Q:

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.

A:

I am posting my comment as an answer (I had no intention of causing any dispute).
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 $K$.

