Here is a
proof that any two
finite fields of the same order
are isomorphic
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 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
in all.
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
is unconvincing.