Every finite field F is isomorphic to GF(pn) for some prime p and natural n. That is to say, F has pn elements, and all fields with the same number of elements are isomorphic. (We then call the field with pn elements GF(pn). This is short for Galois Field.)

Proof: If a field F contains a field K, then F is a vector space over K. If we show that for every F there exists a prime p such that the set of elements of the form 1+1+...+1 is isomorphic to the field GF(p) -- the field defined by taking the addition and multiplication operations modulo p on the integers 0,1,...,p-1 -- then F is a finite vector space over GF(p). It therefore has a dimension n, and thus pn distinct elements. Obviously, two fields F, F' with the same number of elements are isomorphic as vector spaces over GF(p) (all finite dimensional vector spaces over F re isomorphic to Fn) . Such an isomorphism preserves addition (in the fields) as required, but not the fields' multiplication -- only multiplication by one of the p elements 1,1+1,... Choose an isomorphism of this sort which also preserves multiplication (just start by mapping the unit (1) of F to the unit of F' and proceed by extending the map any "legal" way...) and you have an isomorphism between the fields.

We still need to "find" GF(p) inside F. Consider the series 1, 1+1, 1+1+1,... Since F is finite, eventually some element must appear (say at the ith and jth places, i<j). So 1+1+..+1 j-i times is necessarily 0. Let p be the smallest natural number for which 1+1+...+1 (p times) = 0. It follows that if we write "n" for the element 1+1+...+1 (n times), then addition and multiplication on elements of this form are identical to the operations modulo p. p must be prime, else p=ab and then (in the field F) a+a+...+a (b times) = 0, which implies that 1+1+...+1 (b times) = 0, with b

QED

We haven't shown that there actually exists a field with pn elements for every prime p and natural n, but this is true too. One representation of this field is to take polynomials in GF(p), modulo an irreducible polynomial of degree n.

Here is a proof that any two finite fields of the same order are isomorphic1. (The proof in Haggi's writeup is plainly bogus2.)

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.


  1. 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.
  2. 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.

Log in or registerto write something here or to contact authors.