Computing the Galois Group of a Univariate Polynomial (Irreducible or reducible)

1. Jan 15, 2012

joebohr

Is it possible to compute the Galois Group of a polynomial manually (without a computer)? If so, can someone please explain how? I can't seem to find any information (aside from computer algorithms) on how to find a Galois Group or how to factor a polynomial modulo a prime. If it helps to illustrate, use these two examples:

Example 1: x^6-2x^2+3x+1

Example 2: x^2-x-4

2. Jan 15, 2012

zhentil

Where do these polynomials live? You have to fix a base field in order to talk about Galois groups.

3. Jan 15, 2012

zhentil

But the second one is easy. If the polynomial has a root in your field, the Galois group is trivial. If it doesn't, the galois group is the only group with two elements.

4. Jan 16, 2012

morphism

5. Jan 16, 2012

joebohr

I was just talking about the real numbers for my base field. I'm not sure about what you mean by a "trivial" Galois Group, and I'm mostly interested in computing the Galois Group without solving the polynomial, but would the Galois Group of the first example just be isomorphic to the group of second degree separable polynomials

(that is, would Gal(P$_{1}$)$\cong$S$_{2}$? I've figured out some things about Galois Groups since I first posted it, but I still can't seem to make much progress on the second example (unless it's just Gal(P$_{2}$)$\cong$S$_{6}$ but that seems too easy).

6. Jan 16, 2012

morphism

What is "the group of second degree separable polynomials"? Since you write S_2, do you mean the symmetric group on 2 elements?

Anyway, computing Galois groups of polynomials over R isn't interesting. Either the polynomial splits completely into linear factors, in which case the Galois group is trivial (i.e. is the trivial group), or else the polynomial has an irreducible quadratic factor, in which case the Galois group is S_2.

7. Jan 18, 2012

joebohr

Yes, that is exactly what I meant.

Ok, that explains why none of the texts I've read deal with Galois Groups over R. How about we make it interesting and try to find the Galois Groups of the first two examples over the quaternions!

8. Jan 18, 2012

morphism

I'm not sure what you mean by this, as the quaternions don't form a (commutative) field...

9. Jan 18, 2012

joebohr

Oops, I meant to ask "Why don't we find a problem like these two examples that has the quaternion group as its Galois Group." I know these are not the same things

10. Jan 19, 2012

lpetrich

Do you mean the group generated by i, j, k, where i^2 = j^2 = k^2 = i*j*k = -1?

One can easily derive i*j = - j*i = k, j*k = - k*j = i, k*i = - i*k = j
Its elements are {1,-1,i,-i,j,-j,k,-k}, 8 of them
Its conjugacy classes are {1}, {-1}, {i,-i}, and likewise for j and k.
Its nontrivial subgroups are {1,-1}, {1,i,-1,-i}, and likewise for j and k.
All of them are normal; they have quotient groups Z2*Z2, and 3 Z2's.

Since it has 8 elements, it is thus a subgroup of S8. Multiplying by an element -> creating a permutation of elements. Since all the multiplication permutations are even, this group is also a subgroup of A8.

I tried to find out whether it is also a subgroup of S4, S5, S6, or S7 -- as far as I can tell, it isn't. I did that by looking for order-4 elements, finding which pairs of them have an order-4 product, and which pairs a, b of them have a*b*a*b = b*a*b*a. There are some for S8, but not for S7 or smaller index.

So one has to look for a degree-8 polynomial to find one with the quaternion group as the Galois group.

11. Jan 19, 2012

joebohr

Well, I did some research, and according to wikipedia: "As Richard Dean showed in 1981, the quaternion group can be presented as the Galois group Gal(T/Q) where Q is the field of rational numbers and T is the splitting field, over Q, of the polynomial

x8 − 72x6 + 180x4 − 144x2 + 36.

The development uses the fundamental theorem of Galois theory in specifying four intermediate fields between Q and T and their Galois groups, as well as two theorems on cyclic extension of degree four over a field."

Does anyone have any idea how this could be shown. Also, is this the only polynomial with the quaternion group as its Galois Group (Roger Ware has a paper on the subject but I don't have a subscription, the paper is called "www.jstor.org/stable/2047779" [Broken] )? What other groups would be interesting to look at (as Galois Groups) and what other commutative fields would it be interesting to take the Galois Groups of polynomials over?

Last edited by a moderator: May 5, 2017
12. Jan 19, 2012

lpetrich

There's a theorem that states that if F1 is an extension field of F and F2 an extension field of F1, then F2 is also an extension field of F, and their Galois groups are related as follows:

Let G1 = Gal(F1/F), G2 = Gal(F2/F1), and G12 = Gal(F2/F). Then

G1 is a normal subgroup of G12 and G2 is their quotient group.

