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.

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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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

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.

  • $\begingroup$ 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

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