Trying to understand Galois theory book explanation

[jostpuur]

I'm trying to read this:

https://www.amazon.com/dp/1584883936/?tag=pfamazon01-20

On the pages 12-13 it says:

For example, if a cubic polynomial has roots $\alpha_1,\alpha_2,\alpha_3$ and $\omega$ is a primitive cube root of unity, then the expression

$$u = (\alpha_1 + \omega\alpha_2 + \omega^2\alpha_3)^3$$

takes exactly two distinct values. In fact, even permutations leave it unchanged, while odd permutations transform it to

$$v = (\alpha_1 + \omega^2\alpha_2 + \omega\alpha_3)^3$$

It follows that $u+v$ and $uv$ are fixed by all permutations of the roots and must, therefore, be expressible as rational functions of the coefficients. That is, $u$ and $v$ are solutions of a quadratic equation, and can thus be expressed using square roots.

The bold part is where lose the track. Why are quantities, which are invariant under permutations of roots, expressible in some certain way?

 

[Stephen Tashi]

I have a hazy idea of the reason. Someone else may be able to be precise. (Prof. Stong of UVA once said "The half life of Galois Theory is 10 seconds" - meaing it takes 10 seconds to forget half of what you learned about it.)

There are so-called "symmetric functions" of n variables. They are invariant under all permutations of the variables. The set of all symmetric functions of n variables is a vector space since the the sum of scalar multiples of such functions is still a symmetric function. In fact, it's "more than a vector space" since the product of symmetric functions is defined and is a symmetric function. Is it a "ring"?

The coefficients of a polynomial equation in 1 variable are symmetric functions of the roots of the polynomial. For example, for roots of the quadratic r and s, we have (x-r)(x-s) = x^2 - (r+s)x + rs. The constant function 1, the function r+s and the function rs are symmetric functions of the variables r and s.

If you think in terms of vector spaces, when you have two sets of vectors, it is natural to ask whether one set can be expressed using the other one as a "basis". The coefficients of a general polynomial equation are the so-called "elementary" symmetric functions. This suggests the idea that they could be a "basis" for the set of all symmetric functions. So if you have two symmetric functions u and v, it is natural to suspect that u and v can be expressed in terms of elementary functions and these happen to be exactly the coefficients of a polynomial equation.

What is involved in your example is not the statement that u and v are expressible as a linear combination of elementary symmetric functions. As I recall, the idea of "rationally expressibile" is more general than that, but it the result is in the same "spirit" as saying that there is an "elementary' set of things that are a "basis" than can be used to express all other things of the same sort.

See if that book has theorems that say something like "Any symmetric function can be expressed rationally in terms of elementary symmetric functions" and "The coefficients of the general polynomial equation are precisely the elementary symmetric functions".

 

[jostpuur]

I hadn't thought about the polynomial coefficients as symmetric functions of the roots, so your response helped a little bit, but I'm still confused by the claim made by the author. Suppose for example that

$$f_1(x_1,x_2) = x_1 + x_2$$
$$f_2(x_1,x_2) = x_1x_2$$

$$f(x_1,x_2) = e^{x_1 + x_2}$$

If you were given a task to find a rational function $\Phi$ such that

$$f(x_1,x_2) = \Phi(f_1(x_1,x_2),f_2(x_1,x_2))$$

it would be an impossible task. I don't believe that a finite collection of symmetric functions would span the whole collection of all symmetric functions in nearly any sense.

 

[Stephen Tashi]

I see your point. My guess is that "symmetric functions" in the Galois theory of equations refers, by definition, to rational functions that are symmetric, so the problem of $e^{x_1 + x_2}$ is probably "defined away". In the case of the passage you quoted $u$ and $v$ were polynomials.

 

[micromass]

Finally, I think I solved your problem.

Take a monic polynomial $$x^n+a_{n-1}x^{n-1}+...+a_1x+a_0$$ with coefficients in $$\mathbb{Q}$$ (the field doesn't matter much). If $$x_1,...,x_n$$ are the roots of this polynomail, then

$$x^n+a_{n-1}x^{n-1}+...+a_1x+a_0=(x-x_1)...(x-x_n)$$.

By equating the coefficients of both polynomials, we easily get that

$$a_{n-1}=-\sum_{i=1}^n{x_i},~a_{n-2}=\sum_{1\leq i<j\leq n}{x_ix_j},...$$

The point is that every coefficient of the polynomial can be expressed as a polynomial in the roots. But not just polynomials, but elementary symmetric polynomials. (see http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial and http://en.wikipedia.org/wiki/Symmetric_polynomial ).
Now, any symmetric polynomial can be expressed as a polynomial expression of elementary symmetric polynomials (this is the fundamental theorem of symmetric polynomials, see the wikilinks).

Now, what does this have to do with the question. Well, take a quantity a that is invariant under permutations of the roots. We, of course, assume that there exists a polynomial P such that $$a=P(x_1,...,x_n)$$, so a can be expressed as a polynomial expression of the roots.

Since a is invariant under permutation of the roots, it follows that P is a symmetric polynomial. Thus P can be written as

$$P(x_1,...,x_n)=Q(E(x_1,...,x_n),E'(x_1,...,x_n),...)$$

with E, E', ... elementary symmetric polynomials. But we have found that symmetric polynomails of the roots are coefficients of the original polynomial, thus

$$P(x_1,...,x_n)=Q(a_1,...,a_{n-1})$$

For more information, see http://en.wikipedia.org/wiki/Cubic_polynomial under the section Lagrange's method.

I don't think Stewart really meant for readers to understand all of this (I think this because this statement is in the introduction). I guess he wanted to sketch a method. So don't feel bad if you don't entirely get this. This method has nothing to do with what comes later in the book (which he does treat thoroughly, in contrast with this). In fact, he states

Because the formulas are messy and the story is lengthy, the most we can do here is give some flavour of what is involved.

So I don't think Stewart wanted you to grasp everything that's going on. But you should just get an idea of how previous mathematicians did these kind of problems...

 

[jostpuur]

Now, what does this have to do with the question. Well, take a quantity a that is invariant under permutations of the roots. We, of course, assume that there exists a polynomial P such that $$a=P(x_1,...,x_n)$$, so a can be expressed as a polynomial expression of the roots.

$$P(x_1,...,x_n)=Q(a_1,...,a_{n-1})$$

So $a$ is expressible as a polynomial of the coefficients? Was there a reason for Stewart to speak about rational functions?

 

[micromass]

So $a$ is expressible as a polynomial of the coefficients? Was there a reason for Stewart to speak about rational functions?

I understood rational function as polynomial with rational coefficients here. I don't think we actually need the real rational functions here...