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I'm struggling to understand the basics of Galois theory. One of the things I don't understand is how to actually derive automorphisms of a field extension. Let's say you had a simple problem:

$x^2-3$ over $\mathbb{Q}$ has splitting field $\mathbb{Q}(\sqrt{3}$) correct?

Once I have this how would I go about finding the Galois group in this problem? Isn't there an unlimited number of ways I can assign elements of the field? If the Galois group is the set of ALL automorphisms aren't I just limited by my imagination?

user21820
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Roger
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2 Answers2

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The Galois Group of some field extension $E/F$ is the group of automorphisms that fix the base field. That is it is the group of automorphisms $\mathrm{Gal}(E/F)$ is formed as follows: $$ \mathrm{Gal}(E/F) = \{ \sigma \in \mathrm{Aut}(E) \mid \sigma(f) = f \; \forall \; f \in F \} $$ So you are fairly limited actually with regards to this group. First off you can notice that the only elements that can be permuted around are the elements that are adjoined to form the splitting field. From this you can see that if you join $n$ elements, at max you will then have $n!$ elements in this group, for that's the maximum number of permutations of $n$ things you can have. In other words one can notice that if you adjoin $n$ elements then $$ \mathrm{Gal}(E/F) \subset S_n $$ where $S_n$ is the symmetric group of $n$ elements. Naturally you can already see that the maximum number of elements that one would have to adjoin to a base field for a degree $n$ polynomial is $n$ elements, so for any degree $n$ polynomial in a base field $F$ we will have at most $n$ elements adjoined and thus $$ \mathrm{Gal}(E/F) \subset S_n $$ By no means are any of the above proofs for any of these facts, but hopefully they provide intuition. In order to prove these facts you have to dive into how Splitting fields are only unique up to isomorphism. Another neat fact is that if $p(x)$ is a seperable polynomial then $$ \mid \mathrm{Gal}(E/F) \mid = [E : F] $$

Now let's address the question at hand. You're working with a base field of $\mathbb{Q}$ and with $p(x) = x^2 - 3 \in \mathbb{Q}[x]$. Note that this is degree $2$ so that $$ \mathrm{Gal}(E/\mathbb{Q}) \subset S_2 $$ where $E$ is the splitting field of $p(x)$. From this you can see that $\mathrm{Gal}$ has at maximum of $2$ elements. Note that it cannot have $1$ element for you must adjoin at least two elements here (both $\pm \sqrt{3}$) so then $\mathrm{Gal}$ must be identically $S_2$.

Another way to go about this is to notice that $p(x)$ is separable so then we have that $$ \mid \mathrm{Gal}(E/F) \mid = [E : F] = 2 $$ (note showing that showing $[E : F] = 2$ takes some explanation) so then we have the only group with two elements, which is isomorphic to $S_2$.

Lastly from an entirely intuitive perspective we can notice that we need to adjoin two elements to reach the splitting field, namely $\pm \sqrt{2}$ so then the automorphisms can only permute these two elements. Naturally we would have the identity in this group, so some automorphism that sends $$ \sqrt{3} \to \sqrt{3} $$ but notice that we can also have another element that sends $$ \sqrt{3} \to -\sqrt{3} $$ for this last isomorphism preserves the structure of the splitting field (in higher degree polynomials this can in fact fail. This turns out to be an incredibly deep subject, at least in my opinion - finding the galois group for an arbitrary higher degree polynomial. There is a lot to take into account that at the surface don't appear to be an issue).

A few key points to keep in mind that would have helped me when first starting to learn this stuff:

  • $\mathrm{Gal}(E/F)$ must fix the base field. I cannot stress this enough. This appears to be the base misunderstanding you have (based on how you asked the question). It cannot, whatsoever, by definition, permute anything in the base field (ever).

