Criteria for a prime degree field extension to be Galois Let $k$ be a field of characteristic zero, not algebraically closed,
and let $k \subset L$ be a field extension of prime degree $p \geq 3$.

I am looking for an additional condition which guarantees that $k \subset L$ is Galois.

An example for an answer: Here is a nice condition, which says that if $L=k(a)=k(b)$, with $a \neq b \in L$ both having the same minimal polynomial over $k$, then $k \subset L$ is Galois.
(Of course, by the primitive element theorem, there exist $a \neq b \in L$ such that $L=k(a)=k(b)$; the point is that they are conjugate).
(See also this question that asks for a generalization for degree product of two primes).
Thank you very much!

Let $k(\alpha)/k$ a Galois extension of prime degree $[k(\alpha):k] = p$  and $char(k) \ne p$.

If $\zeta_p \not \in k$ then $k(\zeta_p)/k$ as well as $k(\alpha,\zeta_p)/k(\alpha)$ are Galois and (by looking at the possible automorphisms) $[k(\zeta_p):k]=[k(\alpha,\zeta_p):k(\alpha)] = n$ where $n \ |\ p-1$. 
Thus $k(\zeta_p,\alpha)/k$ is Galois of degree $np$ and $k(\zeta_p,\alpha)/k(\zeta_p)$ is Galois of degree $p$, 
by the Kummer theory it means $k(\alpha) = k(\sqrt[p]{a})$ for some $a \in k(\zeta_p)^* /(k(\zeta_p)^*)^p$
and hence $k(\alpha)/k$ wasn't Galois.

Qed. $\zeta_p \in k$ and (Kummer theory) $k(\alpha) = k(\sqrt[p]{a})$ for some $a \in k^* /(k^*)^p$

