2
$\begingroup$

I have just been doing a Galois theory exercise and one part of the exercise requires me to explain why, if $f \in \mathbb{F}_p[X]$ is a polynomial of degree $n$, its splitting field $L$ is a cyclotomic extension of $\mathbb{F}_p$ (i.e. $L/\mathbb{F}_p$ is a cyclotomic extension).

I cannot offer an explanation as to why this is the case. Perhaps it is something to do with there being an element inside the extension with, for example, $x^n = 1$ (thinking of Fermat's little theorem here) but I don't know how to pursue further this line of thinking, or how this is connected to the fact that $L$ is the splitting field of some polynomial.

$\endgroup$
6
  • 4
    $\begingroup$ Hint: every finite field $F$ has a cyclic multiplicative group $F^\times$. Does that give an idea for a polynomial that every element of $F$ must be a root of? $\endgroup$ Commented May 4 at 20:32
  • $\begingroup$ This previous Question Finite extensions of finite fields always generated by cyclotomic type polynomials might help you tie out your intuition. $\endgroup$
    – hardmath
    Commented May 4 at 20:37
  • $\begingroup$ @GregMartin That in, for example, $\mathbb{F}_p$ every element is a $p$-th root of 1? So essentially is it true in more generally that $\mathbb{F}_{p^n}/\mathbb{F}_p$ is a cyclotomic extension (by what you have said) and this implies the result in the exercise I was doing? $\endgroup$ Commented May 4 at 21:06
  • $\begingroup$ See this Question (among others) for the proposition that the multiplicative group of a finite field is cyclic. $\endgroup$
    – hardmath
    Commented May 4 at 21:11
  • 2
    $\begingroup$ Basically this is because every non-zero element of every algebraic extension of $\Bbb{F}_p$ is a root of unity of some order. $\endgroup$ Commented May 4 at 21:14

1 Answer 1

1
$\begingroup$

Since an extension of $\Bbb F_p$ of degree $n$ has order $p^n$ (it's a vector space of dimension $n$ over $\Bbb F_p$), we have that the multiplicative group has order $p^n-1.$

By Lagrange's theorem every element of $\Bbb F_p^*$ is a solution to the equation $$x^{p^n-1}-1=0.$$

Since $L$ is a splitting field, it's the splitting field of $x^{p^n}-x.$

Thus a primitive $p^{n-1}$-th root of unity $\zeta $ will do the trick. That is, $\Bbb F_{p^n}=\Bbb F_p(\zeta) $ is a cyclotomic extension.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .