Proof.
The functors $\mathrm{Hilb}^ d_{X/k}$ are representable by Proposition 44.2.6 (see also Remark 44.2.7) and the fact that $X$ is projective (Varieties, Lemma 33.43.4). The schemes $\underline{\mathrm{Hilb}}^ d_{X/k}$ are separated over $k$ by Lemma 44.2.4. The schemes $\underline{\mathrm{Hilb}}^ d_{X/k}$ are smooth over $k$ by Lemma 44.3.5. Starting with $X = \underline{\mathrm{Hilb}}^1_{X/k}$, the morphisms of Lemma 44.3.4, and induction we find a morphism
\[ X^ d = X \times _ k X \times _ k \ldots \times _ k X \longrightarrow \underline{\mathrm{Hilb}}^ d_{X/k},\quad (x_1, \ldots , x_ d) \longrightarrow x_1 + \ldots + x_ d \]
which is finite locally free of degree $d!$. Since $X$ is proper over $k$, so is $X^ d$, hence $\underline{\mathrm{Hilb}}^ d_{X/k}$ is proper over $k$ by Morphisms, Lemma 29.41.9. Since $X$ is geometrically irreducible over $k$, the product $X^ d$ is irreducible (Varieties, Lemma 33.8.4) hence the image is irreducible (in fact geometrically irreducible). This proves (1). Part (2) follows from the definitions. Part (3) follows from the commutative diagram
\[ \xymatrix{ X^{d_1} \times _ k X^{d_2} \ar[d] \ar@{=}[r] & X^{d_1 + d_2} \ar[d] \\ \underline{\mathrm{Hilb}}^{d_1}_{X/k} \times _ k \underline{\mathrm{Hilb}}^{d_2}_{X/k} \ar[r] & \underline{\mathrm{Hilb}}^{d_1 + d_2}_{X/k} } \]
and multiplicativity of degrees of finite locally free morphisms.
$\square$
Comments (0)