The Galois group seen abstractly for a field extension, and for a concrete polynomial

Question

I am learning Galois theory right now, and I tried to compute a few Galois groups and came upon pecularities.

In my textbook, for a field extension $F \subseteq E$, a $F$-isomorphism is a field isomorphism from $E$ to $E$ that fixes $F$ pointwise, and the Galois group of $E$ over $F$ is the group of all those $F$-isomorphism. Now every $F$-isomorphism maps roots of a polynomial in $F$ to roots, and in my textbook it is proven that if the extension is normal and separable, the Galois group has precisely $[E : F]$ elements.

Now with this knowledge it is quite easy to compute the Galois group for $E = \mathbb Q(\sqrt 2, \sqrt 3)$ and $F = \mathbb Q$, as $E$ is the splitting field of $f = (X^2 - 2)(X^2 - 3)$, and the roots of the factors must be preserved by every $F$-automorphism, and determine the $F$-automorphism, if we number the roots with $r_1 = \sqrt 2, r_2 = -\sqrt 2, r_3 = \sqrt 3, r_4 = -\sqrt 4$ we get $\mathbb Z_2 \times \mathbb Z_2$ as the Galois group which acts as the following permutation group on the roots $$ \{ (), (1~2), (3~4), (1~2)(3~4) $$ and it is not transitive (as $f$ is not irreducible).

So far so good, but the Theorem of the primitive elements, every finite separable field extension is simple, i.e. could be described by adjunction of a single elements. So I thought this might be helpful in computing Galois groups, as then just the image of a single elements matters.

So I did it with the example and the knowledge that $\mathbb Q(\sqrt 2, \sqrt 3) = \mathbb Q(\sqrt 2 + \sqrt 3)$. The element $\alpha = \sqrt 2 + \sqrt 3$ has minimal polynomial $g = X^4 - 10X^2 + 1$ and the four roots $$ \sqrt 2 + \sqrt 3 \qquad \sqrt 2 - \sqrt 3 \qquad -\sqrt 2 + \sqrt 3 \qquad -\sqrt 2 - \sqrt 3. $$ But now the theory so far does not tells me how the Galois group is restricted, or could be computed. I looked up on wikipedia and they do the same example their. They use additional relations (polynomial relations that give elements of the ground field, and hence should stay fixed under every $F$-automorphism). With the knowledge that we must have order four we find that as a permutation group we have the Galois group $$ \{ (), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3) \}. $$ As an abstract group both are isomorphic, but as permutation groups they are very different (one is transitive, the other not).

This seems somewhat odd to me, so the concept of the Galois group might be well-defined as an abstract group, but as a group of permutations it seems to depend on the generators we choose, and this has the implication that we cannot speak of the Galois group of a polynomial as the Galois group of the splitting field, but it is still closely attached to the polynomial (if seen as a permutation group).

Further, the question arises how to find all the relations among the roots that should stay invariant under the action of the Galois group, the sources so far do not answer that for me. As far as I see it, I just can find them by inspection of the roots, and then knowing that I need $[E : F]$ elements in my Galois group.

I found a document Galois groups as permutation groups, in it the author takes an approach as above, he writes:

A Galois group is a group of field automorphisms under composition. By looking at the effect of a Galois group on field generators we can interpret the Galois group as permutations, which makes it a subgroup of a symmetric group

But as said, the field generators (guess he means a minimal set of elements that generate a basis for the field extension?) are not canonically given, and as shown above the permutation representation also not. The author writes:

Although the Galois group of f(T) over K does not have a canonical embedding into Sn in general, its image in Sn is well-defined up to an overall conjugation

But the two permutation representations above are clearly not related by conjugation in $S_4$ (as the cycle structure differes).

• So is it right that the concept of a Galois group as a permutation of roots of a polynomial is a different concept than the Galois group of a field extension (the first is attached to an polynomial), but both are isomorphic as abstract groups
• what is their relation, if I take the second approach to compute the Galois group, how to find all the relations that restrict my Galois group

Answer 1

It seems to me that you are making good progress. Addressing the points where I think I may be able to shed additional light.

1. The statement in your first bullet is correct. The Galois group of a field extension and the Galois group of a polynomial differ in this sense. They are isomorphic as abstract groups, but only the latter is naturally a subgroup of a finite permutation group. Observe that we cannot even conclude that degrees of two polynomials sharing would be equal. For example consider the splitting field $L$ of $p(x)=x^3-2$ over $\Bbb{Q}$. Here $L=\Bbb{Q}(\root3\of2,\omega),\omega=e^{2\pi i/3}$. Undoubtedly you know that $G=Gal(L/\Bbb{Q})$ is a group of order six and, as we can view it as a group of permutations of the roots of $p(x)$, $G$ is isomorphic to $S_3$. But, because $L/\Bbb{Q}$ is normal, the minimal polynomial $m(x)$ of the number $u=\root3\of2+\omega$ also splits into linear factors in $L$. It is not difficult to see that $u$ is a primitive element. In other words $L=\Bbb{Q}(u)$. Therefore $m(x)$ has degree six. So if we view $G$ as a group of permutations of the roots of $m(x)$, then $G$ will be a transitive subgroup of $S_6$. It is still, of course, isomorphic to $S_3$ as an abstract group. Think: Cayley presentation of $S_3$ as permutations of itself, i.e. as a subgroup of $S_6$.
2. When $L/K$ is the splitting field of a polynomial $f(x)\in K[x]$ of degree $n$ we can view the Galois group $G=Gal(L/K)$ as a subgroup of $S_n$. Here fixing an embedding of $G$ into $S_n$ amounts to fixing a numbering of the roots $x_1,x_2,\ldots,x_n$ of $f(x)$ in $L$. Such numberings are totally arbitrary. The embeddings $G\to S_n$ we get by using different numberings of the roots can be gotten from each other by conjugating with the permutation $\in S_n$ that "corrects" the numbering. As you saw, it often happens that $L/K$ is the splitting of field of two different polynomials of same degree. When you use two different polynomials, there is no reason to think that you would get conjugate subgroups of $S_n$.
3. Computing the Galois group $Gal(L/K)$ is a bit tricky in general. If you are lucky you can find simple generators $\alpha_1,\ldots,\alpha_n$ of the extension $L=K(\alpha_1,\alpha_2,\ldots,\alpha_n)$ such that $[L:K]=|G|$ is the product of the degrees of the minimal polynomials of the $\alpha_i$s. Then you know that you can select the images of $\alpha_i$s (among the conjugates of each of them) without constraints. Finding such a set of generators is not alway worth the trouble though. Particularly when you don't yet know $[L:K]$. Dedekind's theorem is my go-to tool in those cases. I have given possibly too many examples of using it on our site.