For the unit-quaternion group, we have

UQ's {1,-1,i,-i,j,-j,k,-k} -(Z2)- Z4 {1,-1,i,-i} -(Z2)- Z2 {1,-1} -(Z2)- identity group {1}

The equation itself has the solution

[noparse]Sqrt[18 + 12*Sqrt[2] + 10*Sqrt[3] + 7*Sqrt[2]*Sqrt[3]][/noparse]

where one can reverse the sign of the overall square root, of sqrt(2), and of sqrt(3). As expected from the Z2 quotient groups, it's all square roots.

The next step is to work out the polynomials that map the equation's solutions onto each other.

13. Jan 20, 2012

lpetrich

I decided to take on a simpler but related problem: a polynomial with a Galois group of Z2*Z2.

This is justified from UQ's -(Z2*Z2)- Z2 {1,-1} -(Z2)- Identity {1}

Handling the problem in general, I created a polynomial with roots

[noparse]c0 + s1*Sqrt[c1] + s2*Sqrt[c2] + s1*s2*c3*Sqrt[c1]*Sqrt[c2][/noparse]

where c0, c1, c2, c3 are in some field F, Sqrt[c1], Sqrt[c2], Sqrt[c1]*Sqrt[c2] are not in F, and s1 and s2 are either -1 or +1. I'm using Mathematica notation.

The polynomial:[noparse]
rtp = (c0^4 - 2*c0^2*c1 + c1^2 - 2*c0^2*c2 - 2*c1*c2 + c2^2 + 8*c0*c1*c2*c3 - 2*c0^2*c1*c2*c3^2 - 2*c1^2*c2*c3^2 - 2*c1*c2^2*c3^2 + c1^2*c2^2*c3^4) - 4*(c0^3 - c0*c1 - c0*c2 + 2*c1*c2*c3 - c0*c1*c2*c3^2)*x + 2*(3*c0^2 - c1 - c2 - c1*c2*c3^2)*x^2 - 4*c0*x^3 + x^4;
[/noparse]
The root-automorphism polynomials:
[noparse]
rtp00 = x;
rtp01 = ((-c0^3 + c0*c1 + 3 c0*c2 - c0^2*c2*c3 - 5 c1*c2*c3 + c2^2*c3 + 3 c0*c1*c2*c3^2 - 2 c0*c2^2*c3^2 + c1*c2^2*c3^3) + (3 c0^2 - 2 c1 - 2 c2 + 2 c0*c2*c3 - 2 c1*c2*c3^2 + c2^2*c3^2)*x - (3 c0 + c2*c3)*x^2 + x^3)/((c2 - c1) (1 - c2*c3^2));
rtp10 = ((-c0^3 + 3 c0*c1 + c0*c2 - c0^2*c1*c3 - 5 c1*c2*c3 + c1^2*c3 + 3 c0*c1*c2*c3^2 - 2 c0*c1^2*c3^2 + c1^2*c2*c3^3) + (3 c0^2 - 2 c1 - 2 c2 + 2 c0*c1*c3 - 2 c1*c2*c3^2 + c1^2*c3^2)*x - (3 c0 + c1*c3)*x^2 + x^3)/((c1 - c2) (1 - c1*c3^2));
rtp11 = ((2 c0 + c0^2*c3 - c1*c3 - c2*c3 + c0^3*c3^2 - 3 c0*c1*c3^2 - 3 c0*c2*c3^2 + 5 c1*c2*c3^3 - c0*c1*c2*c3^4) + (-1 - 2 c0*c3 - 3 c0^2*c3^2 + 2 c1*c3^2 + 2 c2*c3^2 + 2 c1*c2*c3^4)*x + (c3 + 3 c0*c3^2)*x^2 - c3^2*x^3)/((1 - c1*c3^2) (1 - c2*c3^2));
[/noparse]
Giving the roots {s1,s2} values {{-1,-1},{-1,1},{1,-1},{1,1}}
the rtp's make permutations
rtp00 {1,2,3,4}
rtp01 {2,1,4,3}
rtp10 {3,4,1,2}
rtp11 {4,3,2,1}
The coefficients of the rtpnn's are all in F, except for division-by-zero cases. Substituting each rtpnn into each rtpnn and taking the polynomial remainder with respect to rtp gives the right rtpnn for the Galois-group multiplication table.

I evaluated the polynomials that make the other permutations of the roots, and all of them have coefficients outside of F. So the Galois group of rtp is indeed Z2*Z2 and not some larger group that contains it.

14. Jan 20, 2012

joebohr

Wow, this is great work. My only questions are: how did you come up with this and how can it be applied to the problem at hand?

15. Jan 20, 2012

morphism

Nitpick: This is not true unless you assume something about the extensions.

