1

Given $F$ a field of characteristic different from $p$,

$ f(x)=x^p - a \in F[x]$ and $g(x)=x^p - 1 \in F[x]$. $E$ is the splitting field of $g$. If want to show that if $f$ f splits in $E$, then $f(x)$ also has a root in $F$.

If r in E is a root of f, then the set of r times all roots of unity are the p distinct roots of a.

However I don't know how to prove that if $E$ has a root (and therefore p distinct roots) of f, then there is also a root in F.

hereitis
  • 163
  • The roots of $g$ are the roots of unity $\zeta_p$. What are the roots of $f$ ? – reuns May 02 '17 at 00:10
  • Also note that $E(\sqrt[p]{a})$ is the splitting field of $f$. You can compute the degrees of the extensions, and use the fact they are relatively primes – reuns May 02 '17 at 00:18
  • If r in E is a root of f, then the set of r times all roots of unity are the p distinct roots of a. I will think about your second comment! Thank you very much – hereitis May 02 '17 at 01:59
  • $E(\sqrt[p]{a})/E$ is a Kummer extension. If $\sqrt[p]{a} \not \in E$ then I think $x^p-a$ is irreducible and $[E(\sqrt[p]{a}):E] = p$ – reuns May 02 '17 at 02:03
  • And use latex, see the code I typed ${}{}{}$ – reuns May 02 '17 at 02:06
  • Right...I proved this before...I tried to use them but didn't find helpful for this specific question. I will think about these propositions again. Thank you very much! Thanks, I will use learn to use latex! I prove that it is irreducible in E and also in F. But I am trying a second method to prove that it will be irreducible in F if it doesn't have a root there. Thank you very much. – hereitis May 02 '17 at 02:10
  • Related. – Jyrki Lahtonen May 02 '17 at 06:39
  • Thank you so much! I learned how to use some basic latex for the first time, thanks to your help! – hereitis May 02 '17 at 07:26

1 Answers1

1

Here is a stand-alone proof:

Assume $f$ has no root in $F$. Let $b = \sqrt[p]{a} \in E$ be a root of $f$ (which exists by assumption).

$E/F$ is an abelian extension, hence all intermediate fields are again Galois extensions, in particular $F(b)/F$ is. We deduce that $F(b)$ contains some other root of $f$, because it is a normal extension. Any other root has the form $b\zeta$, where $\zeta$ is a $p$-th root of unity (and thus automatically primitive). We obtain $\zeta \in F(b)$, i.e. $E \subset F(b)$ and thus equality. In particular the minimal polynomial of $b$ over $F$ (which is some irreducible factor of $f$) has degree $[E:F]$.

We can do the same with every root of $f$, hence all irreducible factors of $f$ over $F$ have the same degree, namely $[E:F]$. Since those degrees add up to $p$ and $p$ is prime, we deduce that there is only one irreducible factor, i.e. $f$ is irreducible over $F$. This shows $p-1 \geq [E:F] = [F(b):F] = p$, clearly a contradiction.

MooS
  • 31,390
  • $E$ is the splitting field of $g(x)=x^p - 1$ over $F$, the ground field. I am so sorry for the confusion I caused – hereitis May 02 '17 at 06:55
  • Yes, I am aware of this. What precisely do you not understand about my proof? – MooS May 02 '17 at 07:01
  • Maybe $[E:F] = p-1$ and hence I am not sure if it is abelian extension? – hereitis May 02 '17 at 07:03
  • 1
    Only $[E:F] \leq p-1$ (I have not claimed otherwise) holds and this is clear, since $E=F(\zeta)$ and $\zeta$ satisfies a polynomial of degree $p-1$, namely $1+\zeta+ \dotsb + \zeta^{p-1}=0$. And it is an abelian extension, because the Galois group is a subgroup of $(\mathbb Z/p\mathbb Z)^*$. – MooS May 02 '17 at 07:10
  • Thank you so much!! I should have read it more carefully. – hereitis May 02 '17 at 07:14
  • May I ask why it is true that " $F(b)$ contains some other root of $f$, because $F(b)/F$ is a normal extension"? I am confused here because we don't know $f(x)$ is irreducible yet??? I think you are right, just I do not understand this sentence. – hereitis May 02 '17 at 08:06
  • 1
    Yes, we do not know that $f$ is irreducible, but the minimal polynomial of $b$ over $F$ is at least of degree $2$ (we assumed $b$ is not contained in $F$) and hence there is another root, which is also root of $f$. – MooS May 02 '17 at 09:55
  • Thank you so much for your patient explanation!!!!! I understand now!! This is super helpful!! Thank you so much!! – hereitis May 03 '17 at 01:18
  • Hi, sorry.... I still couldn't understand why "Since those degrees add up to $p$ and $p$ is prime, we deduce that there is only one irreducible factor, i.e. $f$ is irreducible over $F$." – hereitis May 04 '17 at 18:26