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.

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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.