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I am new to Galois theory. my problem is

let F be an extension field of a field K the cardinality of Galois group denote by $|Aut_KF|$ is always finite?

I try to find a contradiction on $\Bbb R$

but, since $\Bbb Q$ is the minimal sub field and $|Aut_{\Bbb Q} \Bbb F|=1$ it is impossible on $\Bbb R$

is above statement true or are there any example that contradict it

Bad English
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    Indeed $|Aut_{\mathbb{Q}} \mathbb{R}|=1$, but $\mathbb{R}$ is not a Galois extension of $\mathbb{Q}$ because, for example, it contains only one cube root of $2$. As long as $\mathbb{F}$ is a Galois extension of $\mathbb{Q}$, then you can use the formula in the answer of @ThomasGrubb, that $|Aut_{\mathbb{Q}} \mathbb{F}|$ equals the degree of $ \mathbb{F}$ over $\mathbb{Q}$. – Lee Mosher Nov 16 '17 at 17:47

2 Answers2

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The order of the Galois group is precisely the degree of the field extension $[F:K]$, so one needs to examine an infinite extension to get an infinite Galois group. For instance, if $\overline{\mathbb{F}_p}$ is an algebraic closure of $\mathbb{F}_p$, then $$ \text{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_p)\cong \hat{\mathbb{Z}} $$ which is a profinite group which may be constructed as an inverse limit of $\mathbb{Z}/n\mathbb{Z}$.

CO2
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TomGrubb
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The absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ is not even countable, see

Is the absolute Galois Group of $\Bbb Q$ countable?

Dietrich Burde
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    @user479583 that is not correct; $\overline{Q}$ is the set of all numbers which satisfy a polynomial with $\overline{Q}$ coefficients. For instance, $\sqrt{2}$ satisfies $x^2-2$. However, there are real numbers which satisfy no polynomials; we call them transcendental. For instance, $\pi$ and $e$ are transcendental. And in fact, there are algebraic numbers which are not even real! Perhaps you have heard of the complex number $i$, which satisfies $i^2=-1$. – TomGrubb Nov 16 '17 at 17:50
  • thanks @ThomasGrubb , now I understand $\overline {\Bbb Q} $ means the algebraic closure of $\Bbb Q$ . not the closure we call in topology – Bad English Nov 17 '17 at 04:35
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    @user479583 Equallly $\overline{\mathbb{F}_p}$ is the algebraic closure in in the answer of Thomas, appearing in the absolute Galois group of a finite field. – Dietrich Burde Nov 17 '17 at 10:05