Here is a proof
that any two finite field
s of the same order
. (The proof in Haggi
's writeup is plainly
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 Zp
as a subfield. I claim that the order of F is pn, for some n.
For, if we choose e1,...,en a basis
for F over Zp then
writing a typical element as a linear combination of these
basis elements there are p possibilities for each of
the n coefficients, making pn elements
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=Zp(a) is the simple field extension generated by
a. Now let f(x) be the minimal polynomial of
a over Zp. By the first isomorphism theorem
F is isomorphic to Zp[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)=xq-x, where q=pn.
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 pn-1 we must get 1.
The claim is now clear.
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
Zp[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 pn create a splitting field k
of h(x) (as above) over Zp. 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 pn elements.
A much simpler proof but using deeper technology is to use that a finite
field is a splitting field for h(x)
over Zp (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