2
$\begingroup$

Let : $ \ \mathrm{Gal} : F/ \mathbb{Q} \to \mathrm{Gal} (F / \mathbb{Q} ) $ be the functor, which associate to a Galois extension $ F/ \mathbb{Q} $ of the field $ \mathbb{Q} $, the Galois group $ \mathrm{Gal} ( F/ \mathbb{Q} ) $.

Is the functor : $ \mathrm{Gal} \ : \ F/ \mathbb{Q} \to \mathrm{Gal} (F / \mathbb{Q} ) $ represented by the object $ \overline{ \mathbb{Q} } $ which is the algebraic closure of the field $ \mathbb{Q} $ ?.

If it is not the case, which functor is represented by the object : $ \overline{ \mathbb{Q} } $ ?

Thanks in advance for you help.

$\endgroup$
  • 2
    $\begingroup$ Is this even a functor? I don't see how a field homomorphism $F\to F'$ induces a function between the Galois groups. $\endgroup$ – Arnaud D. Aug 29 '18 at 15:20
  • 1
    $\begingroup$ @ArnaudD, A field homomorphism $F \to F'$ is nothing but a field extension $F' / F$, and to such an extension we can associate a canonical map $\operatorname{Gal}(F' / \mathbb{Q}) \to \operatorname{Gal}(F / \mathbb{Q})$ obtained by restriction. So you can think of it as a (contravariant) functor. $\endgroup$ – Sofie Verbeek Aug 29 '18 at 19:00
  • 2
    $\begingroup$ That said, no matter how you conceive of it as a functor, it probably won't be representable. Let's say for a moment that we see it as a functor $\mathrm{Gal}(\,\cdot\,) : \mathcal{C}^{\mathrm{op}} \to \mathbf{Set}$ from the category $\mathcal{C}$ whose objects are Galois extensions of $\mathbb{Q}$ and whose morphisms are just field maps. Suppose $\operatorname{Gal}(\,\cdot\,,\mathbb{Q})$ were representable, @YoYo, then it would be of the form $\operatorname{Hom}(\,\cdot\,,F)$ for some fixed field $F$, right? Plug some examples in $(\,\cdot\,)$ to find that no $F$ is going to work. $\endgroup$ – Sofie Verbeek Aug 29 '18 at 19:06
  • 1
    $\begingroup$ @sofieverbeek Your canonical map is not well defined, as not all automorphisms of $F'$ restricts to $F$. $\endgroup$ – Arnaud D. Aug 29 '18 at 20:25
  • 2
    $\begingroup$ @ArnaudD., What am I missing? An automorphism $\sigma : F' \to F'$ over $\mathbb{Q}$ preserves roots of a given polynomial. If $F$ is an intermediate field and $F / \mathbb{Q}$ is a Galois extension, then $F$ is a splitting field over some polynomial with coefficients in $\mathbb{Q}$, say $F = \mathbb{Q}(x_1,\ldots,x_n)$ for some roots $x_i$ of a polynomial $f$. As $\sigma$ premutes roots, this $x_i$ will be sent to some element in $F$ because $F$ is a splitting field. $\endgroup$ – Sofie Verbeek Aug 30 '18 at 9:05
1
$\begingroup$

As said in the comments by Sofie Verbeek, if $F/\Bbb Q \to Gal(F/\Bbb Q)$ was representable it would be a functor on the form $F/\Bbb Q \mapsto Hom(F, F_0)$ for some fixed field $F_0$.

We assume that such $F_0$ exists, let $F$ be a field with bigger cardinality than $F_0$, we see now that $Gal(F/\Bbb Q) = \emptyset$, a contradiction.

Tautologically, the functor $F \to Hom(F, \overline{\Bbb Q})$ represents the functor $F \mapsto \{\text{embedding $\sigma : F \to \overline{\Bbb Q}$}\}$.

Edit : This answer is wrong but I can't delete it because it was accepted by the OP. So please moderators if you see this answer I would appreciate if you can delete it.

$\endgroup$
  • $\begingroup$ Thank you Nicolas. :-) $\endgroup$ – YoYo Sep 2 '18 at 14:42
  • $\begingroup$ The question says that the functor associates its Galois group to any Galois extension of $\mathbb{Q}$. Wouldn't a Galois extension necessarily be algebraic over $\mathbb{Q}$, hence countable? $\endgroup$ – Arnaud D. Sep 4 '18 at 10:11
  • $\begingroup$ @ArnaudD. : of course ! My answer is nonsense I'll delete it soon. $\endgroup$ – Nicolas Hemelsoet Sep 7 '18 at 11:45
  • $\begingroup$ I can't delete my answer because it was accepted. If moderators see it please delete this answer. $\endgroup$ – Nicolas Hemelsoet Sep 10 '18 at 8:26
  • $\begingroup$ If you want moderators to see something, you can flag it or maybe post a message in "Math Mod's office" chatroom. Normally, mods don't delete wrong answers, but I don't know if they would do it in such a case. Perhaps the simple option is to ask @YoYo to unaccept the answer since it is not correct? Or maybe edit the question to make the answer acceptable? $\endgroup$ – Arnaud D. Sep 10 '18 at 10:54

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.