\). \(H_2\) is a finite cyclic group since \(x^n = 1\) and $$\begin{align} x^a x^b &= x ^{a + b} \\ x^0 &= 1 \\ \left(x^a\right)^{-1} &= x^{n – a} \end{align}$$
We can also see that \(H \subset H_2\) because \(H\) is the smallest subgroup containing \(x\) and that \(H_2 \subset H\) because every element of \(H_2\) can be written in terms of \(x\), and hence \(H = H_2\).
Also, we can see that the size \(|H| = n\), for if \(x^a = x^b\), we have \(x ^{a-b} = 1\), and hence \(n \mid a – b\) because \(n\) is the minimal (positive) number for which it holds, and we see that \(x^i\) is different for every \(i \in \{ 0, 1, \dots, n – 1 \}\).
This means that for a finite group \(G\), and an element \(x \in G\) that generates the cyclic subgroup \(H = \left\langle x \right\rangle\), we always have $$x ^ {|G|} = x^{|H| [G : H]} = \left(x^{|H|} \right)^{[G:H]} = 1^{[G:H]} = 1 \tag{1}\label{1}$$
Now let us assume that we have a finite field \(E\), and we look at the prime field \(F\) of \(E\) which is the subfield generated by the multiplicative identity \(1\). It is the smallest possible subfield of \(E\). If \(|F| = ab\), we have \((ab) 1 = (a 1)(b 1) = 0\) by the equation \(\eqref{1}\). (Notice that we are using additive syntax here whereas \(\eqref{1}\) uses multiplicative syntax.) Since a field doesn’t have zero divisors (it has multiplicative inverses), we must have that either \(a1\) or \(b1\) is zero in \(F\). But this can’t be possible if both \(a > 1 \text{ and } b > 1\), since then we would have \(ab = |F| < a\) since \(1\) generates the additive group of \(F\). Hence either \(a = 1 \text{ or } b = 1\) and \(|F|\) must be a prime number.
Since \(E\) is a vector space over a scalar field \(F\), we have that $$E = \{ a_0e_0 + \dots + a_ke_k \mid a_i \in F\} $$ for some basis vectors \(e_i \in F\), and by calculating the number of elements in this set we see that \(|F| = p^n\).
To return back to the Wilson’s theorem, let us define a polynomial \(f(x) = x^{q – 1} – 1\) where \(q = p^n\). We can see that every element of \(F^* = F \setminus \{0\}\) is a root of \(f \text{ by equation } \eqref{1}\). Since the size of the multiplicative group is \(\left|F^*\right| = q\, – 1\) is we see that necessarily we have the factorization $$ f(x) = (x – a_0)\dots(x – a_{q\, – 1}) \text{ where }F^*=\{a_0 \dots a_{q\,-1}\}$$
By comparing the constant terms of these two representations of the polynomial \(f(x)\) we end up with Wilson’s theorem (for finite fields): $$ \prod_{a \in F^*} a = -1 $$
As an aside, we can use the polynomial \(f(x) = x^q – x\) to construct the field \(F_q\) as a splitting field of \(F_p\) (though the proof probably requires analysis of the Galois group or something similar) since \(f(x)\) is separable which means that is has no repeated roots. This can be seen from the fact that if \(f(x)=\left(x-a\right)^2g(x)\), the derivative is $$f'(x) = 2(x – a)g(x) + \left(x-a\right)^2g'(x) = (x-a)(2g(x) + (x-a)g'(x))$$ and then \((x-a) \mid \text{gcd}(f(x), f'(x))\). But in this case we have that $$ f'(x) = q x^{q\,-1} – 1 = p^n x^{q\,-1} – 1 = -1$$ since \(p1 = 0\). This implies that \(\text{gcd}(f(x), f'(x)) = 1\), so \(f(x)\) is separable. Also since we have nice isomorphisms between splitting fields, though the actual theorems are pretty technical, any two finite fields of the same size are isomorphic.