$K$ is the splitting field of degree $p$ polynomial $f(x)$. If $[K:F] =tp, t \in \mathbb{N}$, $f(x)$ is irreducible

Question

Let $p$ be a prime and let $f(x) \in F[x]$ have degree $p$. Let $K$ be the splitting field for $f(x)$ over $F$. Suppose $[K:F] =tp$ for some $t \in \mathbb{N}$. Prove that $f(x)$ is irreducible over $F$.

Here's what I have so far.

Suppose not. Suppose instead that $f(x)$ is reducible. Then $f(x) = g(x)h(x)$ for nonconstant $p(x),q(x) \in F[x]$. Since $p$ is prime, then WLOG $\deg(g(x)) = p-1$ and $\deg(h(x)) = 1$. Assume also that $h(x)$ is monic. Then $h(x)=x-\psi$ for some $\psi\in F$. This means $\psi$ is a root of $f(x)$ in $F$. Since $K$ is the splitting field for $f(x)$, we know

$$F \subseteq F(\psi) \subseteq K$$

Therefore,

\begin{equation} [K:F]=[K:F(\psi)][F(\psi): F] = tp \end{equation}

The chain of inclusions is actually strict, i.e.

$$F \subset F(\psi) \subset K$$

and this is because $f(x)$ has more than just $\psi$ as a root, by our reducibility assumption. So neither $K/F(\psi)$ nor $F(\psi)/F$ are trivial extensions. Now,

$$[F(\psi):F] < p \implies [F(\psi):F] \leq t$$

because $p$ is prime. Also, we claim $[K: F(\psi)] \geq p$. To see this, observe that $[K:F(\psi)] < p \implies [F(\psi):F] > t \implies [F(\psi):F] = tp$, which is a contradiction. Hence, $[K:F(\psi)] \geq p$. And so $[F(\psi): F] = t$. And so $[K:F(\psi)] = p$.

Where do I go from here?

Answer 1

As I noted in comments, your argument is incorrect from the start, as you assume that if $f$ is not irreducible, then it must have a linear factor. That is not true.

Claim. If $f(x)\in F[x]$ is separable and has degree $k$, then its splitting field has degree over $F$ dividing $k!$.

Proof. The splitting field is Galois over $F$; and the Galois group acts on the roots of $f$; since the splitting field is generated by the roots of $f$ action of any element of the Galois group is completely determined by the action on the roots. Thus, the Galois group is isomorphic to a subgroup of $S_k$, hence has order dividing $k!$.

Now let $f(x)$ have degree $p$. If $f(x)$ is reducible, $f(x) = g(x)h(x)$ with $\deg(g)=k\lt p$, $\deg(h)=\ell\lt p$, then the splitting field $L$ of $g$ has degree dividing $k!$; and $K$ is the splitting field of $h$ over $L$, so $[K:L]\mid \ell!$. Therefore, $[K:F]=[K:L][L:F]\mid \ell!k!$. In particular, since both $\ell$ and $k$ are strictly smaller than $p$, $p\nmid \ell!k!$.

The result now follows by contrapositive.