Finite Field Extension

how do you type math stuff like that on Veeky Forums

What? You do something really strange.
If you have x and you have to figure out your minimal polynomial you can consider greater and greater powers of x until they become linearly dependent in Q which gives you an exact minimal polynomial for x (non-trivial linear combination of powers of x with coefficients in Q that is zero).

For example, if you have x=sqrt(2)+sqrt(3), you consider:
x^0 = 1
x^1 = sqrt(2)+sqrt(3)
x^2 = 5+2sqrt(6)
x^3 = 5sqrt(2)+4sqrt(3)+5sqrt(3)+6sqrt(2)=11sqrt(2)+9sqrt(3)
they're still linearly independent, but:
x^4=25+20sqrt(6)+24=49+20sqrt(6)
here you have x^4=x^2*10-1. This gives you minimal polynomial for sqrt(2)+sqrt(3)
which is x^4-x^2*10+1.

Oh, I see. I may be taking it too far by not considering linear combinations of powers already found. Thank you, this part is very clear now.

In this root2/root3 example, I already know my basis vectors for representing arithmetic by matrices needs 1, root2, root3, and root6. Is there some way I can easily determine this from the work you've just showed which would translate to an example like [math]\sqrt{3+\sqrt{7}}[/math] or something not obvious by inspection? Do I just look at all the terms generated during this polynomial generation and sort of ignore coefficients from my base field to see what elements are required?

In the x=sqrt(3+sqrt(7)) case it's obvious because you clearly see how to get a rational number out of multiplication and addiction of x powers. x^2=3+sqrt(7), (x^2-3)^2=7 and that's it. You just have to check that (x^2-3)^2-7 is irreducible.

It seems to me that even if your polynomial F has a root, or is otherwise reducible, you can still define multiplication modulo F and represent it as a linear mapping. You won't get a field because you will have a zero divisor, but the machinery will still work.

In OP's case, take the basis 1,x,...,x^7, then there is a 8x8x8 constant tensor A such that \sum_i \sum_j p_i A_{ijk} q_j represents multiplication of the vectors (p_i) and (q_j) (coefficients of the polynomials).

Or you could take A as a 8x16 matrix operating on the 16x1 concatenation of p and q.

I don't think that's right. OP's got 2(a^2+b) and (a^2-b)^2, the squared term has a minus in it.

Check the discriminant when you set y=x^4. It's 16 a^2 b, so all you need to factor it is that b be a square.

Ah, yes, I see, but anyway it's not a minimal polynomial

It always factors?

Yes, because there's a minimal polynomial for x=sqrt(a+sqrt(b)) of degree 4 (it is (x^2-a)^2-b) and minimal polynomial has to divide any polynomial with a root x.

*4 or less