TITLE: Splitting field construction
QUESTION [1 upvotes]: Context: Been teaching myself a little algebra - a lot of it makes more sense than in did years ago when I took that course.
Say $F$ is a field and $p,q\in F[x]$ are irreducible. We want to construct a splitting field for $pq$. I gather we do this: Let $F_1=F[x]/(p)$. If $q$ does not factor into linear factors over $F_1$ then let $F_2=F_1[x]/(q)$.
Question: Is it actually true that either $q$ is irreducible over $F_1$ or $q$ factors into linear factors over $F_1$? If so how do you prove it?
My work so far: I've been trying to show that $q$ is irreduclibe over $F_1$. Just realized there are good reasons I've been unable to prove that, for example $F=\Bbb Q$, $p=x^2+1$, $q=x^2+4$. (Or come to think of  it it could be that $q=p$...)
Come to think of it, I just now realized we don't really need to know that $q$ is either irreducible or splits over $F_1$; if neither we could just factor $q$ into a product of irreducibles over $F_1$ and proceed. But I gather that's not necessary...
Edit: Realized after posting this that I've been assuming without proof that $p$ actually splits over $F_1$. Doesn't really matter, as above, but is that true?

REPLY [1 votes]: For a counterexample, using $F=\mathbb{Q}$, let $p=x^2-2$, and let $q=x^4-2$.

Then $p,q$ are irreducible over $\mathbb{Q}$.

Note that $F_1=\mathbb{Q}[x]/(p)$ is isomorphic to $\mathbb{Q}(\sqrt{2})$.

Regarding $F_1$ as $\mathbb{Q}(\sqrt{2})$, $q$ factors over $F_1$ as
$$q = (x^2+c)(x^2-c)$$
where $c=\sqrt{2}$.

But the roots of $q$ have degree $4$ over $\mathbb{Q}$, so are not in $F_1$.

Hence $q$ does not have a linear factor over $F_1$.

Responding to the edit, here's a similar example, but where $p$ does not split over $F_1$.

Using $F=\mathbb{Q}$, let $p=x^4-2$, and let $q=x^6-2$.

Then $p,q$ are irreducible over $\mathbb{Q}$.

Note that $F_1=\mathbb{Q}[x]/(p)$ is isomorphic to $\mathbb{Q}(c)$, where $c=\sqrt[4]{2}$.

Regard $F_1$ as $\mathbb{Q}(c)$.

Then $p$ factors over $F_1$ as
$$p=(x+c)(x-c)(x^2+c^2)$$
but $p$ does not split over $F_1$ since $F_1\subset\mathbb{R}$, and $p$ has two non-real roots.

Note that $q$ factors over $F_1$ as
$$q = (x^3+c^2)(x^3-c^2)$$

But the roots of $q$ have degree $6$ over $\mathbb{Q}$, so are not in $F_1$.

Hence $q$ does not have a linear factor over $F_1$.
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