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The more I progress, the more contradictions or ambiguity I come by. Probably because Galois group builds up and numerous obscure and abstract concepts and being wobbly and one of them causes tragedy. (I switch from $K$-automorphisms of $L$ to simply "automorphisms" but they're equivalent here)

Anyway, here is the definition of the Galois group

Definition. The Galois Group $\Gamma(L:K)$ of the field extension $L:K$ is the group of $K$-automorphisms of $L$ under the operation of composition of maps.

Question1. Well does it want all possible distinct $K$-automorphisms of $L$? Or only a part of every possible $K$-automorphisms? If so, which ones does it want? What is the condition? How should I systematically decide to pick out which automorphism to include in $\Gamma$ and which ones not to?

Possible Argument. We're talking about roots of a polynomial which we can interchange without losing the validity(apparently. At least, that's what Stewart gets into to introduce Galois Groups in his Galois Thoery 4th Edition). Here's an example (From now on, I will stick to this example)

Example The polynomial $f$ is $f(t)=t^4-4t^2-5$. The roots would be $\alpha = i$, $\beta=-i$, $\gamma=\sqrt{5}$ and $\delta = - \sqrt{5}$. The associated field extension is $\mathbb{Q}(i,\sqrt{5}):\mathbb{Q}$. The interchange of these $4$ roots that are possible are, using permutation, $R=(\alpha\beta)$, $S=(\gamma\delta)$ and $T=(\alpha\beta)(\gamma\delta)$ with the identity $I$. These $4$ permutations form, and are the Galois Group of $\mathbb{Q}(i,\sqrt{5}):\mathbb{Q}$. Any other permutation(interchange of roots) will be invalid in the sense that it will not necessarily satisfy any polynomial equation with rational coefficients that is satisfied with some selection of the $4$ roots.

e.g. $\alpha^2+1=0$, $\delta^2-5=0$, $\gamma+\delta=0$ etc.

So from the example, we want not all $K$-automotphisms (since clearly, there are more automorphisms available in the above but omitted since they don't always satisfy every possible polynomial with some selection of the $4$ roots). For instance,

The permutations above $R,S,T$ are equivalent to the automorphisms,

$$\rho_1: i \to -i, \sqrt{5} \to \sqrt{5}$$ $$\rho_2: i \to i, \sqrt{5} \to -\sqrt{5}$$ $$\rho_3=\rho_1 ○ \rho_2: i \to -i, \sqrt{5} \to -\sqrt{5}$$

We note the elements of $\mathbb{Q}(i,\sqrt{5})$ are written $p+qi+r\sqrt{5}+si\sqrt{5}$ for rational $p,q,r,s$. However, none of the automorphisms $\rho_1,\rho_2,\rho_3$ is the automorphism $p+qi+r\sqrt{5}+si\sqrt{5} \to p-qi-r\sqrt{5}-si\sqrt{5}$.

Question2. Fine, so not all is the answer then, but that's not stated in any formal way in the definition. Am I supposed to experiment every single time to check which automorphisms messes up the polynomials and which ones don't? Just how? In the specific example, it wasn't to hard but there a 100 gabillion polynomials out there over $\mathbb{Q}$.

So, my point is it seems like the Galois Group is not asking for every single automorphisms imaginable, and it's being picky for a given $f$ (and it's associated field extension). That's fine, I ask, "well how would you like me to pick them? Which are the ones you want?" and they throw me back "Those that does allow the roots of $f$ to switch around and still satisfy any possible polynomials built from a selection of the roots" and that's where I go "well, that;s awfully vague and wordy. You asking me to test 10 gabillion polynomials for that? Gimme the tools, what are the tools!!!" $\leftarrow$ I am at this stage now and asking for help.

Tell me if there are any misconceptions but please also elaborate and explain instead of a plain "you"re wrong when you say this-and-that. Ta"

