# Having a separable extension of degree $n$ implies having a Galois extension of degree $n$?

I would like an explanation for the fact stated in the title. To repeat:

Question: How does one prove that if a field has a separable extension of degree $n$, then it has a Galois extension of degree $n$?

Note that the statement might as well be false (though I would not bet on that). [EDIT: I thought I had a proof that the statement was true for $n=3$, but as Emil Jeřábek points out in his comments, this is false as well!]

This might be completely trivial, but I have no idea on how to address this question, and since it kinds of naturally arose in my research, I thought it would be appropriate to ask here.

• That is not true. Let $K$ be the maximal solvable extension of $\mathbb{Q}$ (or $\mathbb{C}(t)$, etc.). There are irreducible quintic polynomials over $K$ (otherwise every quintic over $\mathbb{Q}$ would have a solvable Galois group). Thus, there are separable, degree $5$ extensions of $K$. However, every Galois extension of degree $5$ would have cyclic Galois group. Since $K$ is already the maximal solvable extension of $\mathbb{Q}$, there is no nontrivial, finite cyclic extension of $K$. – Jason Starr Sep 20 '18 at 12:00
• So, you just confirmed my fear that the question is trivial. Thank you very much for your help, that was actually exactly what I was hoping for. – thierry stulemeijer Sep 20 '18 at 12:04
• The reason it holds for $n=3$ is that every transitive subgroup of $S_n$ has a normal subgroup of index $n$, but this is false for $n\ge5$. – Emil Jeřábek Sep 20 '18 at 12:05
• Sorry, you're right, it actually does not. But then I think your property fails already for $n=3$. Let $K/F$ be any non-normal extension of degree 3, and let $L/F$ be its normal closure. We have $[L:K]=2$. Using Zorn's lemma, let $F'$ be a maximal algebraic extension of $F$ such that $F'\cap L=F$, and let $K'$ and $L'$ be the composita of $F'$ with $K$ and $L$, respectively. We still have that $K'$ is a degree-3 extension of $F'$, and $L'$ is its normal closure over $F'$, of degree 2 over $K'$. $Gal(L'/F')\simeq S_3$. Assume for contradiction that $F'$ has a degree-3 Galois extension $H$... – Emil Jeřábek Sep 20 '18 at 14:32
• ... By maximality of $F'$, $H$ has a nontrivial intersection with $L$, thus $H\cap L'$ is a proper extension of $F'$; since it is of prime degree, this implies $H\subseteq L'$. But then $Gal(L'/H)$ is a normal subgroup of $Gal(L'/F')$ of index 3, which as you observed, does not exist. – Emil Jeřábek Sep 20 '18 at 14:37

That is not true. Let $K$ be the maximal solvable extension of $\mathbb{Q}$ (or $\mathbb{C}(t)$, etc.). There are irreducible quintic polynomials over $K$ (otherwise every quintic over $\mathbb{Q}$ would have a solvable Galois group). Thus, there are separable, degree 5 extensions of $K$. However, every Galois extension of degree 5 would have cyclic Galois group. Since $K$ is already the maximal solvable extension of $\mathbb{Q}$, there is no nontrivial, finite cyclic extension of $K$.