Counterexample to Proposition of Granville related to abc conjecture

Question

Looks like there is counterexample to Proposition related to abc conjecture. Confusion is likely.

From RATIONAL AND INTEGRAL POINTS ON QUADRATIC TWISTS OF A GIVEN HYPERELLIPTIC CURVE, Andrew Granville

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p. 11, Proposition 2 b

Suppose that $G(x,y) \in \mathbb{C}[x,y]$ is homogeneous without any repeated factors. For any coprime polynomials $r(t),s(t) \in \mathbb{C}[t]$, we have

$$ \#\{\alpha \in \mathbb{C}: G(r(\alpha),s(\alpha))=0\} \ge \max\{\deg(r),\deg(s)\}(\deg(G)-2) + 2. $$

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$\#\{\alpha \in \mathbb{C}: G(r(\alpha),s(\alpha))=0\}$ counts the distinct zeros and equals the degree of the radical of $G(r(t),s(t))$.

Explicit counterexample.

Let $G(x,y)=x^4+xy^3,r(t)=8t^3 + 64,s(t)=t^4 - 64t$

We have:

$$ G(r(t),s(t))=\left(8\right) \cdot (t + 2) \cdot (t^{2} - 2 t + 4) \cdot (t^{2} + 4 t - 8)^{2} \cdot (t^{4} - 4 t^{3} + 24 t^{2} + 32 t + 64)^{2} $$

So $G(r(t),s(t))$ have $9$ distinct zeros.

By the Proposition $9 \ge (( (4\cdot(4-2)+2)=10)$ which is false.

Q1 Is this really counterexample?

The Proposition is unconditional and this doesn't appear to contradict abc.

The errata of the paper doesn't address this.

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Andrew Granville ask for other $G$. There are constructions.

Here is example in computer readable form with t=x:

G=x^3*y + x*y^3 + 8*y^4
r=x^16 - 40*x^14 - 4352*x^13 + 348*x^12 + 1024*x^11 + 189416*x^10 + 14080*x^9 + 622022*x^8 + 4485120*x^7 + 910312*x^6 + 13647104*x^5 + 65163612*x^4 + 3943424*x^3 + 46235608*x^2 + 134216960*x - 1050623
s=16*x^15 + 176*x^13 + 5248*x^12 + 400*x^11 + 30976*x^10 + 433584*x^9 - 4224*x^8 + 343472*x^7 + 486912*x^6 - 392816*x^5 - 4060288*x^4 - 16662352*x^3 + 1313024*x^2 + 8413200*x + 33685632

Answer 1

I checked the proof of Granville. The proof only yields the bound $\max \{ \deg(r),\deg(s)\}(\deg(G)-2)\}+1$ which covers your counterexample.

To be more detailed: The polynomials $r,s$ yield a morphism $\mathbb{P}^1\to \mathbb{P}^1$. The set $G=0$ consists of $\deg(G)$ points. The set of $\alpha$ with $G(r(\alpha),s(\alpha)=0$ is the set of points in the preimage of this set with finite $t$-coordinate. The complete set consists of at least $\max \{ \deg(r),\deg(s)\}(\deg(G)-2)\}+2$ points, but one of them may be the point at infinity and hence you have to subtract one. This is precisely what is happening in your example.

Answer 2

Thanks to Remke for telling me about this. The proposition should have read

$$ \#\{\alpha \in \mathbb{C}\cup \{ \infty\}: G(r(\alpha),s(\alpha))=0\} \ge \max\{\deg(r),\deg(s)\}(\deg(G)-2) + 2. $$

ie you have to include the possible "root at infinity" (appropriately interpreted). However, it is probably easiest to change the statement of the result the way Remke suggests, so as to avoid worrying about the correct interpretation of a root at infinity. Sorry for the confusion.

It is a challenge to construct such examples in order to obtain equality for other $G$ (or even arbitrary $G$).