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I am confused as to why the Galois group is as follows for my problem.

Find the splitting field for $f(x)=x^2+1$ and find the Galois group $Gal(f)$.

Now, The splitting field is $\Sigma= \mathbb{Q}(\pm i)=\mathbb{Q}(i)=\{a+bi:a,b \in \mathbb{Q}\}$. Okay.

As I understand, this Galois group I need is $Aut_{\mathbb{Q}}(\mathbb{Q}(i))$. So in other words, I am looking for the different $\mathbb{Q}$-automorphisms on $\mathbb{Q}(i)$. Well, there's the identity map $id(p) = p$ for any $p \in \mathbb{Q}$. Issue is, I can't find anymore but the answer is apparently $id$ AND $*$ the map that sends and element to the complex conjugate.

Namely, I am told the answer is $Gal(f)=\{id,*\}$ where $*:\mathbb{C} \rightarrow \mathbb{C}$. I understand what a complex conjugate is sure, but how is this different in the case of $\mathbb{Q} \rightarrow \mathbb{C}$?(Well, technically, $\mathbb{Q}(i) \subset \mathbb{C}$ but nonetheless...)

Isn't $*(p)=p$ anyway for any $p \in \mathbb{Q}$ since it's real? How is this effectively different from the identity map? Do I need to list up ALL automorphisms even if they are essentially the same thing, same map in the field considered? I am confused.

Can someone tell me why the answer is like so?

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  • $\begingroup$ You are looking just at your maps restricted to $\Bbb Q$, you want to look at the map applied to all of $\Bbb Q(i)$. $\endgroup$ – TokenToucan Mar 1 '16 at 19:33
  • $\begingroup$ wait so hang on, is it like "look for maps that 1.sends elements of $\mathbb{Q}(i)$ to $\mathbb{Q}(i)$ in a homomorphic manner $+$ 2. that acts like an identity map for elements of $\mathbb{Q}$" is that what I am looking for? That might make more sense... $\endgroup$ – John Trail Mar 1 '16 at 21:55
  • $\begingroup$ Yup! Any automorphism is an isomorphism of an object with itself, so you want the isomorphisms $\phi: \Bbb Q(i) \rightarrow \Bbb Q(i)$ which behave as the identity when restricted to $\Bbb Q$. $\endgroup$ – TokenToucan Mar 1 '16 at 23:43
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Indeed, complex conjugation acts as the identity on $\mathbb{Q}$, as does any field automorphism. But it does not act trivially on $\mathbb{Q}(i)$, and what you are looking for are automorphism of $\mathbb{Q}(i)$.

In your case there are exactly 2 automorphisms of $\mathbb{Q}(i)$, they both act trivially on $\mathbb{Q}$, so they make up $Aut_\mathbb{Q}(\mathbb{Q}(i))$.

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  • $\begingroup$ Wait, I'm sorry I don't get it. Doesn't conjugation act trivially in our case here? I still don't get why there's a distinguishment. $\endgroup$ – John Trail Mar 1 '16 at 20:35
  • $\begingroup$ The element $i$ belongs in $\mathbb{Q}(i)$, and is not fixed by complex conjugation. To "act trivially" on some set means that all the elements are fixed points. Here you may easily find points that are not fixed by complex conjugation (indeed, any element of $\mathbb{Q}(i)$ that is not in $\mathbb{Q}$). $\endgroup$ – Captain Lama Mar 2 '16 at 12:32
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The whole point of a Galois group is that every automorphism in the group fixes the base field, so I think you might need to review the fundamental definitions if you're surprised by the fact that conjugation fixes the rationals. Note that Q(i) is a vector space over the rationals, so every element is of the form a + bi for rational a and b. You're looking for an automorphism that fixes the rationals, so for any such automorphism f, f(a + bi) = a + bf(i). What are the possible images of i under an automorphism?

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We are to consider automorphisms of $\mathbb{Q}(i)$ over $\mathbb{Q}$, that is to say, the automorphisms have to restrict to the identity map in $\mathbb{Q}$: if $\phi\in Aut_{\mathbb{Q}}(\mathbb{Q}(i))$, then $\phi|_{\mathbb{Q}}\equiv \text{id}_{\mathbb{Q}}$. So, complex conjugation must satisfy this if it is supposed to be in the Galois group, and it does. It is, of course, not the identity in the extension.

Well, as for why these two are all the automorphisms: The degree of the extension is of course 2. The characteristic of $\mathbb{Q}$ is $0$, so the separable degree is automatically equal to the degree*, and is hence $2$. Well, the automorphisms take roots to roots. We've already found two, so there can't be any more.

*Why? Consider, purely algebraically, the derivative of a polynomial. Any double root must also be a root of the derivative. Check that that's not possible if the characteristic is $0$.

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$\newcommand{\Q}{\mathbb{Q}}$The elements of the Galois group are maps on $\Q(i)$. Now the identity takes $i$ to $i$, but complex conjugation takes $i$ to $-i$, so on $\Q(i)$ they are different maps.

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