Htykuut

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)sright)\
&=&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 $a,b,cinBbb{C}[x_2,x_3,x_4]$ 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

1. there exists some $dinBbb{C}[x_2,x_3,x_4]$ with $deg d>0$ that divides $a$, $b$ and $c$; or


2. there exist some $alpha,beta,gamma,deltainBbb{C}[x_2,x_3,x_4]$ such that $f=(alpha x_1+beta)(gamma x_1+delta)$;

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
$$alphagamma=x_4,qquad
alphadelta+betagamma=3x_4^2-x_2x_3,qquad
betadelta=x_2^3+x_3^3+9x_4^3-6x_2x_3x_4,$$
so from the first equality, without loss of generality $alpha=1$ and $gamma=x_4$. The last two equalities tell us
$$deg(alphadelta+betagamma)=2
qquadtext{ and }qquad
degbeta+degdelta=3,$$
which implies that $degbeta=1$ and $deg delta=2$. Plug in $beta=ux_2+vx_3+wx_4$ to find that
$$delta=alphadelta=(3x^2-x_2x_3)-betagamma=(3-w)x_4^2-x_2x_3-ux_2x_4-vx_3x_4,$$
but this contradicts the fact that the product $betadelta$ 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 $ginBbb{C}[x_1,x_2,x_3,x_4]$ 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 $ginBbb{C}[x_1,x_2,x_3,x_4]$ with $g(p,q,r,s)=0$ must then be in the kernel of the transpose of this $10times10$-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 $3times3$-identity matrix, so $f$ is irreducible.