# IMO 2017/6 via arithmetic geometry

The (very nice) final problem of IMO 2017 asked contestants to show:

If $S$ is a finite set of lattice points $(x,y)$ with $\gcd(x,y)=1$, then there is a nonconstant homogeneous polyonmial $f \in \mathbb Z[x,y]$ such that $f(x,y) = 1$ for all $(x,y) \in S$.

It's claimed in this forum post that the above IMO problem is a special case of Lemma 7.3 of arXiv:16040.01704. The former post phrases the lemma as follows:

If $X$ is a finite scheme over $\operatorname{Spec} \mathbb Z$ then $\operatorname{Pic}(X)$ is finite.

Being unknowledgable as I am, I do not see how to deduce the IMO problem from the lemma. Can someone make the connection explicit?

The set $S$ gives rise to a subscheme (which let's also denote by $S$) of $\mathbb{P}^1_{\mathbb{Z}},$ because relatively a prime pair $(x,y)$ corresponds to a section of $\mathbb{P}^1_{\mathbb{Z}}\rightarrow\operatorname{Spec}\mathbb{Z}$, and we take the union of these divisors in $\mathbb{P}^1_\mathbb{Z}$.
Now, we have a map $\mathcal{O}(d)_{\mathbb{P}^1}\rightarrow\mathcal{O}(d)_S,$ and it will suffice to show that for some $d$ some element in the image of the induced map on global sections is nowhere zero. (If one composes this map with the map $\mathcal{O}(d)_S\rightarrow\mathcal{O}(d)_{\operatorname{Spec}\mathbb{Z}}$ corresponding to a section $(x,y)$, one gets at the level of global sections the evaluation map $\mathbb{Z}[x,y]_d\rightarrow\mathbb{Z}$ on degree $d$ homogeneous polynomials.)
For this, it suffices that one can find $d$ so that this map is surjective on global sections and so that $\mathcal{O}(d)_S$ is trivial. The map will be surjective on global sections for large enough $d$ by ampleness of $\mathcal{O}(1)$, and the triviality of a power of $\mathcal{O}(1)_S$ follows from the finiteness of $\operatorname{Pic}(S)$.
• Here the subscheme $S$ is a finite coproduct of copies of $\mathrm{Spec}(\mathbb{Z})$, so its Picard group is actually trivial (so lemma $7.3$ is not needed here). – js21 Jul 26 '17 at 6:35
• @js21: Not so! Every irreducible component of $S$ is a copy of $\operatorname{Spec}{\mathbb{Z}}$, but they may be joined at some finite primes. As I see it, this is the essential source of difficulty for the elementary solutions to this problem. – dhy Jul 26 '17 at 6:56
• This produces an $f$ with $f(x,y) \in \mathbb Z^{\times} = \{\pm 1\}$ for all $(x,y) \in S$. To get $f(x,y) = 1$, take the square. – R. van Dobben de Bruyn Jul 26 '17 at 11:31