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

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


\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?

  • 1
    $\begingroup$ 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? $\endgroup$ Mar 11 at 2:49
  • $\begingroup$ @ArturoMagidin Oh that's a good point. Maybe there's a better way to go about showing this? $\endgroup$ Mar 11 at 2:52
  • $\begingroup$ 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 $(\deg(f)-1)!=(p-1)!$ $\endgroup$ Mar 11 at 2:53
  • $\begingroup$ $K=\mathbb{Q} (\omega)$ is the splitting field of $x^3-1$ over $F=\mathbb{Q}$. Here $[K:F] = 2$, a prime. But $x^3-1$ is reducible. $\endgroup$
    – Yathi
    Mar 11 at 3:07
  • $\begingroup$ @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 $p=3$. $\endgroup$ Mar 11 at 3:09

1 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.

  • $\begingroup$ This clears up so much confusion. Such a clear and concise answer. $\endgroup$ Mar 11 at 13:51

You must log in to answer this question.

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