Lemma 44.2.4. Let $X \to S$ be a morphism of schemes. If $X \to S$ is separated and $\mathrm{Hilb}^ d_{X/S}$ is representable, then $\underline{\mathrm{Hilb}}^ d_{X/S} \to S$ is separated.

**Proof.**
In this proof all unadorned products are over $S$. Let $H = \underline{\mathrm{Hilb}}^ d_{X/S}$ and let $Z \in \mathrm{Hilb}^ d_{X/S}(H)$ be the universal object. Consider the two objects $Z_1, Z_2 \in \mathrm{Hilb}^ d_{X/S}(H \times H)$ we get by pulling back $Z$ by the two projections $H \times H \to H$. Then $Z_1 = Z \times H \subset X_{H \times H}$ and $Z_2 = H \times Z \subset X_{H \times H}$. Since $H$ represents the functor $\mathrm{Hilb}^ d_{X/S}$, the diagonal morphism $\Delta : H \to H \times H$ has the following universal property: A morphism of schemes $T \to H \times H$ factors through $\Delta $ if and only if $Z_{1, T} = Z_{2, T}$ as elements of $\mathrm{Hilb}^ d_{X/S}(T)$. Set $Z = Z_1 \times _{X_{H \times H}} Z_2$. Then we see that $T \to H \times H$ factors through $\Delta $ if and only if the morphisms $Z_ T \to Z_{1, T}$ and $Z_ T \to Z_{2, T}$ are isomorphisms. It follows from the very general More on Flatness, Lemma 38.23.4 that $\Delta $ is a closed immersion. In the proof of Lemma 44.2.3 the reader finds an alternative easier proof of the needed result in our special case.
$\square$

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