# irreducible polynomial is still irreducible on purely inseparable extension

I wish to prove that

$\alpha \in C$ where $C$ is algebraically closed field of $F$, $f(t) \in F[t]$ is minimal monic irreducible, separable polynomial having $\alpha$ as a root. Then let $S_{\alpha} = \{\sigma : F(\alpha) \to C$ where $\sigma$ is $F-$homomorphism$\}$ Show $|S_{\alpha}| = deg(f)$.

Actually I think $|S_{\alpha}| = deg(f)!$, not $deg(f)$.

What I already know is that $\sigma$ permutes roots of the polynomial in $f$. But isn't it possible that there is two homomorphism which permutes $\sigma$ to the same (other) root but permute other root to distinct another roots? What I mean is that, if $\alpha_1 , \alpha_2$ is another root of $f$, then there exist $F-$homomorphism $\sigma, \sigma'$ such that$\sigma(\alpha) = \sigma'(\alpha) = \alpha_1$ but $\sigma(\alpha_1)=\alpha_{2}$, $\sigma(\alpha_{1}) = \alpha$. This implies $|S_{\alpha}| = deg(f)!$. Could you agree with my argument or is there some point I've missed?

## 1 Answer

There is no typo, you should try to show $|S_\alpha|=deg(f)$.

If $\sigma(\alpha)=\sigma'(\alpha)$, then $\sigma=\sigma'$ as $F$-homomorphisms $F(\alpha)\rightarrow C$.

I think the point you may be confused on is that $F(\alpha)$ need not contain all roots of $f$. For instance, if you adjoin to $\mathbb{Q}$ a real root of the polynomial $x^4-2$, this will certainly not contain any imaginary roots.

In other words, $F(\alpha)$ is not necessarily the splitting field of $f$.