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I would like to ask the following:

Is $\mathbb{C}$ the Galois extension of $\mathbb{Q}$?

I considered the Galois extension in a finite field, thus I am stuck here.

I think $\mathbb{C}$ is not the finite extension of $\mathbb{Q}$, because it is not an algebraic extension. ($\pi$ is not algebraic over $\mathbb{Q}$, and we have the proposition "finite extension is algebraic extension").

Please help me.

jvdhooft
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Chanr13
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  • Well, ${\Bbb R}$ is an infinite extension of $\Bbb Q$, since $\Bbb R$ is not denumerable but $\Bbb Q$ is. – Wuestenfux Apr 27 '20 at 15:57
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    What's your definition of Galois? If it is "finite, normal and separable", the answer is "no", if it is "normal and separable", the answer is "yes". – Thorgott Apr 27 '20 at 16:05
  • For the definition of Galois extension in Wikipedia see here. – Dietrich Burde Apr 27 '20 at 16:55
  • I know that Normality must be algebraic... but C is not algebraic extension of Q... Do i have the incorrect definition of normality? – Chanr13 May 01 '20 at 12:40

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I think the current answer to this question is quite wrong. Every source I'm aware of defines a Galois extension to be algebraic, and Galois theory does not work without this assumption. $\mathbb{C}$ is not an algebraic extension of $\mathbb{Q}$, so it is not Galois. If there are sources that don't use this assumption then I think they're just wrong.

In some places you might see the notation $\text{Gal}(L/K)$ used to refer to the automorphism group of $L$ over $K$ even if $L$ is not Galois. In my opinion this notation is also wrong. It should be $\text{Aut}(L/K)$ in the general case and $\text{Gal}$ should be reserved for Galois extensions.

Qiaochu Yuan
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An extension doesn't need to be algebraic in order to be Galois (at least, not if it hasn't been explicitly required in the definition). There are different conventions regarding that, which give different answers.

Suppose a Galois extension is one that is both separable and normal. Then, every field extension over a field of characteristic zero (like the rationals) is separable, so you only need to check normality, which means every irreducible polynomial in $ \mathbf Q[X] $ which has a root in $ \mathbf C $ splits completely over $ \mathbf C $. This is trivially satisfied in this case because $ \mathbf C $ is algebraically closed - every polynomial in $ \mathbf Q[X] $ splits completely in $ \mathbf C[X] $. As such, the extension $ \mathbf C/\mathbf Q $ is Galois, by definition, even though it's not algebraic.

Ege Erdil
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  • @DietrichBurde It's certainly possible, but I've seen far too many references to $ \operatorname{Gal}(\mathbf C/\mathbf Q) $ to use this convention. – Ege Erdil Apr 27 '20 at 16:51
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    Yes, this is possible. Although some authors denote by $Gal(L/K)$ just the automorphism group, also for non-Galois extension. For example, $Gal(\Bbb Q(\sqrt[3]{2})/\Bbb Q)=1$. – Dietrich Burde Apr 27 '20 at 16:58
  • It seems interesting to relate this answer to the following one, where it is shown that $\mathbb{C}^{\mathrm{Gal}(\mathbb{C}/\mathbb{Q})} = \mathbb{Q}$: https://math.stackexchange.com/questions/3653358/the-galois-group-rm-gal-bbb-c-bbb-q-rm-aut-bbb-c-bbb-q – Matematiflo Dec 20 '22 at 14:24