  • $\mathrm{Gal}(E/F) \subset S_n$ it is not necessarily the case that $\mathrm{Gal}(E/F) = S_n$. For me, at first this seemed a little weird; why can't we just permute any of the roots? A good example to think about would be the polynomial $(x^2 -3)(x^2-2)$. We cannot place $\sqrt{3} \to \sqrt{2}$ for then the intermediate fields here would become messed up; that is, $\sqrt{2}$ is a root of $x^2-2$, and in some sense when you send $\sqrt{3} \to \sqrt{2}$ you're implying that $\sqrt{3}$ is also a root, but it actually is not so you cannot have this automorphism in your Galois Group.

  • This is a hard topic in abstract algebra. It may appear like a simple question at first, but it takes some thought! Don't get too bogged down by all the notation and definitions. At some point I'm sure you'll gain an intuitive understanding for this very difficult subject.

Hopefully this helps, let me know if you need more explanation!

DanZimm
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  • I am having the same problem as OP, but I particularly get lost in degrees > 2. I understand your example of $(x^2 - 3)(x^2 - 2)$ but that polynomial is clearly reducible, and I thought it only made sense to apply this theory to irreducible polynomials? Is the reducibility the thing that makes it so the Galois group isn't all of $S_n$? TIA! – William Dec 18 '14 at 14:48
  • @William what do you mean that this polynomial is reducible? Over which field? Certainly not over the rationals. Finding the Galois Group for higher degree polynomials is fairly difficult and involves a LOT of different types of problems that can come up. In other words saying "I get lost in degrees > 2" is a bit vague, since each polynomial you study can be a beast in its own regard. – DanZimm Dec 18 '14 at 23:38
  • I may have oversimplified, I mainly meant that it's the product of two quadratics, so to me doesn't that mean it's reducible? I thought $P$ is reducible if there is some $P_1, P_2$ such that $P = P_1 P_2$. – William Dec 18 '14 at 23:44
  • @William Oop! My apologies, been a while since I looked at this, you're right that is a reducible polynomial. Anyhow, you're interested in irreducible polynomials? Do you have a specific example? – DanZimm Dec 18 '14 at 23:49
  • thank you so much for your help. Mainly I'm trying to understand if there's a criterion under which the Galois group of a polynomial is the entire symmetric group, and was postulating that (perhaps) you were suggesting an irreducible polynomial will be the whole symmetric group while a reducible one will have some permutation that doesn't work.... also, on that subject, these permutations must also respect the field operations -- is that a material constraint typically, or do all perms of roots work by def'n of the extension? Confused there too :( – William Dec 19 '14 at 14:55
  • @William yes field operations are one of the restraints that cause the Galois Group to be smaller sometimes - I'll try to come up with an example of that. As for irreducibility, I'm pretty sure you can have an irreducible polynomial with a Galois Group that isn't $S_n$, again I'll try to come up with an example. – DanZimm Dec 19 '14 at 22:05
  • @William As for an irreducible polynomial that has Galois Group smaller than $S_n$ consider $x^4 - 2$. Note that $\pm i\sqrt[4]{2} \not\to \pm \sqrt[4]{2}$ since it would mess up the intermediate splitting fields. – DanZimm Dec 19 '14 at 23:04
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It is important to remember that these automorphisms fix the base field. In particular, they fix the polynomial $x^2 - 3$, so they must send roots of that polynomial to other roots.

They are also field maps and, since they fix the base field, they are in particular linear over the base field - Q, in your example . Since the splitting field is generated by $1$ and $\sqrt(3)$ over $Q$ (as a vector space), knowing where $1$ and $\sqrt(3)$ are sent determines the entire map. But $1$ must be sent to $1$ and $\sqrt(3)$ to another root of your polynomial. So you can see that there are only a few possibilities.

It is not always the case that you can just send roots to other ones - there may be subtle relationships among the roots that prevent every map that permutes roots from being automorphisms. One trick for determining the automorphisms (which happens frequently enough to be worth remembering) is that Galois theory tells you that in the Galois case there are exactly as many automorphism as the degree of the minimal polynomial ( which is degree 2 in this case, since you can factor 1 out of $x^3 - 1$). If you can show that there are exactly that many possible automorphisms by the consideration in the previous paragraph and exhibit them (here the identity and $\sqrt(3) \to -\sqrt(3)$), then you are done.

Elle Najt
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