Let $p,q,r,s \in \mathbb C[x,y,z]$ be defined as \begin{eqnarray*} p(x,y,z)&=&x^3+y^3+z^3,\\ q(x,y,z)&=&x^2y+y^2z+z^2x,\\ r(x,y,z)&=&xy^2+yz^2+zx^2,\\ s(x,y,z)&=&xyz. \end{eqnarray*}
Then how to find an irreducible polynomial (if it exists) $f(x_1,x_2,x_3,x_4) \in \mathbb{C}[x_1,x_2,x_3,x_4]$ such that $$f(p(x,y,z), q(x,y,z),r(x,y,z),s(x,y,z))=0,$$ i.e. $f(p(x,y,z), q(x,y,z),r(x,y,z),s(x,y,z))$ is the zero polynomial in $\mathbb C[x,y,z]$ ?
May be it has something to do with Hilbert Nullstellensatz, but I'm not sure.
Please help.
Note that $p$, $q+r$ and $s$ are homogeneous symmetric polynomials, so they are polynomials in the elementary symmetric polynomials $$e_1:=x_1+x_2+x_3,\qquad e_2:=x_1x_2+x_1x_3+x_2x_3,\qquad e_3:=x_1x_2x_3.$$ It is not hard to find expressions explicitly; $$p=e_1^3-3e_1e_2+3e_3,\qquad q+r=e_1e_2-3e_3,\qquad s=e_3.$$ Moreover $q-r=(x-y)(x-z)(y-z)$ is alternating and so $(q-r)^2$ is also symmetric; we have $$(q-r)^2=e_1^2e_2^2-4e_2^3-27e_3^2-4e_1^3e_3+18e_1e_2e_3.$$ From these we can isolate $e_1e_2$, $e_1^3$ and $e_2^3$ to find \begin{eqnarray*} e_1e_2&=&q+r+3s.\\ e_1^3&=&p+3q+3r+6s\\ e_2^3&=&-\frac{1}{4}\left((q-r)^2-(e_1e_2)^2+27s^2+4e_1^3s-18(e_1e_2)s\right)\\ &=&qr-ps+3qs+3rs+3s^2.\\ \end{eqnarray*} Plugging this back into the relation $(e_1e_2)^3-e_1^3e_2^3=0$ yields $$(q+r+3s)^3-(p+3q+3r+6s)(qr-ps+3qs+3rs+3s^2)=0.$$ This can be expanded to give $$9s^3+3ps^2-6qrs+p^2s+q^3+r^3-pqr=0.$$
It remains to check irreducibility. Let $a,b,c\in\Bbb{C}[x_2,x_3,x_4]$ be such that $$f =9x_4^3+3x_1x_4^2-6x_2x_3x_4+x_1^2x_4+x_2^3+x_3^3-x_1x_2x_3 =ax_1^2+bx_1+c.$$ Then $f$ is reducible if and only if either
or both.
In the first case, comparing coefficients shows that if such a $d$ exists, it is a factor of $$a=x_4,\qquad b=3x_4^2-x_2x_3,\qquad c=x_2^3+x_3^3+9x_4^3-6x_2x_3x_4,$$ which immediately implies that $\deg d=0$, a contradiction.
In the second case, comparing coefficients shows that $$\alpha\gamma=x_4,\qquad \alpha\delta+\beta\gamma=3x_4^2-x_2x_3,\qquad \beta\delta=x_2^3+x_3^3+9x_4^3-6x_2x_3x_4,$$ so from the first equality, without loss of generality $\alpha=1$ and $\gamma=x_4$. The last two equalities tell us $$\deg(\alpha\delta+\beta\gamma)=2 \qquad\text{ and }\qquad \deg\beta+\deg\delta=3,$$ which implies that $\deg\beta=1$ and $\deg \delta=2$. Plug in $\beta=ux_2+vx_3+wx_4$ to find that $$\delta=\alpha\delta=(3x^2-x_2x_3)-\beta\gamma=(3-w)x_4^2-x_2x_3-ux_2x_4-vx_3x_4,$$ but this contradicts the fact that the product $\beta\delta$ is cubic in $x_2$ and $x_3$. Hence $f$ is irreducible.
An alternative way to show that $f$ is irreducible is by showing that there is no quadratic $g\in\Bbb{C}[x_1,x_2,x_3,x_4]$ such that $g(p,q,r,s,y)$. Note that $g$ must be homogeneous. Expressing all products of pairs from $\{p,q,r,s\}$ on the basis of monomials in $\Bbb{C}$ of degree $6$ yields $$\begin{matrix} &x^6 &x^5y &x^5z &x^4y^2&x^4z^2&x^4yz &x^3y^3&x^3y^2z&x^3yz^2&x^2y^2z^2\\ p^2& 1 & & & & & & 2 & & & \\ q^2& & & & 1 & & & & & 2 & \\ r^2& & & & & 1 & & & 2 & & \\ s^2& & & & & & & & & & 1 \\ pq & & 1 & & & 1 & & & 1 & & \\ pr & & & 1 & 1 & & & & & 1 & \\ ps & & & & & & 1 & & & & \\ qr & & & & & & 1 & 1 & & & 3 \\ qs & & & & & & & & 1 & & \\ rs & & & & & & & & & 1 & \end{matrix}$$ Here we used the fact that $p$, $q$, $r$ and $s$ are invariant under cyclic shifts of $x$, $y$ and $z$. The coefficients of any quadratic $g\in\Bbb{C}[x_1,x_2,x_3,x_4]$ with $g(p,q,r,s)=0$ must then be in the kernel of the transpose of this $10\times10$-matrix, so it suffices to show that its determinant is nonzero.
Laplace expansion along the first three columns and subsequently along the rows $s^2$, $ps$, $qs$ and $rs$ shows its determinant is the same as that of thr $3\times3$-identity matrix, so $f$ is irreducible.