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


by joebohr
Tags: computing, galois, irreducible, polynomial, reducible, univariate
joebohr
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#1
Jan15-12, 06:08 PM
P: 57
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
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zhentil
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#2
Jan15-12, 10:10 PM
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Where do these polynomials live? You have to fix a base field in order to talk about Galois groups.
zhentil
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#3
Jan15-12, 10:11 PM
P: 491
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.

morphism
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#4
Jan16-12, 01:02 PM
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Computing the Galois Group of a Univariate Polynomial (Irreducible or reducible)


It's not too hard to compute the Galois group of a poly of deg <= 4 (at least over Q, and certain other fields) by hand. See for example http://www.math.uconn.edu/~kconrad/b...bicquartic.pdf

On the other hand, deg >= 5 is a lot trickier.
joebohr
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#5
Jan16-12, 03:12 PM
P: 57
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[itex]_{1}[/itex])[itex]\cong[/itex]S[itex]_{2}[/itex]? 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[itex]_{2}[/itex])[itex]\cong[/itex]S[itex]_{6}[/itex] but that seems too easy).
morphism
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#6
Jan16-12, 07:56 PM
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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.
joebohr
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#7
Jan18-12, 01:36 PM
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Quote Quote by morphism View Post
What is "the group of second degree separable polynomials"? Since you write S_2, do you mean the symmetric group on 2 elements?
Yes, that is exactly what I meant.

Quote Quote by morphism View Post
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.
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!
morphism
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#8
Jan18-12, 02:00 PM
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Quote Quote by joebohr View Post
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!
I'm not sure what you mean by this, as the quaternions don't form a (commutative) field...
joebohr
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#9
Jan18-12, 07:45 PM
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Quote Quote by morphism View Post
I'm not sure what you mean by this, as the quaternions don't form a (commutative) field...
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
lpetrich
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#10
Jan19-12, 12:57 AM
P: 514
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.
joebohr
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#11
Jan19-12, 01:31 PM
P: 57
Quote Quote by lpetrich View Post
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.
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 A Note on the Quaternion Group as Galois Group )? 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?
lpetrich
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#12
Jan19-12, 11:35 PM
P: 514
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

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

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.
lpetrich
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#13
Jan20-12, 11:48 AM
P: 514
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

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

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:
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;

The root-automorphism polynomials:

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));

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.
joebohr
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#14
Jan20-12, 01:48 PM
P: 57
Quote Quote by lpetrich View Post
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

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

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:
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;

The root-automorphism polynomials:

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));

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.
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?
morphism
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#15
Jan20-12, 10:18 PM
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Quote Quote by lpetrich View Post
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.
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/si...oisInverse.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.
joebohr
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#16
Jan21-12, 09:17 AM
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Quote Quote by morphism View Post

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/si...oisInverse.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.
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.
morphism
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#17
Jan21-12, 02:33 PM
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Quote Quote by joebohr View Post
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?
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. :)

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.
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 [itex]\overline{\mathbb Q} / \mathbb Q[/itex] (here [itex]\overline{\mathbb Q}[/itex] is the field of algebraic numbers, so this extension is infinite).
joebohr
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#18
Jan21-12, 04:09 PM
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[QUOTE=morphism;3720969]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. :)[QUOTE]

Great, I found a couple of papers on this (http://www.math.uiuc.edu/Algebraic-Number-Theory/0136/ and www.arxiv.org/pdf/1112.1522). These should help.

Quote Quote by morphism View Post

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 [itex]\overline{\mathbb Q} / \mathbb Q[/itex] (here [itex]\overline{\mathbb Q}[/itex] is the field of algebraic numbers, so this extension is infinite).
Ok, can we explore some example problems in which these extensions would come up?

Also, how would absolute Galois Groups fit into this?


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