# Rules to compute with elementary symmetric polynomials

I'm given a polynomial of degree $4$ and its roots, let's call them $r_1,$ $r_2$, $r_3$ and $r_4$. I'm asked to show what is the value of the expression of $\sum r_1^2r_2$, that is, the sum of all different monomials one can form with permutations of the variables on the expression $r_1^2r_2$.

My approach

I know the relation between a polynomial an its roots. So, I know what is the value of the elementary symmetric polynomials $s_i$ and I just need to express all the polynomial $\sum r_1^2r_2$ in terms of $s_i$. There is a generic procedure to do so which I take from Cox's book "Galois theory".

My problem

However, I see that computing the value by hand apparently takes a lot of time. So for the first summand I should take the symmetric polynomial $f-s_1s_2$. Then I should repeat the process until I get to a zero polynomial.

I think there is a way to speed up the computations based in a form of symbolic computation. For instance in page 33, Cox reasons in the following way:

Given $f = \sum_4 x_1^3x_2^2x_3$ the leading term of $f$ is $x_1^3x_2^2x_3$ and therefore we should use polynomial $s_1s_2s_3$. Then $f_1 = f - s_1s_2s_3$ and here comes the magic:

$\sum s_1s_2s_3 = \sum_4 x_1^3x_2^2x_3+3\sum_4x_1^3x_2x_3x_4+3\sum x_1^2x_2^2x_3^2+ 8 \sum_4x_1^2x_2^2x_3x_4$

It is true that Cox claims that maybe a computer is helpful for this case. But how could I calculate this expression in order to avoid to directly calculate all the monomials and doing all the calculations?

It's not that hard. So, what you have is (I shall use $a$, $b$, $c$, and $d$, instead of $r_1$, $r_2$, $r_3$, and $r_4$)$$a^2b+a^2c+a^2d+b^2a+b^2c+b^2d+c^2a+c^2b+c^2d+d^2a+d^2b+d^2c.\tag{1}$$The term in which $a$ has the greatest exponent is $a^2b$ (and $a^2c$ and $a^2d$). So, from $(1)$ you subtract $(ab+ac+ad+bc+bd+cd)(a+b+c+d)$, thus getting$$-3(abc+abd+acd+bcd)$$and you're done!
$$\sum_{sym}r_1^2r_2=2\sum_{cyc}r_1^2(r_2+r_3+r_4)=\frac{1}{2}\sum_{cyc}r_1\sum_{sym}r_1r_2-6\sum_{cyc}r_1r_2r_3.$$
Actually, let $r_1=a$, $r_2=b$, $r_3=c$ and $r_4=d$.
Thus, $$\sum_{sym}a^2b=2\sum_{cyc}a^2(b+c+d)=$$ $$=2(a^2(b+c+d)+b^2(a+c+d)+c^2(a+b+c)+d^2(a+b+c));$$ $$\sum_{cyc}a=a+b+c+d;$$ $$\sum_{sym}ab=4(ab+ac+bc+ad+bd+cd);$$ $$\sum_{cyc}abc=abc+abd+acd+bcd.$$ In our case $$\sum_{sym}a^2b=2\sum_{cyc}a^2(b+c+d)=$$ $$=2(a+b+c+d)(ab+ac+ad+bc+bd+cd)-6(abc+abd+acd+bcd).$$
• can i ask what does $\sum_{cyc}$ means exactly? you sum only on cycle permutation is it? – Rodrigo Sep 21 '17 at 21:57
• @Rodrigo $\sum\limits_{cyc}$ it's a cyclic sum. – Michael Rozenberg Sep 22 '17 at 1:08