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Related to this question.

For polynomial $f$, let $rad(f)$ denote the radical of $f$, the product of irreducible factors.

Suppose that $G(x,y) \in \mathbb{C}[x,y]$ is homogeneous without any repeated factors.

Let $r,s \in \mathbb{C}[u_1,\ldots,u_n] $ be coprime and at least one of them depends on at least two variables. Let $f \in \mathbb{C}[u_1,\ldots,u_n],R,S \in \mathbb{C}[t]$.

A "bad" identitiy is identity of the form $r=R(f),s=S(f)$.

Q1 Is it true that $\deg{(rad(G(r,s)))}\ge \max\{\deg(r),\deg(s)\}(\deg(G)-2) + 2$, except for "bad" identities?

Pasten's nice comment about linear substitution showed bad identities can't be excluded, here is the construction.

From the linked question, let $G(x,y)=x^4+xy^3$.

Certain univariate $r,s$ are counterexample.

The radical is divisible by $t+2$. Set $t=(u+v)^2-2$.

Explicitly:

r=8*u^6 + 48*u^5*v + 120*u^4*v^2 + 160*u^3*v^3 + 120*u^2*v^4 + 48*u*v^5 + 8*v^6 - 48*u^4 - 192*u^3*v - 288*u^2*v^2 - 192*u*v^3 - 48*v^4 + 96*u^2 + 192*u*v + 96*v^2
s=u^8 + 8*u^7*v + 28*u^6*v^2 + 56*u^5*v^3 + 70*u^4*v^4 + 56*u^3*v^5 + 28*u^2*v^6 + 8*u*v^7 + v^8 - 8*u^6 - 48*u^5*v - 120*u^4*v^2 - 160*u^3*v^3 - 120*u^2*v^4 - 48*u*v^5 - 8*v^6 + 24*u^4 + 96*u^3*v + 144*u^2*v^2 + 96*u*v^3 + 24*v^4 - 96*u^2 - 192*u*v - 96*v^2 + 144

The degree of the radical is $17$, which replaces $+2$ with $+1$.

I suspect $abc$ implies this, so $abc$ for multivariate polynomials might solve this.

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    $\begingroup$ Except for the "+2" (which requires finer analysis on pencils in $\mathbb{P}^2$) a bound of this sort might be proved as follows: a general linear substitution $u=at+b$, $v=ct+d$ will preserve the degree of $r,s$, the coprimality of $r,s$, and cannot increase the degree of $rad(G(r,s))$. $\endgroup$
    – Pasten
    Commented Apr 13, 2015 at 23:37
  • $\begingroup$ @Pasten Many thanks! You essentially solved this (check my answer). Added another question here. $\endgroup$
    – joro
    Commented Apr 14, 2015 at 6:44
  • $\begingroup$ @Pasten I am not satisfied by my answer, though it appears correct. If you have less trivial answer to Q1 please answer. $\endgroup$
    – joro
    Commented Apr 14, 2015 at 10:40
  • $\begingroup$ What my comment hints is a way to prove the bound with $+1$ (using what one knows in the 1-variable case). You understood something different, but this misundertanding is actually interesting (unless I misunderstood my own comment). $\endgroup$
    – Pasten
    Commented Apr 15, 2015 at 15:15
  • $\begingroup$ @Pasten Maybe I misunderstood your first comment, but it lead to something of interest to me :-) $\endgroup$
    – joro
    Commented Apr 15, 2015 at 16:32

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