Finite Field Extension

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>Thanks.

You're welcome.

>>8345188
1,x,x^2,x^3,....,x^7.

>>8345188
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 >>8345415
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 >>8345415.

Just as easy as this. I think it should be in any book with Galois theory

>>8345470
>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.
>>8345415
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?

>>8345654
>Given that I found a polynomial which has this term as a zero
Which term?

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.

>>8345681
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?

>>8345883
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 >>8345470 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].

>>8345997
*your field is Q[x]/((x^2-a)^2-b)

>>8345188
how do you type math stuff like that on 4chan

>>8345920
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.

>>8346044
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?

>>8346083
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.

>>8345997
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.

>>8346230
Ah, yes, I see, but anyway it's not a minimal polynomial

>>8346240
It always factors?

>>8346281
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.

>>8346314
*4 or less

So let's see if you understood this OP (and other people having trouble with this in here), what's the degree of field Q(sqrt(9+sqrt(80))) over Q?

>>8346801
Well I get [eqn](x^2-9)^2-1280[/eqn] so it's degree four, and I guess my basis vectors are [math]1,~ \sqrt{5},~ \sqrt{9+16\sqrt{5}},~ and~ \sqrt{45+80\sqrt{5}}[/math].

You mean (x^2-9)^2-80? That polynomial indeed has sqrt(9+sqrt(80)) as a root, but to prove that it's minimal polynomial, you have to prove it's irreducible. (Hint: it's not)

>>8346851
Yes, algebra mistake. Never do math before your first cup of coffee. Then the polynomial would be [math]x^4-18x^2+1=0[/math]. If this had a rational root it would have to be 1, but 1 is not a root, so this seems irreducible to me.

If I understand, this would make the basis vectors [math]1,~ \sqrt{5},~ \sqrt{9+4\sqrt{5}},~ \sqrt{45+20\sqrt{5}}[/math] as these are all the terms not in Q that I encountered in my expansion from the term to the polynomial.

>>8346882
A polynomial not having rational roots does not mean it's irreducible over rational numbers. For example, the polynomial x^4-2x^2-3 has no rational roots, and yet it factors as (x^2+1)(x^2-3).

So no, that's not a basis. To give you a hint why not, I'm going to show that that third element in your "basis" is a linear combination of the first two. We claim that a+bsqrt(5) = sqrt(9+4sqrt(5)) for some rational numbers a and b, i.e. (a+bsqrt(5))^2=9+4sqrt(5). Multipliying the left hand side out we get

(a^2+5b^2)+2absqrt(5)=9+4sqrt(5), so
a^2+5b^2=9
2ab=4, and we can immediately notice that a=2, b=1 solves this equation. That means that
(2+sqrt(5))^2=9+4sqrt(5), i.e. 2+sqrt(5)=sqrt(9+4sqrt(5)).

Now you might want to try the analogous procedure for sqrt(9+sqrt(80)).

>>8346910
Unexpected!

But 9+4root5 is 9+root80 so you did my work for me. This is very interesting.

So if I suppose this fourth order factors into quadratics, then [math](x^2+ax+b)(x^2+cx+d) = x^4+(a+c)x^3+(b+d+ac)x^2+(bc+ad)x+bd[/math] and we have that a+c=0, b+d+ac=18, bc+ad=0, and b*d=1. Since c=-a, the second condition is b+d-a^2=18, and the third is a(d-b)=0, so either a is zero or b=d. Suppose b=d, 2b - a^2 = 18 has -1, 4 as a solution.

So [math]x^4-18x^2+1 = (x^2+4x-1)(x^2-4x-1)[/math] which are definitely irreducible quadratics. This suggests the extension field I need is [math]\mathbb{Q}(\sqrt{5})[/math].

>>8346974
And if I shift those polynomials by 2 (respectively -2) then they show that they're both just [math]x^2-5[/math].