The Stacks project

Proof. Equivalence of (2) and (3). This follows from Divisors, Lemma 31.18.9 if we can show the equivalence of (2) and (3) when $S$ is the spectrum of a field $k$. Let $x \in X$ be a closed point. As $X$ is smooth of relative dimension $1$ over $k$ and we see that $\mathcal{O}_{X, x}$ is a regular local ring of dimension $1$ (see Varieties, Lemma 33.25.3). Thus $\mathcal{O}_{X, x}$ is a discrete valuation ring (Algebra, Lemma 10.119.7) and hence a PID. It follows that every sheaf of ideals $\mathcal{I} \subset \mathcal{O}_ X$ which is nonvanishing at all the generic points of $X$ is invertible (Divisors, Lemma 31.15.2). In other words, every closed subscheme of $X$ which does not contain a generic point is an effective Cartier divisor. It follows that (2) and (3) are equivalent.

If $S$ is Noetherian, then any locally quasi-finite morphism $D \to S$ is locally of finite presentation (Morphisms, Lemma 29.21.9), whence (3) is equivalent to (4).

If $X \to S$ is proper (and $S$ is arbitrary), then $D \to S$ is proper as well. Since a proper locally quasi-finite morphism is finite (More on Morphisms, Lemma 37.44.1) and a finite, flat, and finitely presented morphism is finite locally free (Morphisms, Lemma 29.48.2), we see that (1) is equivalent to (2). $\square$

Proof. By Lemma 44.3.1 we see that $D_1$ and $D_2$ are relative effective Cartier divisors on $X/S$. Thus $D = D_1 + D_2$ is a relative effective Cartier divisor on $X/S$ by Divisors, Lemma 31.18.3. Hence $D \to S$ is locally quasi-finite, flat, and locally of finite presentation by Lemma 44.3.1. Applying Morphisms, Lemma 29.41.11 the surjective integral morphism $D_1 \amalg D_2 \to D$ we find that $D \to S$ is separated. Then Morphisms, Lemma 29.41.9 implies that $D \to S$ is proper. This implies that $D \to S$ is finite (More on Morphisms, Lemma 37.44.1) and in turn we see that $D \to S$ is finite locally free (Morphisms, Lemma 29.48.2). Thus it suffice to show that the degree of $D \to S$ is $d_1 + d_2$. To do this we may base change to a fibre of $X \to S$, hence we may assume that $S = \mathop{\mathrm{Spec}}(k)$ for some field $k$. In this case, there exists a finite set of closed points $x_1, \ldots , x_ n \in X$ such that $D_1$ and $D_2$ are supported on $\{ x_1, \ldots , x_ n\} $. In fact, there are nonzerodivisors $f_{i, j} \in \mathcal{O}_{X, x_ i}$ such that

Then we see that

From this one sees easily that $D$ has degree $d_1 + d_2$ over $k$ (if need be, use Algebra, Lemma 10.121.1). $\square$

Proof. This proof is almost exactly the same as the proof of Lemma 44.3.2. By Lemma 44.3.1 we see that $D_1$ and $D_2$ are relative effective Cartier divisors on $X/S$. By Divisors, Lemma 31.18.4 there is a relative effective Cartier divisor $D \subset X$ such that $D_2 = D_1 + D$. Hence $D \to S$ is locally quasi-finite, flat, and locally of finite presentation by Lemma 44.3.1. Since $D$ is a closed subscheme of $D_2$, we see that $D \to S$ is finite. It follows that $D \to S$ is finite locally free (Morphisms, Lemma 29.48.2). Thus it suffice to show that the degree of $D \to S$ is $d_2 - d_1$. This follows from Lemma 44.3.2. $\square$

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

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. $\square$

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). $\square$

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

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

and multiplicativity of degrees of finite locally free morphisms. $\square$