Lemma 109.5.1. The diagonal of $\mathcal{C}\! \mathit{urves}$ is separated and of finite presentation.
Proof. Recall that $\mathcal{C}\! \mathit{urves}$ is a limit preserving algebraic stack, see Quot, Lemma 99.15.6. By Limits of Stacks, Lemma 102.3.6 this implies that $\Delta : \mathcal{P}\! \mathit{olarized}\to \mathcal{P}\! \mathit{olarized}\times \mathcal{P}\! \mathit{olarized}$ is limit preserving. Hence $\Delta $ is locally of finite presentation by Limits of Stacks, Proposition 102.3.8.
Let us prove that $\Delta $ is separated. To see this, it suffices to show that given a scheme $U$ and two objects $Y \to U$ and $X \to U$ of $\mathcal{C}\! \mathit{urves}$ over $U$, the algebraic space
\[ \mathit{Isom}_ U(Y, X) \]
is separated. This we have seen in Moduli Stacks, Lemmas 108.10.2 and 108.10.3 that the target is a separated algebraic space.
To finish the proof we show that $\Delta $ is quasi-compact. Since $\Delta $ is representable by algebraic spaces, it suffices to check the base change of $\Delta $ by a surjective smooth morphism $U \to \mathcal{C}\! \mathit{urves}\times \mathcal{C}\! \mathit{urves}$ is quasi-compact (see for example Properties of Stacks, Lemma 100.3.3). We choose $U = \coprod U_ i$ to be a disjoint union of affine opens with a surjective smooth morphism
\[ U \longrightarrow \textit{PolarizedCurves} \times \textit{PolarizedCurves} \]
Then $U \to \mathcal{C}\! \mathit{urves}\times \mathcal{C}\! \mathit{urves}$ will be surjective and smooth since $\textit{PolarizedCurves} \to \mathcal{C}\! \mathit{urves}$ is surjective and smooth by Lemma 109.4.2. Since $\textit{PolarizedCurves}$ is limit preserving (by Artin's Axioms, Lemma 98.11.2 and Quot, Lemmas 99.15.6, 99.14.8, and 99.13.6), we see that $\textit{PolarizedCurves} \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is locally of finite presentation, hence $U_ i \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is locally of finite presentation (Limits of Stacks, Proposition 102.3.8 and Morphisms of Stacks, Lemmas 101.27.2 and 101.33.5). In particular, $U_ i$ is Noetherian affine. This reduces us to the case discussed in the next paragraph.
In this paragraph, given a Noetherian affine scheme $U$ and two objects $(Y, \mathcal{N})$ and $(X, \mathcal{L})$ of $\textit{PolarizedCurves}$ over $U$, we show the algebraic space
\[ \mathit{Isom}_ U(Y, X) \]
is quasi-compact. Since the connected components of $U$ are open and closed we may replace $U$ by these. Thus we may and do assume $U$ is connected. Let $u \in U$ be a point. Let $Q$, $P$ be the Hilbert polynomials of these families, i.e.,
\[ Q(n) = \chi (Y_ u, \mathcal{N}_ u^{\otimes n}) \quad \text{and}\quad P(n) = \chi (X_ u, \mathcal{L}_ u^{\otimes n}) \]
see Varieties, Lemma 33.45.1. Since $U$ is connected and since the functions $u \mapsto \chi (Y_ u, \mathcal{N}_ u^{\otimes n})$ and $u \mapsto \chi (X_ u, \mathcal{L}_ u^{\otimes n})$ are locally constant (see Derived Categories of Schemes, Lemma 36.32.2) we see that we get the same Hilbert polynomial in every point of $U$. Set
\[ \mathcal{M} = \text{pr}_1^*\mathcal{N} \otimes _{\mathcal{O}_{Y \times _ U X}} \text{pr}_2^*\mathcal{L} \]
on $Y \times _ U X$. Given $(f, \varphi ) \in \mathit{Isom}_ U(Y, X)(T)$ for some scheme $T$ over $U$ then for every $t \in T$ we have
\begin{align*} \chi (Y_ t, (\text{id} \times f)^*\mathcal{M}^{\otimes n}) & = \chi (Y_ t, \mathcal{N}_ t^{\otimes n} \otimes _{\mathcal{O}_{Y_ t}} f_ t^*\mathcal{L}_ t^{\otimes n}) \\ & = n\deg (\mathcal{N}_ t) + n\deg (f_ t^*\mathcal{L}_ t) + \chi (Y_ t, \mathcal{O}_{Y_ t}) \\ & = Q(n) + n\deg (\mathcal{L}_ t) \\ & = Q(n) + P(n) - P(0) \end{align*}
by Riemann-Roch for proper curves, more precisely by Varieties, Definition 33.44.1 and Lemma 33.44.7 and the fact that $f_ t$ is an isomorphism. Setting $P'(t) = Q(t) + P(t) - P(0)$ we find
\[ \mathit{Isom}_ U(Y, X) = \mathit{Isom}_ U(Y, X) \cap \mathit{Mor}^{P', \mathcal{M}}_ U(Y, X) \]
The intersection is an intersection of open subspaces of $\mathit{Mor}_ U(Y, X)$, see Moduli Stacks, Lemma 108.10.3 and Remark 108.10.4. Now $\mathit{Mor}^{P', \mathcal{M}}_ U(Y, X)$ is a Noetherian algebraic space as it is of finite presentation over $U$ by Moduli Stacks, Lemma 108.10.5. Thus the intersection is a Noetherian algebraic space too and the proof is finished. $\square$
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