# Finding roots in an extension field

1. May 16, 2016

### PsychonautQQ

1. The problem statement, all variables and given/known data
Let q be a root of p(x) = x^3 + x^2 + 1 in an extention field of Z2 (integers modulus 2). Show that Z2(q) is a splitting field of p(x by finding the other roots of p(x)

hint: this question can be greatly simplified by using the frobenius automorphism to find these zero's

2. Relevant equations
frobenius automorphism : f(a) = a^p

3. The attempt at a solution
So we are given one root and need to prove that this is enough to when adjoined to the original field to get the other two roots...

so p(q) = 0
and p(x) = (x-q)m(x) where m(x) is of degree 2

I'm a little confused on how to use the frobenius automorphism here or even why it would help at all...
anyone have any insight for me?

2. May 17, 2016

### Staff: Mentor

Have you tried to compute $f(q)$? Is it a root again? You know $q≠1$.

3. May 17, 2016

### PsychonautQQ

warning: i use p(x) to express the polynomial i'm trying to find the roots of and also f(q) = q^p, where p is the prime that the frobenius automorphism maps to. Sorry for unnecessary confusion >.<...

wouldn't q^p = q in Z mod 2? I mean, if q is odd then q = q^p = 1, if q is even then q = q^p = 0... perhaps these roots are not integers, i probably should have figured that >.< haha.

Does the solution have to do with the fact that if we apply the frobenius autmorphism to p(x), it will fix the coefficients of each term?

f(q) = q^p. I am not sure if this is a root or not. I suppose it must be for the reason I just stated, if i apply the frobenius morphism as so: f(0) = f(p(q)) = p(q^p), so yes, it appears that q^p is a root of p(x)... but how do we know that q^p doesn't equal q? Also, it's still a bit foggy to me how this frobenius morphism is going to help me prove the main question, i don't understand the connection very clearly. Thanks for being so patient with me!

4. May 17, 2016

### Staff: Mentor

Well, you have everything at hand, resp. said already. (again ...)

Your $p$'s are really confusing. Let us denote the prime by $2$ and the minimal polynomial $p(x) = x^3 +x^2 + 1 \in ℤ_2[x]$.
The good news is we don't have to bother signs and $a+a=0$ for any $a$.

You've told me last time that automorphisms map roots to roots.
Since $f(q) = q^2$ is a root, we obstinately could divide $p(x)$ by $(x+q)$ and next by $(x+q^2)$ and find the third root. With $p(q)=0$ we can get rid of the higher potentials of $q$. If there will be a remainder somewhere, then we must have done something wrong.

Edit: the "again" above has been meant as: You know more than you think you know.

Last edited: May 17, 2016
5. May 17, 2016

### PsychonautQQ

Automorphisms map roots to roots, how interesting. No matter what polynomial in what subfield, the roots will be mapped to roots. Should this be obvious or are automorphisms amazing? I realize that automorphisms are all about preserving basically everything, but still, why is it obvious that a counter example to this would be impossible? Like i understand that automorphisms should always map roots to roots if it fixes every coefficient in the polynomial, but even if these coefficients arn't fixed it still maps roots to roots? cool.

6. May 17, 2016

### Staff: Mentor

There's another hint: If $f$ maps roots to roots, what does $f^2$? (I admit I stubborn calculated the third root and found out afterwards that it is $f^2(q)$. Silly me.)

7. May 17, 2016

### Staff: Mentor

No, I don't think so. Please forget this at once. The Frobenius automorphism you mentioned isn't any automorphism! And you can easily check whether $q^2$ is a root of $p(x)$ by dividing $p$ by $x+q^2$
Edit: ... or easier whether $(q^2)^3 + (q^2)^2 + 1 = 0$.

Last edited: May 17, 2016
8. May 19, 2016

### Staff: Mentor

Let us be more careful here.

If we have a field extension $\mathbb{F} \subset \mathbb{E}$, a polynomial $p(x) \in \mathbb{F}[x]$ and an automorphism $σ \in Aut(\mathbb{E} / \mathbb{F})$, i.e. $σ(a) = a$ for all $a \in \mathbb{F}$, then for every zero $z \in \mathbb{E}$ of $p(x)$ we get: $$0 = p(z) ⇒ 0 = σ(0) = σ(p(z)) = p(σ(z))$$ So the crucial point here is the invariance of $\mathbb{F}$ under $σ$ that allows us the "root hopping" with automorphisms.

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