# A concept about normal extension and splitting field

Let $$K$$ be an field extension of $$F$$. Then I know there is a theorem that says: $$K$$ is a splitting field of some polynomial over $$F$$ iff $$K$$ over $$F$$ is finite and normal. (Here the normal means every polynomial in $$F[x]$$ has a root then have all roots.)

Now, $$\Bbb Q(\sqrt{2})$$ does not have all roots of $$x^3-2$$, so it is not normal. Is it possible that $$\Bbb Q(\sqrt{2})$$ has all roots of some polynomial $$p(x)$$, but these roots are all not in $$\Bbb Q$$?

• Is your question about a general $K$, or specifically $K=\mathbb Q(\sqrt2)$? – BallBoy Nov 19 '18 at 14:39
• Specifically $K=\Bbb Q(\sqrt{2})$. let me edit – Eric Nov 19 '18 at 14:57
• $\Bbb Q(\sqrt{2})/\Bbb Q(\sqrt{2})$ is a (trivial) normal extension, the splitting field of $x \in \Bbb Q(\sqrt{2})[x]$. For extensions of $\Bbb Q$ when not specified we mean $K/\Bbb Q$ is a normal extension, thus the splitting field of some $p(x) \in \Bbb Q[x]$. – reuns Nov 19 '18 at 15:11

First, such a polynomial $$p(x)$$ would have to be irreducible over $$\mathbb Q$$. If it wasn't, it would split into factors at least one of which would be linear. This would yield a root in $$\mathbb Q$$.
If $$p(x)$$ is such an irreducible polynomial, then we have that $$\mathbb Q (^3 \! \sqrt2)$$ contains the splitting field of $$p(x)$$. But this extension has no intermediate extensions. (By multiplicity of degrees of extensions).
So $$\mathbb Q (^3 \! \sqrt2)$$ becomes the splitting field for an irreducible polynomial over $$\mathbb Q$$, which makes it a normal extension. But we know that this isn't the case - The polynomial $$x^3 - 2$$ has only one root here, out of the three possible.
They'd all have to be $$\notin \mathbb Q$$, as otherwise $$p$$ would be reducible.