# Does every field have a non-trivial Galois extension?

Clearly a field $F$ with no Galois extensions must have only non-separable elements in any extension (otherwise, take the minimal polynomial of some separable element over $F$ - its splitting field will be a Galois extension of $F$). This argument reduces the possible examples to non-perfect fields, or in other words - infinite fields of positive characteristic. However, I can't come up with an example of such a field with no separable elements over it.

Are there any such examples?

EDIT: I should have been clearer - I'm looking for a field that has non-trivial algebraic extensions, but has no non-trivial Galois extensions.

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Do you mean, a nontrivial Galois extension? (Every field is Galois over itself) –  Arturo Magidin Sep 29 '11 at 19:11
What's your definition of "Galois" for an extension that is not algebraic? –  Chris Eagle Sep 29 '11 at 19:12
If you mean nontrivial Galois extension, no algebraically closed field has one. –  Qiaochu Yuan Sep 29 '11 at 19:13
@Arturo - Yes, of course (maybe I should have been clearer in formulating the question). –  Pandora Sep 29 '11 at 19:13
@Qiaochu - I meant a field that does have other algebraic extensions. –  Pandora Sep 29 '11 at 19:15
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Let $F$ be any field. The compositum of separable extensions of $F$ contained in the algebraic closure $\overline{F}$ of $F$ will itself be separable, and so there is a largest separable extension of $F$ contained in $\overline{F}$ (namely, the compositum of all separable extensions). This is called the separable closure of $F$ in $\overline{F}$. (See for example Lang's Algebra, revised 3rd Edition, Theorem 4.5 and discussion following, pp. 241f).
Now start with a non-perfect field; for example, take $\mathbb{F}_p(x)$, the field of rational functions with coefficients in the field of $p$ elements. Let $K$ be the separable closure of $F$ as above; because $F$ is not perfect, $K$ cannot equal $\overline{F}$. In particular, $K$ is not algebraically closed.
However, every nontrivial algebraic extension of $K$ is not separable (in fact, it will be purely inseparable): because if $L$ is an algebraic separable extension of $K$, then $L$ is also an algebraic separable extension of $F$, hence must be contained in $K$, so $L=K$.
Thus, no nontrivial algebraic extension of $K$ is separable, so no nontrivial algebraic extension of $K$ is Galois over $K$; and yet there are nontrivial algebraic extensions of $K$, since $K$ is not algebraically closed.