Maple and finite fields

Maple and finite fields PhL 1.23.13 I have just discovered the Maple does in fact know something on this subject, at first I thought it did not. Consider GF(23). We know that p(x) = x3 + x + 1 is a prim poly and there is some primitive element α such that p(α) = 0. This α is a root of x3 + x + 1. We can tell Maple this by saying Now consider the following: This command is telling us the 8 roots of x7-1 in terms of 1 and powers of α. Thus, we have a listing here of all the non-zero elements of GF(23) each in terms of α. This is pretty important I think. The second item is just multiplicity, so ignore it. This we have α2+α 1 α+1 α α α2 α2 α+1 1 α2+α α2+1 α2+α + 1 α2+α + 1 α2+1 I can reorder them as shown on the right, and these are exactly my list in (6.11). Comment: This give me a quick way to make this list probably for any GF(pm) field. I know how to do it manually, but this is certainly faster. Question: Does every GF(2) polynomial have such a set of roots? I think the answer is no. For example, Although x2+x+1 has two roots in the complex numbers, I think the above says that it has no roots in GF(8). I would conclude that this could not be a minimum polynomial for GF(8)! So for the first time I seem to have a way to test polynomials. Let's try the two prim polys of GF(8), As expected, each of these prim polys has three roots in GF(8). Let's advance now to GF(16) which is my interest at the moment. I know a prim poly from my table 6.2 so here we go There are your 15 non-zero elements of GF(16), very good. Now try the prim poly itself and as expected there are four roots. Now we are in a position to test all the 4th degree polys I piled up in (5.31). Here is the first one x4 + x3 + x2 + x + 1 This DOES have four roots! Come back to this one later. x4 + x3 + x2 + 1 I think this is saying that have only one root in GF(16). What exactly does that mean? I think it means this cannot be a min poly because any min poly has all its roots in GF(16) ! Continue on x4 + x3 + x + 1 If I call the first and last elements a and b, this seems to say x4 + x3 + x + 1 = (x-1)2(x-a)(x-b) = (x+1) [(x+1)(x-a)(x-b)] = (x3+1)(x+1) Since a root is repeated, this is not a min poly. I can reduce it as shown. I would then claim that the squre bracket shows the roots of x3 + 1 and Maple verifies that fact, Let's keep on truckin' : x4 + x2 + x + 1 Again, this is like x4 + x3 + x2 + 1 above, it has only 1 root, it cannot be a min poly. Next This is a surprise to me. There are four roots, this is a min poly. I cannot reduce it in GF(p), it has coefficients in GF(p), it has at least one root in GF(q) so it is a min poly! This must be the third line of (5.27) ! That is what I was looking for! Question: How can I have Maple evaluate one of those factored min polys from Ch 5? For example, for the GF(24) case we have p1(x) = (x - α)(x - α2)(x - α4) (x - α8)