Suppose $f(x)$ is irreducible over $\mathbb{F}_p[X]$ and let $\alpha$ be a root of $f$ in some extension field.

I want to prove the following claim. I have included thoughts below, but I am very confused as to how this can be shown.

Claim: $\mathbb{F}_p(\alpha)$ is a splitting field for $f$ over $\mathbb{F}_p$.

Now, if we suppose that $\text{deg}(f)=n$, then $\mathbb{F}_p(\alpha)$ contains $p^n$ elements, so $\mathbb{F}(\alpha) \cong \mathbb{F}_{p^n}$.

By consider the Frobenius map $\theta:\alpha \rightarrow \alpha^p$, which is an automorphism, we can show that $\text{Gal}(\mathbb{F}_{p^n} : \mathbb{F}_p) \cong C_n$, the cyclic group of order $n$.

$\text{Gal}(\mathbb{F}_{p^n} : \mathbb{F}_p) \cong C_n$ permutes the roots of $f$ (Is this true?) and so all of the roots are of the form $\alpha^t$ for some $t$ and so contained in $\mathbb{F}_{p^n}$. So any splitting field would have to be a subfield of order dividing $p^n$.

But $p$ is prime, so the only other subfields are those of size $p^s$ where $s |n$. (Why can none of these be splitting fields?)

  • 3
    $\begingroup$ Yes, the Galois group permutes the roots: this is how Galois thought of his groups (not as automorphisms of fields: that point of view was introduced later) . No strict subfield of $\mathbb{F}_p(\alpha)$ is a splitting field because, hey, the least obligation of a splitting field is to contain the root $\alpha$ of $f$ ! $\endgroup$ – Georges Elencwajg Apr 21 '14 at 18:02

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