Let $f(x) \in \mathbb{F_p}[x] $ be an irreducible polynomial of degree $n$.

Let $L$ be the splitting field of $f$.

Prove $[L:\mathbb{F_p}]=n$.

If $a_1,...,a_n$ are the roots of $f(x)$, then $L=\mathbb{F_p}(a_1,...,a_n)$.

It is obvious that $[L:\mathbb{F_p}] \le [\mathbb{F_p}(a_1,...,a_n):\mathbb{F_p}(a_1,...,a_{n-1})]...[\mathbb{F_p}(a_1):\mathbb{F_p}] \le n^n$.

Hence $L/\mathbb{F_p}$ is a finite extension of a finite field - thus separable, and simple,So $L=\mathbb{F_p}(\alpha)$, and $L$ is a finite field.

We can also see that it is the splitting field of $x^{p^m}-x$ ($m$ is such that $|L|=p^m$), so this is a Galois Extension and $[L:\mathbb{F_p}]=|Gal(L/\mathbb{F_p})$|.

So if I could find the size of the galois group... I would be done...

I don't know how to find it though :)

Please help

  • $\begingroup$ you are right. It is a different number. n is some natural such that $|L|=p^n$ , I will fix it. $\endgroup$ – Daniel Oct 4 '15 at 18:11

If you adjoin just one root of $f$ to $\mathbb{F}_p$, you already get an extension of degree $n$, so what you're really trying to show is that $K=\mathbb{F}_p[x]/(f(x))$ is the splitting field of $f$. Instead of trying to show directly that $K$ is the splitting field of $f$, it suffices to show that it is the splitting field of some polynomial (and hence is normal). Can you show this?


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