Anyway, I feel as though I should mention that the problem of constructing a Galois extension E/Q with a prescribed finite group G as its Galois group is very difficult in general. In fact, it is a notoriously difficult open problem whether it is even possible to do so for an arbitrary G: this is known as the Inverse Galois Problem.

If you look at Section 5.2.5 in http://csag.epfl.ch/files/content/sites/csag/files/travdipl/GaloisInverse.pdf, you'll see a sketch of why the Galois group of x^8 − 72x^6 + 180x^4 − 144x^2 + 36 over the rationals is the quaternion group.

16. Jan 21, 2012

joebohr

I had no idea that the problem was so difficult or that it has not yet been solved. That pdf you posted a link to was great and very interesting, though. It seems like a very interesting field. I wonder which specific topics have been worked out within it. Furthermore, how can the method used by lpetrich above be used for similar Inverse Galois Problems. I understand the general procedure of his method but I'm not sure exactly which steps would work for a general problem.

I have a few more questions that will hopefully spark some for discussion. Firstly, for which groups has the Inverse Galois Problem been solved and which of these groups would be interesting to discuss? Also, which fields would be interesting to study the Galois Groups of polynomials over (specifically for the example problems). I am asking these questions because I thought they would spark some interest and I want to enhance my understanding of Galois Theory.

17. Jan 21, 2012

morphism

This is not my area of expertise, so I can't really say much. I do know however that solvable groups are known to be realizable (as the Galois group of a Galois extension) over Q. This is a really deep theorem of Shafarevich, and its proof uses a lot of interesting number theory (and class field theory in particular). Not sure what else is known. I think it's still unknown if every simple group is realizable over Q. You're much better off asking google for more information. :)

Lots of fields, e.g. finite extensions of Q, finite fields, ...

But honestly, I think it's a lot more interesting to study the Galois group of an extension rather than the Galois group of a specific polynomial (i.e. let's sidestep the issue of figuring out what the splitting field is). For instance a lot of number theory nowadays is concerned with understanding Galois groups of such extensions as $\overline{\mathbb Q} / \mathbb Q$ (here $\overline{\mathbb Q}$ is the field of algebraic numbers, so this extension is infinite).

18. Jan 21, 2012

joebohr

Last edited by a moderator: May 5, 2017
19. Jan 21, 2012

mathwonk

my class notes on the topic are on my web page. look at algebra class notes 843-2.

there are some computations on pages 43-50.

http://www.math.uga.edu/~roy/

basically, the galois group is based on how different the roots of a polynomial are from each other.

e.g. if whenever you adjoin one root of a polynomial to the previous base field, you never get any other roots for free, then the group is large as possible.

but if when you adjoin just one root, then the other roots are automatically contained in the first extension field, then the group is as small as possible.

e.g. if the polynomial is X^5 - 1, then just adjoining one root, a primitive root of 1, gives all the other roots as powers of that one. hence the field extension of degree 4 obtained by adjoining that one root already has all the roots in it, so the group has order 4.

then you have to figure out which group of order 4 it is.

more precisely, the galois group depends not just on how many other roots you get each time you adjoin another one, but on how the polynomial factors over the various successive adjunction fields.

the group is big as possible if each time you adjoin another root, it only splits off that one linear factor from the polynomial, and the rest remains irreducible.

Last edited: Jan 21, 2012
20. Jan 21, 2012

morphism

The story is a bit too complicated to tell here, but one basic example comes from trying to answer the question: what are the primes that can be written in the form x^2 + ny^2 (with x, y in Z)? Whenever you write p=x^2+ny^2, this amounts to saying that p can be "factored" as p=(x-sqrt(-n)y)(x+sqrt(-n)y) in Q(sqrt(-n)). This kind of factorization can be encoded in the Galois group of Q(sqrt(-n))/Q.

This is just one way field extensions show up in number theory.

Last edited: Jan 21, 2012
21. Jan 22, 2012

joebohr

Ok, this clears some things up. I was wondering how you could factor over Q(√2) but this seems to explain the basics. However, I am still confused as to how you could solve p=x^2+ny^2 to determine if it is factorable.

22. Jan 23, 2012

morphism

That's not so easy to figure out. In fact, you could write a 400-page book on this problem.

Note that the case of n=1 is essentially Fermat's theorem on the sum of two squares.

23. Jan 23, 2012

joebohr

How would this change if I were factoring over another extension field, say Q(√i)?

Sorry for asking so many difficult questions. Could someone at summarize the results in the book or show how to solve this problem for an easier field? Thanks.

24. Jan 23, 2012

morphism

Like I said, the case of n=1 (i.e. factoring in Q(i)) is Fermat's theorem on the sum of two squares. This is treated in most algebra books (and there's plenty of relevant stuff on the internet).

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