Symmetric polynomial written in elementary polynomials

Question

Consider the symmetric polynomial in three variables $x, y, z$ $$|x^2y+y^2z+z^2x-xy^2-yz^2-zx^2|.$$ A Theorem says that it can be written in elementary symmetric polynomials $$s_1=x+y+z, \quad s_2=xy+yz+zx, \quad s_3=xyz.$$ I am trying to find that expression but I can't. How can we find such representation?

Answer 1

Let $s_1=3u$, $s_2=3v^2$ and $s_3=w^3$.

Hence, $$\left|\sum_{cyc}(x^2y-x^2z)\right|=\left|(x-y)(x-z)(y-z)\right|=\sqrt{\prod_{cyc}(x-y)^2(x-z)^2(y-z)^2}=$$ $$=\sqrt{\sum_{cyc}(x^4y^2+x^4z^2-2x^3y^3-2x^4yz+2x^3y^2z+2x^3z^2y-2x^2y^2z^2)}=$$ $$=\sqrt{27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)}=$$ $$=\sqrt{s_1^2s_2^2-4s_2^3-4s_1^3s_3+18s_1s_2s_3-27s_3^2}.$$

Answer 2

Someone already stated in the comments that the expression is not a polynomial. This is because of the absolute value. But $$|x^2y+y^2z+z^2x-xy^2-yz^2-zx^2|^2=(x^2y+y^2z+z^2x-xy^2-yz^2-zx^2)^2$$ and the right-hand side is a polynomial. So it can be expressed by elemntary symmetric polynomials. We have to do some calculations with polynomials: multiplication and addition/subtractions. I use the ouput of a CAS (MAXIMA) to show you the intermediate results of the calculation

The first command tels MAXIMA how to format the output. $$\begin{array}\\ (\%\text i1)\; \text{display2d}:\text{false};\\ (\%\text o1)\; \text{false} \end{array}$$

Westore the symmtetric polynomials into the variables $s1, s2, s3$: $$\begin{array}\\ (\%\text i2)\; s_1:x+y+z;\\ (\%\text o2)\; z+y+x\\ (\%\text i3)\; s_2:xy+yz+zx;\\ (\%\text o3)\; yz+xz+xy\\ (\%\text i4)\; s_3:xyz;\\ (\%\text o4)\; xyz\\ \end{array}$$

We store the squared expression in the variable $e1$. The keyword expandtells MAXIMA that it should actually multiply out the expression in parenthesis.

$$\begin{array}\\ (\%\text i5)\; e_1:(x^2y+y^2z+z^2x−xy^2−yz^2−zx^2)^2,\text{expand};\\ (\%\text o5)\; y^2z^4-2xyz^4+x^2z^4-2y^3z^3+2xy^2z^3+2x^2yz^3-2x^3z^3 +y^4z^2+2xy^3z^2-6x^2y^2z^2+2x^3yz^2+x^4z^2-2xy^4z +2x^2y^3z+2x^3y^2z-2x^4yz+x^2y^4-2x^3y^3+x^4y^2\\ \end{array}$$ Note the order how the terms of the expression are displayed. It is the lexicographical order:

A term $x^{a_3}y^{a_2}z^{a_1}$ is printed before $x^{b_3}y^{b_2}z^{b_1}$

• if $a_1>b_1$
• or if $a_1=b_1$ and $a_2>b_2$
• or if $a_1=b_1$ and $a_2=b_2$ and $a_3>b_3$ The term with the highest exponent in regards to the lexicographical order is the leftmost term of line $\%\text o 5$: $y^2z^4$. If the leftmost term is $Ax^{a_3}y^{a_2}z^{a_1}$ the algorithm requires to subtract $$A{s_1}^{a_1-a_2}{s_2}^{a_2-a_3}{s_3}^{a_3}$$ This guarantees the the highest term disappears and the remaining expression has terms with lower exponents in regards go the lexicographical order. Als the degree of the difference will not be higher than the degree of the starting polynomial.

$$\begin{array}\\ (\%\text i6)\; e_2:e_1-{s_1}^2{s_2}^2,\text{expand};\\ (\%\text o6)\; (-4xyz^4)-4y^3z^3-6xy^2z^3-6x^2yz^3-4x^3z^3-6xy^3z^2 -21x^2y^2z^2-6x^3yz^2-4xy^4z-6x^2y^3z -6x^3y^2z-4x^4yz-4x^3y^3\\ \end{array}$$

We will repeat this process and will finally end with a difference $0$.

$$\begin{array}\\ (\%\text i7)\; e_3:e_2+4{s_1}^3{s_3},\text{expand};\\ (\%\text o7)\; (-4y^3z^3)+6xy^2z^3+6x^2yz^3-4x^3z^3+6xy^3z^2+3x^2y^2z^2 +6x^3yz^2+6x^2y^3z+6x^3y^2z-4x^3y^3\\ (\%\text i8)\; e_4:e_3+4{s_2}^3,\text{expand};\\ (\%\text o8)\; 18xy^2z^3+18x^2yz^3+18xy^3z^2+27x^2y^2z^2+18x^3yz^2 +18x^2y^3z+18x^3y^2z\\ (\%\text i9)\; e_5:e_4-18s_1s_2s_3,\text{expand};\\ (\%\text o9)\; -27x^2y^2z^2\\ (\%\text i10)\; e_6:e_5+27{s_3}^2,\text{expand};\\ (\%\text o10)\; 0\\ \end{array}$$

From this we see that

$$e_1={s_1}^2{s_2}^2-4{s_1}^3s_3-4{s_2}^3+18 s_1s_2s_3-27{s_3}^2$$

and therefore

$$|x^2y+y^2z+z^2x-xy^2-yz^2-zx^2|=\sqrt{{s_1}^2{s_2}^2-4{s_1}^3s_3-4{s_2}^3+18s_1s_2s_3-27{s_3}^2}$$

A detailed description of the proof of the algorithm can be found here.