Splitting field construction

Question

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?

quasi · Accepted Answer · 2018-03-27T17:26:15.313

1

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 $F_1$.

edited Mar 27 '18 at 17:26

answered Mar 27 '18 at 16:40

quasi

58,772

Thanks. So I gather we do factor $q$ into irreducibles over $F_1$ and then mod one more out. This doesn't seem consistent with what I read on wikipedia... What's the answer to the question I added in the edit? – David C. Ullrich Mar 27 '18 at 16:43
$p$ has a root in $F_1$, but $F_1$ need not be the splitting field of $p$. Of course, in the case where $\text{deg}(p)=2$, then in that case, it's automatic that $p$ splits over $F_1$. – quasi Mar 27 '18 at 16:46
Cool. Simple example? – David C. Ullrich Mar 27 '18 at 16:47
Are you asking for a counterexample where $p$ doesn't split over $F_1$? – quasi Mar 27 '18 at 16:51
Yes.(..........) – David C. Ullrich Mar 27 '18 at 16:58
Coming up . . . – quasi Mar 27 '18 at 17:07
I edited in an example which I think answers your new query. – quasi Mar 27 '18 at 17:24
Yes indeed, thanks. – David C. Ullrich Mar 27 '18 at 17:31

Splitting field construction

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