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Express the element $(a-b)^2(a-c)^2(b-c)^2$ In terms of the symmetric elementary polynomials.

I read the proof using Galois Theory, that any symetric polynomials can be written in terms of the symmetric polynomials. I was doing some explicit examples. I know that there exist an algorithm , but I don't want to read it , I only want to compute some examples. But I don't now how to express the above polynomial D:

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the algorithm shows you how to do the computation. – user51427 Dec 5 '12 at 17:45
up vote 1 down vote accepted

As it happens, that expression is the discriminant $\Delta(a, b, c)$ of the polynomial $$(x-a)(x-b)(x-c) = x^3 - s_1 x^2 + s_2 x - s_3$$ where $s_1, s_2, s_3$ are symmetric polynomials in $a, b, c$ of degree $1, 2, 3$. The discriminant indicates when this polynomial has a multiple zero, or in other words, when the polynomial and its derivative have a common root. This is expressed by a multiple of their resultant and after checking a particular value (e.g. $a=-1, b = 0, c=1$) we find

$$ \begin{eqnarray} \Delta(a,b,c) &=& -\operatorname{resultant}(x^3-s_1x^2+s_2x-s_3, 3x^2-2s_1x+s_2)\\ &=& s_1^2 s_2^2-4 s_2^3-4 s_1^3 s_3+18 s_1 s_2 s_3-27 s_3^2 \end{eqnarray} $$

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Well, start by writing down the products of elementary symmetric polynomials on $3$ variables that have the correct degree, take a linear combination of them with coefficients $t_1, t_2, \ldots$, expand everything out, and see what the coefficients have to be in order to make this match your polynomial.

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Set $c=0$.

$(a-b)^2a^2b^2 = ((a+b)^2-4ab)^2 ( ab) ^2 = (e_1^2 - 4e_2)e_2^2$

Subtract this expression from the full polynomial:

$$(a-b)^2(a-c)^2(b-c)^2 - ((a+b+c)^2-4ab-4bc-4ac) (ab+bc+ca)^2 \\ = abc \cdot (\text{polynomial of degree $3$}) $$

Repeat the procedure for the polynomial of degree 3.

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You can use the fact that $(x_3-x_1)(x_3-x_2)(x_2-x_1)$ is the determinant of the Vandermond matrix $V\in GL_3(\mathbb{R})$ having $V_{ij}=x_j^i$. The determinant of $V\, V^T$ is clearly the square of the determinant of $V$, and every element of $W=V\,V^T$ is a symmetric polynomial in $(x_1,x_2,x_3)$, in particular: $$W_{ij}=x_1^{i+j}+x_2^{i+j}+x_3^{i+j},$$ so $\det W$ is a symmetric polynomial, too.

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