Likely a mistake, but got very large exceptional set in Vojta's more general abc conjecture.

In A more general abc conjecture, p. 7 Paul Vojta conjectures:

If $x_0,\ldots x_{n-1}$ are nonzero coprime integers satisfying $x_0 + \cdots x_{n-1}=0$

$$ \max\{|x_0|,\ldots |x_{n-1}|\} \le C \prod_{p\mid x_0 \cdots x_{n-1}}p^{1+\epsilon}\qquad (1) $$

for all $x_0 , \ldots, x_{n-1}$ as above outside a proper Zariski-closed subset.

Define the quality $q(x_0 \ldots x_{n-1})=\frac{\log(\max(|x_i|))}{\log\operatorname{rad}(x_0 \cdots x_{n-1})}$. The conjecture implies $\limsup q(x_0 \ldots x_{n-1})=1$ outside a proper Zariski-closed subset.

Consider the set of all genus $0$ curves parametrized by two quadratics. $x=a t^2+b t + c,y=d t^2+ e t +f$. Assume $a,b$ are nonzero and $|a|\ge |d$ and $x,y$ are coprime over $\mathbb{Q}[t]$.

The explicit form of the curve is

$$ d^{2} x^{2} + (-2 a d) x y + a^{2} y^{2} + (-2 c d^{2} + b d e - a e^{2} + 2 a d f) x + (- b^{2} d + 2 a c d + a b e - 2 a^{2} f) y + c^{2} d^{2} - b c d e + a c e^{2} + b^{2} d f - 2 a c d f - a b e f + a^{2} f^{2}=0 \qquad(1)$$

Fix $a,b,c,d,e,f$. Take $x_i$ the monomials in $x,y$ of (1), i.e. $x_0 =d^{2} x^{2} ,x_1= (-2 a d) x y$. For infinitely many of them we get quality $4/3-o(1)$ infinitely often for $x,y$ on (1).

$\max|x_i|=a^2 y^2 \sim a^2 d t^4$. The radical is $poly(a,b,c,d,e,f) x y \sim t^4$.

One can make the radical of the quadratic $x$ only $O(t)$ in at least two ways: (a) there exist integers $k$ such that $a t^2+b t + c=k r^2$ has infinitely many integral solutions and (b) find integer $p$ such that for arbitrary large $n, a t^2+b t + c \equiv 0 \pmod{p^n}$ via Hensel lifting. The gcd of $x_i$ is bounded by the resultants and clearing bounded gcd doesn't decrease the quality.

Both (a) and (b) make the radical only $O(t^3)$ for arbitrary large $t$, which gives quality $4/3-o(1)$ and we can get rid of the $o(1)$.

So (1) is in exceptional Vojta's set. By gross abuse of notation, "almost all" $a,b,c,d,e$ give quality $4/3-o(1)$ so we can treat them as variables in (1). This in general makes the dimension of (1) $7$, which is greater than the dimension of $x_0 + \cdots + x_5=0$.

I believe higher dimensional variety can't be subvariety of lower dimensional variety.

Why isn't this construction counterexample?

  • $\begingroup$ Your interpretation that $\limsup q = 1$ outside a Zariski closed subset is incorrect. For every $\epsilon > 0$ there is a Zariski closed subset but it depends on $\epsilon$. As $\epsilon \to 0$ the union of these Zariski closed subsets may well be Zariski dense. I haven't looked at your example to see if in fact that's what you are getting. $\endgroup$ Dec 27, 2014 at 11:45
  • $\begingroup$ @FelipeVoloch Are you sure? I am taking logs(), which kills all epsilons. If in abc you get limsup bigger than one, this will certainly disprove abc. $\endgroup$
    – joro
    Dec 27, 2014 at 12:01
  • $\begingroup$ @FelipeVoloch from the abc records database: math.leidenuniv.nl/~desmit/abc "The ABC conjecture says that the limsup of the quality when we range over all ABC triples, is 1" $\endgroup$
    – joro
    Dec 27, 2014 at 12:03
  • $\begingroup$ $q < 1 + \epsilon$ outside of $Z_{\epsilon}$ so $q \le 1$ outside of the union of all $Z_{\epsilon}$. The original ABC which I guess is $n=3$ in your notation doesn't have an excepcional set, you only need it for $n>3$. Meanwhile Michael answered your question. $\endgroup$ Dec 27, 2014 at 18:00
  • $\begingroup$ @FelipeVoloch Indeed the question is answered. If you disagree with limsup, check the argument about t^4 vs t^(3+o(1)) which I believe still stands. $\endgroup$
    – joro
    Dec 27, 2014 at 18:23

1 Answer 1


Your construction is not a counterexample because your tuples $(x_0,\ldots,x_5)$ (obviously) satisfy the relation $x_1^2 - 4x_0x_2 = 0$, so they all lie in the proper Zariski closed subset defined by this equation.


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