Section on $\mathcal{X}(1)$ induced by a cusp of $X(1)$

Question

I'm reading the article Generalized arithmetic intersection numbers of U. Kuehn. At the beginning of section 4.12 "Modular forms over $\mathbb{Z}$" we are in the following situation.

Let $\Gamma(1) \simeq \mathrm{SL}_2(\mathbb{Z})$ be the full modular group, we consider the complex modular curve $X(\Gamma(1)):= \overline{\Gamma(1)\backslash \mathbb{H}} \simeq \mathbb{P}^1_{\mathbb{C}}$ and its model $\mathcal{X}(1)\simeq \mathbb{P}^1_\mathbb{Z}$ defined over $\mathrm{Spec}(\mathbb{Z})$. Then the author claims that the cups $\mathrm{S}_{i \infty} \in X(\Gamma(1))$ induces a section $s_\infty : \mathrm{Spec}(\mathbb{Z}) \rightarrow \mathcal{X}(\Gamma(1))$.

How does this induction work? In specific, is it true for general congruence subgroups $\Gamma$ that a cusp $S_c \in X(\Gamma)$ induces a section $s_c : \mathrm{Spec}(\mathcal{O}_E) \rightarrow \mathcal{X}(\Gamma)$? (Where $\mathcal{O}_E$ is the ring of integers of the number field $E$ of minimal degree over which a model of $X(\Gamma)$ exists.)

Thank you in advance!

Answer 1

This answer is just an expansion of the hints given by Damian Rössler and David Loeffler in the comments. If I made any mistake please point it out!

In the notation of the question we have that $X(\Gamma(1))$ has a rational smooth, connected and compact model $X(\Gamma(1))_\mathbb{Q}$. By Belyi's Theorem this is true, replacing $\mathbb{Q}$ by a suitable number field $E(\Gamma)$, for any generalized modular curve $X(\Gamma)$.

Now the cusp $S_{i \infty} \in X(1)(\mathbb{C})= X(1)_\mathbb{Q}(\mathbb{C})$ induces a rational point $S_{i \infty, \mathbb{Q}} \in X(1)_\mathbb{Q}(\mathbb{Q})$. Indeed the objects parametrized by $X(1)$ close to the cusp $S_{i \infty}$ are elliptic curves of the form $\mathbb{C}^*/q^{\mathbb{Z}}$ for $|q|<1$, and the object parametrized by the cusp corresponds to $q=0$ (which is not an elliptic curve). Explicit Weierstrass equations can be given for those elliptic curves, and for $q=0$ the equation becomes $y^2 +xy =x^3$, which is a rational curve. As David pointed out this is where things get messy with arbitrary modular curves.

Finally we have a rational point $S_{i \infty}: \mathrm{Spec}(\mathbb{Q})\rightarrow X(1)_\mathbb{Q}$, which can be composed with the inclusion map $X(1)_\mathbb{Q} \hookrightarrow \mathcal{X}(1)$ to give a rational point on the latter space. We also observe that the natural map $\mathcal{X}(1)\simeq \mathbb{P}^1_\mathbb{Z} \rightarrow \mathrm{Spec}(\mathbb{Z})$ is proper. Applying the valuative criterion of properness, as suggested by Damian, this gives a section $s_\infty : \mathrm{Spec}(\mathbb{Z}) \rightarrow \mathcal{X}(1)$.