0

Let $f$ be an irreducible polynomial over $\mathbb{F}_p$ of degree n. How many elements does the splitting field of $f$ over $\mathbb{F}_q$ has? With $q=p^m$ and $m<n$.

I am a bit stuck with this problem. I understand that the the splitting field of $f$ over the original field $\mathbb{F}_p$ has cardinality $p^n$. Now, irreducibility of $f$ over $\mathbb{F}_p$ also implies irreducibility over $\mathbb{F}_q$ but I don't know if the splitting field would be the same or $\mathbb{F}_(q^n)$

Nevado
  • 9
  • 1
    See this question. – Dietrich Burde Jun 07 '18 at 13:37
  • I can't see how this answer the question. Could you help me? – Nevado Jun 07 '18 at 13:42
  • Think about if $m$ divides $n$ that will mean you'll have a tower $\Bbb F_{p}$, $\Bbb F_{p^m}$,$\Bbb F_{p^n}$. If $m$ doesn't divide $n$ think about two separate branches going up $\Bbb F_p$ one to $\Bbb F_{p^m}$ and one to $\Bbb F_{p^n}$ you'd need to find an$\Bbb F_{p^l}$ containing both. – N8tron Jun 07 '18 at 13:43
  • And therefore this $\Bbb F_{p^l}$ would be the splitting field right? – Nevado Jun 07 '18 at 13:49
  • Only if it's the smallest such $l$. Also you should note that for a degree $n$ polynomial $f$ being irreducible in $\Bbb F_p$ does not force irreducibility in $\Bbb F_{p^m}$ when $m<n$ for instance take $f=x^4+x^3+1$ over $\Bbb F_2$ it's irreducible and $\Bbb F_{2^4}\cong \Bbb F[x]/(f)$ but over $\Bbb F_{2^2}[x]$ $f$ factors as $(x^2 + \omega x + \omega) * (x^2 + (\omega + 1)*x + \omega + 1)$ where $\Bbb F_4=\Bbb F_2(\omega) \cong \Bbb F_2[x]/(x^2+x+1)$ – N8tron Jun 07 '18 at 13:59
  • Oh, I see it now, thank you very much – Nevado Jun 07 '18 at 15:26

0 Answers0