Irreducibility in Galois/non Galois Extensions

Question

Let $k$ be a field and $\alpha$ algebraic over $k$. Let $K$ be the Galois closure of $k(\alpha)$ (obtained by adding all conjugates of $\alpha$). If $f(x) \in k[x]$ is irreducible over $k[\alpha]$ is $f(x)$ also irreducible over $K$ ?

Answer 1

No: let $k=\mathbb Q$, $\alpha=\sqrt[3]{2}$ and $f(x)=x^2+x+1$. Then $f(x)$ is irreducible over $k[\alpha]=\mathbb Q(\sqrt[3]{2})$ because such a field is totally real, but not over the algebraic closure of $k[\alpha]$ which is $\mathbb Q(\sqrt[3]{2},\zeta_3)$, where $\zeta_3$ is a primitive cubic root of $1$.