Lemma 44.3.5. 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 schemes $\underline{\mathrm{Hilb}}^ d_{X/S}$ are smooth over $S$ of relative dimension $d$.

Proof. We have $\underline{\mathrm{Hilb}}^0_{X/S} = S$ and $\underline{\mathrm{Hilb}}^1_{X/S} = X$ thus the result is true for $d = 0, 1$. Assuming the result for $d$, we see that $\underline{\mathrm{Hilb}}^ d_{X/S} \times _ S X$ is smooth over $S$ (Morphisms, Lemma 29.34.5 and 29.34.4). Since $\underline{\mathrm{Hilb}}^ d_{X/S} \times _ S X \to \underline{\mathrm{Hilb}}^{d + 1}_{X/S}$ is finite locally free of degree $d + 1$ by Lemma 44.3.4 the result follows from Descent, Lemma 35.14.5. We omit the verification that the relative dimension is as claimed (you can do this by looking at fibres, or by keeping track of the dimensions in the argument above). $\square$

## Comments (3)

Comment #4950 by SDIGR on

First sentence of the Proof: $\underline{\mathrm{Hilb}_{X/S}^0}=S\mapsto\underline{\mathrm{Hilb}_{X/S}^d}=S$.

Comment #4951 by SDIGR on

Sorry, I mean the other way around.

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