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  • $\begingroup$ Issue is, I haven't learned Galois Extensions yet and the book introduces Galois Groups before mentioning it. Normal and separable have been touched on but nothing more. Do you think I should skip Galois Groups and go to sections that focus on those ideas and come back later to look at Galois groups?? $\endgroup$ – John Trail Mar 28 '16 at 18:13
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    $\begingroup$ I don't understand when you say "clearly, there are more automorphisms available in the above but omitted since they don't always satisfy every possible polynomial with some selection of the 4 roots" : no, there are not. There are more permutations of the roots, yes, but not more $K$-automorphisms. Not all permutations of the roots can be extended to $K$-automorphisms. $\endgroup$ – Captain Lama Mar 28 '16 at 18:15
  • $\begingroup$ @JohnTrail yes, go learn about Galois extensions. Galois groups are only defined for such extensions. Your book may be using the same terminology for some automorphism groups, but the community as a whole has determined not to call those "Galois groups" $\endgroup$ – Adam Hughes Mar 28 '16 at 18:18
  • $\begingroup$ Hi Captain Lama, so correct me if I am wrong. Automorphisms are those that maps elements of a field to elements of the same field. So, $L \to L$. $K$-automorphisms of $L$ for $K \subseteq L$ is, an automorphisms $L \to L$ BUT for elements in $K$, it will necessarily act as an identity map. If my understandings is right, in the context of the $\textbf{example}$, the $\mathbb{Q}$-automorphism will map $p \to p$ in $p+qi+r\sqrt{5}+si\sqrt{5}$. Now, $p+qi+r\sqrt{5}+si\sqrt{5} \to p-qi-r\sqrt{5}-si\sqrt{5}$ should also be a $\mathbb{Q}$-automorphism, yes? But how's this possibl with the $3$ $\rho$? $\endgroup$ – John Trail Mar 28 '16 at 18:32
  • $\begingroup$ Note that $\rho_3(p+qi+r\sqrt{5}+si\sqrt{5}) = \rho_3(p) + \rho_3(qi) + \rho_3(r\sqrt{5}) +\rho_3(si\sqrt{5}) = p+q\rho_3(i) + r\rho_3(\sqrt{5}) +s\rho_3(i)\rho_3(\sqrt{5}) = p-qi-r\sqrt{5}+si\sqrt{5}$. When you replace + by - in the last term, you do not get an automorphism! $\endgroup$ – Maestro13 Mar 28 '16 at 19:55
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I think you are forgetting that $K$-automorphisms need to be field homomorphisms. Your example of the assignment $$p+qi+r\sqrt{5}+si\sqrt{5} \to p-qi-r\sqrt{5}-si\sqrt{5}$$

is not a homomorphism (It is, however an isomorphism of $K$ vector spaces). Indeed, since $\sqrt 5$ maps to $-\sqrt 5$ and $i$ maps to $-i$, this forces $\sqrt5i$ to map to $(-\sqrt 5 )(-i)=\sqrt 5 i$ but your assignment carries $\sqrt5i$ to $-\sqrt5i$.

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  • $\begingroup$ A clear example like this is always the best way to correct a misapprehension. Plus one. $\endgroup$ – Lubin Mar 28 '16 at 22:40
  • $\begingroup$ Ah, okay, so the fact that if i let $i\sqrt{5} \to -i\sqrt{5}$ be possible...that won't qualify as a homomorphism hence not an automorphism? $\endgroup$ – John Trail Mar 28 '16 at 22:41
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    $\begingroup$ An automorphism needs to be consistent with multiplication. $\endgroup$ – Lubin Mar 28 '16 at 22:44
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An automorphism of $L$ over $K$ is a bijection from $L$ to $L$ which fixes elements of $K$ pointwise, and which distributes over addition and multiplication in $L$. The Galois group $G$ of $L/K$ is the group of all such automorphisms.

If $f$ is a polynomial with coefficients in $K$, and with (let's say distinct) roots $r_1, ... , r_t$ in some algebraically closed field containing $K$, and if $L$ is the field $K[r_1, ... , r_t]$, then any automorphism $\sigma \in G$ is completely determined by its effect on $r_1, ... , r_t$. That is because the elements of $L$ are all of the form $h(r_1, ... , r_t)$, where $h \in k[X_1, ... ,X_t]$, and $\sigma$ fixes $K$.

Also, $\sigma(r_i)$ is still a root of $f$. This shows that $\sigma$ induces a permutation of the elements $r_1, ... , r_t$. And, if you know a permutation of $r_1, ... , r_t$, then you will know the unique element of $G$ which induces this permutation, if it exists.

The thing is, you may choose some permutation $\tau$ of the $r_i$; there need not be an element of $G$ inducing that permutation. If there isn't, and you try to define a function $\sigma: L \rightarrow L$ by the formula $$\sigma(h(r_1, ... , r_t)) = h(\tau(r_1), ... , \tau(r_t))$$ then you are not going to get a $K$-automorphism of $L$, because this map will not be well defined.

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