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

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,

$$$$[K:F]=[K:F(\psi)][F(\psi): F] = tp$$$$

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?

• You cannot assume "without loss of generality" that $f$ has a linear factor. Why could it not be that, say, a polynomial of degree $5$ is the product of an irreducible quadratic and an irreducible cubic? Mar 11 at 2:49
• @ArturoMagidin Oh that's a good point. Maybe there's a better way to go about showing this? Mar 11 at 2:52
• If it really was the case that $f$ already has a root in $F$, then the splitting field of $f$ would be the same as the splitting field of $g$, which has degree dividing $(\deg(f)-1)!=(p-1)!$ Mar 11 at 2:53
• $K=\mathbb{Q} (\omega)$ is the splitting field of $x^3-1$ over $F=\mathbb{Q}$. Here $[K:F] = 2$, a prime. But $x^3-1$ is reducible. Mar 11 at 3:07
• @YathirajSharma That's not the question. There $[K:F]$ may be prime, but it is not of the form $3t$; note that $p=\deg(f)$ in the question, so for your example that would be $p=3$. Mar 11 at 3:09

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 $$p$$, $$p\nmid \ell!k!$$.