Finite Fields, The Automorphisms of a Finite Field

The Automorphisms of a Finite Field

Let F be a finite field of order pⁿ. Let b generate the multiplicative group of F. In other words, the powers of b cover the nonzero elements of F. If an automorphism maps b to b^j, then b^k becomes b^j to the k, which is b^jk, or b^kj, or b^k raised to the j. Thus an automorphism raises all elements to the j power, for some j.

Restrict attention to Z_p, and remember that its multiplicative group has a generator. Let r be such a generator. If we raise r to the p-1 power we get 1. We must raise r to the p to get r back again. Raising r to a different power gives another integer, but all integers must map to themselves. That means every automorphism raises all elements to the j power, where j is some multiple of p.

The obvious choice is the frobenius automorphism, which raises everything to the p power. This is indeed a field automorphism, and by the above paragraph, it generates all possible automorphisms.

If c() is the frobenius automorphism, it can be invoked again and again, until we reach cⁿ, which raises everything to the pⁿ. Remember that the powers of b cover the nonzero elements of F, all pⁿ-1 of them, so when we raise b to the pⁿ, we get b back again, giving the trivial automorphism. The group of automorphisms is a cycle of length n.

If K is a subfield of F it is finite, with order p^r for some r. If K and L are two subfields of order p^r, with K ≠ L, then there are more than p^r elements satisfying x^{(p^r)} = x, which is impossible. There is at most one subfield of order p^r. Remember that p^r - 1 must divide pⁿ - 1, hence r must be a factor of n. You can show this by synthetic division. To illustrate, divide p²+p+1 (which is p cubed minus 1) into p⁴+p³+p²+p+1 (which is p to the fifth minus 1). The first term of the quotient is p², giving a remainder of p+1, which is smaller than the divisor. It doesn't work because 3 does not divide 5. Thus r must divide n.

For any r dividing n, let d = c^r. Thus d is the automorphism produced by raising everything to the p^r. The field fixed by this automorphism is precisely the elements satisfying x^{(p^r)} = x. Aside from 0, the roots of this equation are generated by b^{(p^n-r)}. this establishes the subfield of order p^r.

The subfields of F are precisely the fields fixed by the various subgroups of the group of automorphisms on F; one subfield and subgroup for each r dividing n.

If K is an intermediate extension of order p^s, the field extensions between K and F are the subfields of F that contain all of K. These are the subfields of order p^r, where s divides r and r divides n. Once again these subfields are fixed by the subgroups of the group of automorphisms of F that fix K.

All this may seem confusing for now, but once you are familiar with Galois extensions, it will seem straightforward. We are really saying every finite extension of a finite field is Galois.

Conjugates and the Circular Shift

Let u be a root of an irreducible polynomial q(x) of degree n over Z_p, so that Z_p adjoin u produces a finite field extension of dimension n. The field of order pⁿ is born. The images of u under the automorphisms are u to the p, u to the p², and so on up to u to the p^n-1. Raise the entire polynomial q to the p power, using the frobenius automorphism, and the coefficients do not change. Thus each image of u is a root of q. The conjugates of q are the roots of q, which are u to the p, u to the p², and so on. Each automorphism performs a circular shift on these conjugates. In other words, the galois group is a circular shift of the roots of q, when theyare are arranged properly.

Let's use this example to dispel a common myth. In a nice galois extension, the conjugates need not span the extension as a vector space. Adjoin u²+1 to Z₃. Since 2 is not a square, this is irreducible. The sum of the roots is given by the second coefficient, which is 0. Thus u+u³ = 0. The conjugates are not linearly independent, and cannot span the entire vector space of dimension 2.

Let's explode another myth - the adjoined element need not be primitive. In the above example, u⁴ = 1. Thus u does not have order 8. A primitive element might be u+2, whose square is u.