Lemma 44.3.4. Let $X \to S$ be a smooth morphism of schemes of relative dimension $1$ such that the functors $\mathrm{Hilb}^ d_{X/S}$ are representable. The morphism $\underline{\mathrm{Hilb}}^ d_{X/S} \times _ S X \to \underline{\mathrm{Hilb}}^{d + 1}_{X/S}$ is finite locally free of degree $d + 1$.

Proof. Let $D_{univ} \subset X \times _ S \underline{\mathrm{Hilb}}^{d + 1}_{X/S}$ be the universal object. There is a commutative diagram

$\xymatrix{ \underline{\mathrm{Hilb}}^ d_{X/S} \times _ S X \ar[rr] \ar[rd] & & D_{univ} \ar[ld] \ar@{^{(}->}[r] & \underline{\mathrm{Hilb}}^{d + 1}_{X/S} \times _ S X \\ & \underline{\mathrm{Hilb}}^{d + 1}_{X/S} }$

where the top horizontal arrow maps $(D', x)$ to $(D' + x, x)$. We claim this morphism is an isomorphism which certainly proves the lemma. Namely, given a scheme $T$ over $S$, a $T$-valued point $\xi$ of $D_{univ}$ is given by a pair $\xi = (D, x)$ where $D \subset X_ T$ is a closed subscheme finite locally free of degree $d + 1$ over $T$ and $x : T \to X$ is a morphism whose graph $x : T \to X_ T$ factors through $D$. Then by Lemma 44.3.3 we can write $D = D' + x$ for some $D' \subset X_ T$ finite locally free of degree $d$ over $T$. Sending $\xi = (D, x)$ to the pair $(D', x)$ is the desired inverse. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).