In relation to my previous question (which I will get back), I am rather very confused with the various explanations people and textbooks have.

Firstly, I noticed there are $2$ different Galois Groups(or are they?),

$1.$ $Gal_{\mathbb{A}}(f)$ where $f$ is a polynomial over $\mathbb{A}$.

$2.$ $Gal_{\mathbb{K}}(L)$ for a field extension $L:K$ where $K,L$ are fields.

One is about polynomials over a field, another is talking about field extensions. Sure, an extended field might be a solution to a particular polynomial(i.e. splitting field) but so, are the $2$ Galois Groups the same? Different? No textbook states they're different but the trouble I have is, neither do they make any clear statement relating them. What on Earth are they?

So, what is a Galois group? Here are the $2$ "intuitive" explanations I have been given.

$1.$ A group of automorphisms; group of $K$-automorphisms over $L$ "for an extension $K:L$$(okay, so they're not talking about polynomials here...)

$2.$ A group of symmetries (switching) of solutions to a polynomial $f$ that does NOT change the relationship between them with coefficient in $K$. So, if $K=\mathbb{Q}$ and $a=i,b=-i,c=\sqrt{5},d=-\sqrt{5}$ are solutions to some $f$, $a+b=0$, $ad-bc=0$ are some examples. Switching $a,b$ and $cd$ does not change the validity of these relations.

Now, what confuses me is that this seemingly tells me there are $2$ different Galois group depending on the definition(approach I take). Here is an example

Let $f(t)=t^4-4t^2-5=0$. Solutions are $a=i,b=-i,c=\sqrt{5},d=-\sqrt{5}$ and has field extension $\mathbb{Q}(i,\sqrt{5}):\mathbb{Q}$.

Now, if I think of $Gal_{\mathbb{Q}}(\mathbb{Q}(i,\sqrt{5})$(i.e. about the $\textbf{field extension}$), I reason as follow

I am looking for $\mathbb{Q}$-automorphisms over $\mathbb{Q}(i,\sqrt{5})$. Any element of $\mathbb{Q}(i,\sqrt{5})$ has the form $p+q\sqrt{5}+ri$ for $p,q,r \in \mathbb{Q}$. So, I want maps $p+q\sqrt{5}+ri \rightarrow p'+q'\sqrt{5}+r'i$ but also if any element is rational, it will map to itself(hence an automorphism).

We obviously have the identity $id:p+q\sqrt{5}+ri \rightarrow p+q\sqrt{5}+ri$. One element of Galois group found. Second how about $F_1:p+q\sqrt{5}+ri \rightarrow p-q\sqrt{5}+ri$? This still maps $p \rightarrow p$ and is not the identity. Similarly, $F_2:p+q\sqrt{5}+ri \rightarrow p+q\sqrt{5}-ri$,$F_3:p+q\sqrt{5}+ri \rightarrow p-q\sqrt{5}-ri $ would do. So my group is $\{id, F_1,F_2,F_3\}$.

Then if I switch and focus on $\textbf{ the polynomial $f(t)$}$, namely if I look for $Gal_{\mathbb{Q}}(f(t))$

I have $4$ solutions $a,b,c,d$. By experimenting, it seems like I can interchange $a,b$ and $c,d$ but not otherwise. So, in terms of permutation, $id,\rho_1=(ab),\rho_2=(cd),\rho_3=(ab)(cd)$ is the element of groups I get.

Well, didn't I just get $2$ $\textbf{different}$ Galois groups? Depending on whether I consider automorphisms of $\textbf{fields}$ involved or $\textbf{polynomials}$ in the field extension? I can hardly say the $2$ groups above are the same. One maps elements in $\mathbb{Q}(i,\sqrt{5}) \rightarrow \mathbb{Q}(i,\sqrt{5})$ and the latter maps $\{\text{solution set of $f(t)$}\} \rightarrow \{\text{solution set of $f(t)$}\}$. The mere link I see is that each solutions are in $\mathbb{Q}(i, \sqrt{5})$. But there are vast other elements in $\mathbb{Q}(i, \sqrt{5})$ too, which we consider also in the automorphism case.

I am very confused as to what this Galois Group is; again, people mainly describe it to me either in one of the $2$ ways above, which seem to be very different. Is this one of those "oh it depends on the question, try to judge aptly from what you're being asked in each question"-things? People tel me it's very interesting and tells us a lot about polynomials and fields and it's impressive, but I am not growing any liking or interest to Galois Groups so far because of its vast ambiguity! Just what on Earth are you Galois Group?

  • 2
    $\begingroup$ When you say "A group of symmetries (switching) of solutions to a polynomial $f$", do you realize that implicitly, you construct a field extension $L$ of $K$ where these solutions live ? This is the link. $\endgroup$ – Captain Lama Mar 6 '16 at 18:07
  • $\begingroup$ I would probably need an extensive explanation on that. I've been thinking there must be something, but I just cannot make clear sense of it on my own. Going through examples just keep confusing me with the reason above I have asked in the question body. $\endgroup$ – John Trail Mar 6 '16 at 18:09
  • $\begingroup$ I don't understand why your example illustrates your confusion: in both cases the Galois group you computed is (isomorphic to) the Klein 4 group $\Bbb Z_2\times\Bbb Z_2$. We can see this since they both have 4 elements and all non-identity elements have order 2. Yes, the groups are not literally equal, but an experienced algebraist like yourself shouldn't be concerned with that. So I feel I am missing your point. $\endgroup$ – Eric Stucky Mar 6 '16 at 19:28

Take a polynomial $f\in K[X]$. Then consider a field extension $L$ of $K$ that is generated by the roots of $f$ (for instance, if you have an algebraic closure $\overline{K}$ of $K$, and in $\overline{K}$ the roots of $f$ are $\alpha_1,\dots, \alpha_r$, then take $L=K(\alpha_1,\dots,\alpha_r)$. This is called a splitting field of $f$.

Now if $\sigma$ is an automorphism of $L$ that fixes $K$, then $\sigma$ permutes the $\alpha_i$, since $P(\sigma(\alpha_i)) = \sigma(P(\alpha_i)) = 0$ so $\sigma(\alpha_i)$ must be some $\alpha_j$. So any element of $Gal(L/K)$ acts by permutation on the roots.

Conversely, if you have $\sigma$ in what you call $Gal_K(f)$, by definition it acts on the $\alpha_i$, and since any element of $L$ is a linear combination of products of the $\alpha_i$, you can extend it uniquely to an automorphism of $L$ (the fact that it does extend to an automorphism precisely expresses that $\sigma$, as you put it, "does NOT change the relationship between them with coefficient in $K$"). For instance, you will send $\alpha_i\alpha_j - \lambda \alpha_k^2$ to $\sigma(\alpha_i)\sigma(\alpha_j) - \lambda \sigma(\alpha_k)^2$.

So those two groups are canonically the same. Actually, I don't really know how you can rigourously define $Gal_K(f)$ in an other way than $Gal(L/K)$ where $L$ is a splitting field of $f$, because "preserving relations between the solutions" is a nice intuitive way to think, but does not give a real definition (actually, it is perfectly formalized by the fact that they give automorphisms of the splitting field fixing $K$). And anyway, you need a splitting field just to make sense of "symmetries between the solutions", because you need to construct such solutions.


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