13
$\begingroup$

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.

$\endgroup$
6
  • 3
    $\begingroup$ Do you mean, a nontrivial Galois extension? (Every field is Galois over itself) $\endgroup$ Commented Sep 29, 2011 at 19:11
  • $\begingroup$ What's your definition of "Galois" for an extension that is not algebraic? $\endgroup$ Commented Sep 29, 2011 at 19:12
  • 3
    $\begingroup$ If you mean nontrivial Galois extension, no algebraically closed field has one. $\endgroup$ Commented Sep 29, 2011 at 19:13
  • $\begingroup$ @Arturo - Yes, of course (maybe I should have been clearer in formulating the question). $\endgroup$ Commented Sep 29, 2011 at 19:13
  • $\begingroup$ @Qiaochu - I meant a field that does have other algebraic extensions. $\endgroup$ Commented Sep 29, 2011 at 19:15

1 Answer 1

Reset to default
22
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.