Group Theory-solution to polynomial equations

1. Jan 24, 2008

Himanshu

How does Group Theory help to find the solutions of the polynomial equations of degree five or greater. How did Galios proved that polynomial equations of degree five or greater can't be solved in an algebraic way. A formal answer would be helpful.

2. Jan 24, 2008

HallsofIvy

Staff Emeritus
It's hard to believe you are serious! There are whole books and whole semester courses devoted to that topic. How much abstract algebra have you had? Do you know what a "field extension" is? Do you know what a "normal subgroup" is? I would consider those pre-requisites.

3. Jan 25, 2008

Himanshu

I know the basics of group theory. I know what is a group, a subgroup, rings though I have not explored too much of it, multiplication table, a dihedral group, normal subgroup, a little bit of representation theory, and its application in the quark model. I studied all these in "Classical Mechanics" by-Herbert Goldstein. Group theory is added as an appendix to this book. I haven't studied much of abstract algebra in a mathematics book.

I have also read about Evastre Galios that had invented Group Theory to find the solutions of the polynomial equations of degree five or greater. Therefore I assume that he used the 'infant' Group theory to solve his problems. Later it was extended by Lie, and was realised that Group Theory is an important tool for Physics.

4. Jan 25, 2008

HallsofIvy

Staff Emeritus
Very good. (It is "Galois", by the way, not "Galios".) There is simply not room here (and this isn't even a "margin") to go into it in depth, but here is a very general, very rough, outline. You start with "field extensions". Given a number, r, Q(r) is defined as the 'smallest field containing all rational numbers (Q) and the number r. For example, if r= $\sqrt{3}$, one can show that Q($\sqrt{3}$) consists of all number of the form a+ b$\sqrt{3}$, where a and b are rational numbers. If r= $^3\sqrt{2}$, one can show that Q($^3\sqrt{2}$) consists of all numbers of the form a+ b$^3\sqrt{2}$+ c$^3\sqrt{4}$ where a, b, c are rational numbers. Given several numbers, r1, r2, etc, the we define Q(r1, r2, ...) in the obvious way: the smallest field containing all rationa numbers and each of those numbers.

Given a polynomial, p(x), with integer coefficients, we can define Q(p) as "the smallest field containg all rational numbers and all solutions to p(x)= 0. If p(x)= x2- 3, then Q(p) is exactly the same as Q($\sqrt{3}$), all numbers of the form a+ b$\sqrt{3}$. However, if p(x)= x3- 2, Q(p) is not the same as Q($^3\sqrt{2}$) since that contains only real numbers and p(x)= x3- 2= 0 has complex number solutions. We have to include $\omega_3$, the "principal third root of unity".

That leads to a precise definition of what we mean when we say we can have a formula for a polynomial equation:
A polynomial is said to be "solvable by roots" if and only if there exist a finite sequence of numbers, x1[/sup], x2, ..., xn, together with a corresponding sequence of positive integers, k1, k2, ..., kn, such that:
Q(p) is a subfield of Q(x1, x2,... , xn) (Typically, you can take the x's to be the solutions to p(x)= 0 and Q(p) is equal to that, but it is not necessary) and
For each i, either
1) xiki is contained in Q(x1[/sup, ...,xi-1[/itex] or
2) xi= $\omega$ki
Do you see what that means? Each x is either a complex "root of unity" or it is a root of a number which is a root,...

Once you have that, you can look at automorphisms on those fields. That is, isomorphisms from the field to itself. Since the automorphism are from one field to another, we can "chain" them- define an operation, o, by f o g= f(g(x)). Of course, the identity function e(x)= x is an automorphism and acts as an identity under "composition". Also, since each function is an isomorphism, it has an inverse. The set of all automorphisms on a field is a group with composition as the operation. Given a field extension of the rational numbers, that is, a field F which contains Q, the "Galois group" of F, G(F), is defined as the set of all automorphisms on F that "fix" members of Q: if r is in Q, then f(r)= r. Clearly the identity function is in that set and so is the inverse of any such functions: G(F) is a subgroup of the group of automorphisms on F. If p(x) is a polynomial with integer coefficients, Q(p) is such a field and G(Q(p)) is the "Galois group of p(x)", called G(p).

