^{1}. (The proof in Haggi's writeup is plainly bogus

^{2}.)

Let *F* and *F'* be two finite fields
of the same order. These fields must have
characteristic *p*, for a prime p, that is
they contain **Z**_{p}
as a subfield. I claim that the order of *F* is *p ^{n}*, for some

*n*. For, if we choose

*e*a basis for

_{1},...,e_{n}*F*over

**Z**

_{p}then writing a typical element as a linear combination of these basis elements there are

*p*possibilities for each of the

*n*coefficients, making

*p*elements in all.

^{n}
We know that the group of units of *F* is cyclic (see
a proof that the multiplicative group of a finite field is cyclic).
Let *a* in *F* be some generator. Then clearly
*F= Z_{p}(a)* is the simple field extension generated by

*a*. Now let

*f(x)*be the minimal polynomial of

*a*over

**Z**

_{p}. By the first isomorphism theorem

*F*is isomorphic to

**Z**

_{p}

*[x] / (f(x))*. So our aim is to show that

*F'*has the same property.

I claim that the elements of *F'* are exactly the roots of
the polynomial *h(x)=x ^{q}-x*, where

*q=p*. To see this, consider an element of the field that is not zero. It is an element of the group of units so if we raise it to the power

^{n}*p*we must get 1. The claim is now clear.

^{n}-1
Now factorise *h(x)* into a product of irreducibles. Since
our original *a* is a root of *h(x)* it must be that
*f(x)* is one of the irreducible factors of *h(x)*.
Choose *b* in *F'* that is a root of *f(x)*.
Then, again by the first isomorphism theorem, we have that
**Z***(b)* is isomorphic to
**Z**_{p}*[x] / (f(x))*. This
subfield of *F'* therefore has as many elements as *F* (and
so *F'*) and so this subfield equals *F'*. We are done.

On the other hand, in order to construct a finite field of
order *p ^{n}* create a splitting field

*k*of

*h(x)*(as above) over

**Z**

_{p}. One can show easily that the zeroes of

*h(x)*form a subfield of

*k*. The only tricky part is that if

*a,b*are zeroes then so is

*a-b*, but this follows because the appropriate binomial coefficients are divisible by

*p*(cf Frobenius endomorphism). Since this polynomial is separable (it shares no roots with its derivative) these zeroes are distinct. So this subfield is a field with

*p*elements.

^{n}-
A much simpler proof but using deeper technology is to use that a finite
field is a splitting field for
*h(x)*over**Z**_{p}(this was shown in the proof above) and then appeal to the fact that splitting fields are unique. - The problem with Haggis's argument is that he just shows that there is a vector space isomorphism between the two fields. His "argument" that you can choose such a vector space isomorphism that preserves multiplication is unconvincing.