For simple extensions of Q like [math]\sqrt{N}[/math] I know how to determine the matrix representation of multiplication by an element in the new field. But if I have a more complicated expression like [eqn]\sqrt{a+\sqrt{b}}[/eqn] which is a root of [eqn]x^8 -2(a^2+b)x^4 + (a^2-b)^2[/eqn] how do I find my basis vectors? Obviously I need 8 of them since the resulting polynomial is of degree 8.
Also is there a good web reference for this kind of thing? Representation theory / finite field extension stuff?
Thanks.
Carter Smith
>Thanks.
You're welcome.
Levi Baker
1,x,x^2,x^3,....,x^7.
Brayden Gomez
You should remember that your field is just Q[x]/(x^8-2(a^2+b)x^4+(a^2-b)^2) And the basis is just Since x^8=2(a^2+b)x^4-(a^2-b)^2 you can reduce any x^n with n>7 to some polynomial of degree 7 or less. And this is how you get your multiplication matrix for .
Just as easy as this. I think it should be in any book with Galois theory
James Cruz
>Since x^8=2(a^2+b)x^4-(a^2-b)^2 you can reduce any x^n with n>7 to some polynomial of degree 7 or less. Thanks guys, this makes sense.
Given that I found a polynomial which has this term as a zero, is there some way I can know that this is the right polynomial? Some way I can be sure that there isn't some sneaky 7th order polynomial for this term?
I got it by squaring, subtracting, etc, until I got zero, and then just substituted the variable x in the steps I took. Is this a sound way to always generate the proper order of a term so I know how to extend my base field?
Carter Stewart
>Given that I found a polynomial which has this term as a zero Which term?
Dominic Carter
ITT: Brainlets. I'm on phone so can' be bothered to write much, but there is a 4 degree polynomial which has that as a root. But that doesn't have to be minimal ploynomial, e.g. take a=1 and b=9, in which case it's Q, or a=2 and b=9, in which case it generates extension of degree 2.
Isaiah Diaz
Whichever.
For instance, if I take a term like [math]\sqrt{2}+\sqrt{3}[/math] I know in reality I need the basis vectors [math]1,~ \sqrt{2},~ \sqrt{3},~ \sqrt{6}[/math], but if I take [eqn]x=\sqrt{2}+\sqrt{3}[/eqn] then I end up with a polynomial like [eqn]x^4-10x^2+1[/eqn] and simply taking powers of [math]\sqrt{2}+\sqrt{3}[/math] doesn't seem to yield the right basis vectors.
Well, root 2 plus root 3 is a straightforward extension so this seems like a lot of busywork, but when I have a slightly more complicated expression such as [math]\sqrt{a+\sqrt{b}}[/math] then how do I know what the right extension field is?
Brody Price
Yep, he's right, OP. x^8 - 2(a^2+b)x^4+(a^2-b)^2 = (x^4-(a^2+b))^2 obviously, lol. So it's not a minimal polynomial since it's not irreducible in Q[x]. But neither x was a root of this polynomial (check it for a=2, b=1). x is a root of (x^2-a)^2-b.
So isn't right at all too. Your field is Q[x]/((x^4-a)^2-b) but only if (x^2-a)^2-b is irreducible in Q[x].
