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. $r(u,v),s(u,v) \in \mathbb{C}[u,v]$ are
Let $r,s \in \mathbb{C}[u_1,\ldots,u_n] $ be coprime. At and at least one of them depends on both $u,v$, iat least two variables.e Let $f \in \mathbb{C}[u_1,\ldots,u_n],R,S \in \mathbb{C}[t]$.
A "bad" identitiy is not univariateidentity of the form $r=R(f),s=S(f)$.
Q1 Is it true that $\deg{(rad(G(r(u,v),s(u,v))))}\ge \max\{\deg(r),\deg(s)\}(\deg(G)-2) + 2$$\deg{(rad(G(r,s)))}\ge \max\{\deg(r),\deg(s)\}(\deg(G)-2) + 2$, except for "bad" identities?
If we drop the restriction to depend on both $u,v$ thisPasten
's nice comment about linear substitution
showed bad identities can't be excluded, here is falsethe construction.
The bound is attainableFrom the linked question, let $G(x,y)=x^4+xy^3$.
Added Certain univariate $r,s$ are counterexample.
Pasten's comment essentially disproved Q1The radical is divisible by $t+2$. Set $t=(u+v)^2-2$.
LetExplicitly:
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 $r,s \in \mathbb{C}[u_1,\ldots,u_n] $ be coprime$17$, which replaces $+2$ with $+1$.
Q2 Is it true that $\deg{(rad(G(r,s)))}\ge \max\{\deg(r),\deg(s)\}(\deg(G)-2) + 1$?
I suspect $abc$ implies this, so $abc$ for multivariate polynomials might solve this.