Criteria for a prime degree field extension to be Galois

Question

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!

Answer 1

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

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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 $a \in k^* /(k^*)^p$

Answer 2

At this point of the discussion between the two users, 237522 and 1952009, the original question could be slightly "bent" and brought back to the following: "Given a field $k$ of characteristic $\neq p$ (where $p$ is an odd prime), classify the Galois extensions of $k$ of degree $p$". Such an extension is cyclic of degree $p$ and will be called a $C_p$-extension, where $C_p$ denotes the cyclic group of order $p$. I extract from https://math.stackexchange.com/a/2093431/300700 the following answer :

Denote by $G_k$ the absolute Galois group of $k$, so that any $C_p$-extension of $k$ is the fixed field of a subgroup of index $p$ of $G_k$, and any such subgroup is the kernel of a non trivial homomorphism $G_k \to C_p$, so that the $C_p$-extensions of $k$ are classified by the non trivial elements of the group $Hom(G_k, C_p)$. To progress further, introduce the field $K = k(\mu_p)$, where $\mu_p$ is the group of $p$-th roots of unity, and consider the restriction map $Hom(G_k, C_p) \to Hom(G_K, C_p)$. For any Galois module $X$, $\Delta=Gal(K/k)$ acts on $Hom (G_K , X)$ in the usual way, more precisely via $(\delta, f) \in \Delta \times Hom(G_K, X) \to f^{\delta}$ given by $f^{\delta}(x)= \delta (f(\delta^{-1}(x))$ for $x \in X$ . Note that this action is the only one which is functorial; here $\Delta$ acts trivially on $C_p$. Since $\Delta$ has order prime to $p$, it is classically known that $Hom(G_k, C_p)\cong Hom(G_K, C_p)^{\Delta}$, where the superscript $(.)^{\Delta}$ denotes the invariants under $\Delta$. Since $G_K$ acts trivially on $\mu_p$, one has $Hom(G_K, C_p)\cong Hom(G_K,\mu_p)$ as groups, but not as $\Delta$-modules because of the canonical action of $\Delta$ defined above. Actually, one can check easily that $Hom(G_K, C_p)\cong Hom(G_K,\mu_p)(-1)$, where $(.)(-1)$ denotes the so called "Tate twist", which means that the Galois action on $Hom(G_K,\mu_p)$ has been replaced by a new action defined by $\delta^{new} (f)= \kappa(\delta)^{-1}.\delta^{old}(f)$, where $\kappa$ is the mod $p$ character defined by $\delta(\zeta)=\zeta^{\kappa(\delta)}$ for $\zeta \in \mu_p$ . Consequently, $Hom(G_k, C_p)$ can be identied with the elements $f\in Hom(G_K,\mu_p)$ such that $\delta (f)=\kappa(\delta).f$ (these can be viewed as "eigenvectors" corresponding to the "eigenvalues" $\kappa(\delta)$). Now, by Kummer theory over $K$, $Hom(G_K,\mu_p)\cong (K^{*}/K^{*})^p$ , and this isomorphism is $\Delta$-equivariant, so we are done, at least theoretically.