Since any field containing the rational numbers must have an infinite number of members, you might think those "Galois groups" would be infinite. That is not in general true. Take p(x)= x2- 3. We have already seen that any member of Q(p) can be written as $a+ b\sqrt{3}$. Apply a function, f, of G(p), to such a number: Since f is an isomorphism, f($a+ b\sqrt{3}$)= $f(a)+ f(b)f(\sqrt{3})$. Since f 'fixes' rational numbers, that is $a+ bf(\sqrt{3})$. Can $f(\sqrt{3})$ be anything?

No, $\sqrt{3}$ satisfies $(\sqrt{3})^2= 3$. f($\sqrt{3}^2$)= (f($\sqrt{3})^2= f(3)= 3. That is, f([itex]\sqrt{3}$) must satisfy the same equation. There are only two such numbers, $\sqrt{3}$, and $-\sqrt{3}$. If e($\sqrt{3}$)= $\sqrt{3}$, then e(a+ b$\sqrt{3}$)= a+ be($\sqrt{3}$)= a+ b$\sqrt{3}$, the identity function. If f($\sqrt{3}$)= $-\sqrt{3}$, then f(a+ b$\sqrt{3}$)= a- b$\sqrt{3}$ and fof= e. With p(x)= x2- 3, G(p) is the group with 2 elements.

Essentially, for any polynomial p, functions in G(p) permute solutions to p(x)= 0. What Galois did was not develop all of Group theory but show, and use, the properties of groups of permutations (without using the word "group" of course). All of Group theory has arisen from that.

Now, another definition: A group, G, is called "solvable" if and only if there exist a finite sequence of groups, G0, G1, ..., Gn, such that G0 is the "trivial" group containing only the identity, Gn= G and, for all i> 0, Gi-1 is a normal subgroup of G and the quotient group, Gi/Gi-1 is a commutative group.

Notice how the "extension fields" in the definition of "solvable by radicals" matches up with the subgroups. The main theorem, which I can surely not give here, is "A polynomial, p(x), with integer coefficients, is 'solvable by radicals' if and only if its Galois group, G(p), is a 'solvable group' ".

The fact that there cannot be a "formula" for solving all polynomials of a degree greater than or equal to 5" follows from a few more facts:
1. Given any positive integer, n, there exist a polynomial, of degree n, whose Galois group is the symmetric group Sn.
2. If n is greater than or equal to 5, Sn is NOT a solvable group.

One relatively elementary textbook that goes into this is "Abstract Algebra a concrete introduction" by Robert Redfield, published by Addison Wesley Longman

5. Jan 28, 2008

Himanshu

Thank you very much for that help. It took me some time to read and understand. I have few questions.

Can't we have a feild that contains imaginary numbers or quaternions or octonions. Or is it that in the above case the feild is defined only for rational numbers. I guess if the feild would be defined for imaginary numbers too then Q(p) must be same as Q($$\sqrt[3]{2}$$).

It seems I find those sentences conflicting. An automorphism is an isomorphism from a mathematical object to itself. But the third sentence "Since the automorphism are from one field to another..." is conflicting the second

That was an interesting point. At first I really thought that a "Galois group" would be infinite. The above point elucidates it.

Yes I thought that must be true. How could Galois develop all of Group theory as a teenager!

Last edited: Jan 28, 2008
6. Jan 28, 2008

mathwonk

since roots of complex numbers form a cyclic pattern, if all roots of an equation can be obtained from the coefficient field by successively adjoining roots, then the structure of the field containing the roots has a symmetry which decomposes into cyclic structures.

i.e. the "automorphism group", or symmetry group, of the root field , has a tower of relatively normal subgroups, with cyclic quotients.

galois was strong and insightful, and he read the works of the masters who came before him.