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I am trying to self-read through Ian Stewart's Galois theory and am stuck at this paragraph. I quote - "Lagrange observed that all methods for solving polynomial equations by radicals involve constructing rational functions of the roots that take a small number of values when the roots $\alpha_j$ are permuted. Prominent among these is the expression

$$\delta = \prod_{j< k} (\alpha_j - \alpha_k)$$

which takes just two values, $\pm \delta$, plus for even permutations and minus for the odd ones. Therefore, $\Delta = \delta^{2}$ is a rational function of the coefficients. This gets us started and it yields a complete solution for the quadratic."

My question is as follows: Why should an expression that is unchanged by permuting the roots of a polynomial be expressible as a rational function of the polynomial's coefficients? Is this statement even true or have I misunderstood the book?

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up vote 8 down vote accepted

This is a straightforward consequence of the Fundamental Theorem on Symmetric Functions. Indeed, the coefficients of the polynomial are the evaluation of the so called elementary symmetric functions at the roots.

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Thanks for your answer. I will read the theorem and post the complete reasoning as an answer once I understand it. –  Dinesh Oct 23 '10 at 22:04
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@Dinesh: the first proof of the fundamental theorem at Wikipedia is, in my opinion, unreadable. I prefer the one labeled "an alternative proof," which I believe is equivalent to the one I wrote up here: qchu.wordpress.com/2009/08/20/… –  Qiaochu Yuan Oct 23 '10 at 22:38
    
@QIaochu, it would take quite a bit of inventiveness to outworst that proof! –  Mariano Suárez-Alvarez Oct 23 '10 at 23:39
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