# Normal field extension implies splitting field

I feel like this fact should be easy but I'm struggling to see it. If I have a polynomial $f \in K[x]$ which is irreducible and has roots $\alpha$, $\beta$ in some finite normal (over $K$) extension field $L$, is it true that $L$ is then the splitting field of $f$ over $K(\alpha)$?

I know that in general a finite extension is normal if and only if it is the splitting field of SOME polynomial. I also know that if $L$ is normal over $K$ then it is normal over $K(\alpha)$, hence $f \in K(\alpha)[x]$ must split in $L$, but how do we know there is no intermediate field $A$ between $K(\alpha)$ and $L$ where $f$ also splits?

Thanks

• Yes I did, thanks for the corrections – Wooster Nov 14 '14 at 12:12

## 1 Answer

In fact you don't know that there are no intermediate fields between $K(\alpha)$ and $L$. Take as an example $K = \mathbb{Q}$, $L =$ splitting field of $x^4+1$. Then $f=x^2+1 \in \mathbb{Q}[x]$ is irreducible, it factors inside $L$, but $L$ is NOT the splitting field of $f$ over $\mathbb{Q}$.

• Are the splitting fields of $x^4 +1$ and $x^2 + 1$ not the same? Being $\mathbb{Q}(i)$? – Wooster Nov 14 '14 at 12:55
• No. $x^4+1$ has as a root something whose square is $i$ (e.g. $e^{\frac{i \pi}{4}}$) – Crostul Nov 14 '14 at 13:29
• Ah yes, I was thinking of $x^4 -1$ for some reason. Thanks – Wooster Nov 14 '14 at 13:39
• Could it possibly the the case that in fact $L$ is the splitting field for $f$ as considered a polynomial in $K(\alpha)$? – Wooster Nov 15 '14 at 15:49