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Write $D = D_1 - D_2$ with $D_1, D_2$ effective. By Riemann–Roch, we have $$\chi(X,\mathcal O(D_i)) = \int_X \operatorname{ch}(\mathcal O(D_i)) \cdot \operatorname{td}(T_X).$$ The point is that the right hand side is defined purely in terms of intrinsic geometry of $X$ and the intersection behaviour of $D_i$. Indeed, for a line bundle $\mathcal L$, we have $$\operatorname{ch}(\mathcal L) = 1 + c_1(\mathcal L) + \frac{c_1(\mathcal L)^2}{2} + \ldots + \frac{c_1(\mathcal L)^n}{n!};$$ we cup it with some fixed thing $\operatorname{td}(T_X)$, and we integrate (i.e. take the degree of the component in dimension $0$). This only depends on the intersection behaviour of $c_1(\mathcal L)$. We conclude that $$\chi(X,\mathcal O(D_1)) = \chi(X,\mathcal O(D_2)),$$ since the intersections of $D_1$ and $D_2$ with any curve agree.

If $H$ is ample, replacing $D_i$ by $-D_i + nH$ and using additivity in exact sequences, we similarly get $$\chi(X,\mathcal O_{D_1}(nH)) = \chi(X,\mathcal O_{D_2}(nH)),$$ so the Hilbert polynomials of $D_1$ and $D_2$ agree.

When we fix the Hilbert polynomial, the Hilbert scheme is projective. In particular, it has finitely many components, so a multiple of $D_i$ has to land in the identity component $\operatorname{Pic}^0_{X/k}$. Choose $n$ such that $nD_i \in \operatorname{Pic}^0_{X/k}$ for $i \in \{1,2\}$ (it actually suffices that they land in the same component; not necessarily the identity component). Choosing a curve connecting these two points in $\operatorname{Pic}^0_{X/k}$ gives a family of effective divisors from $nD_1$ to $nD_2$, showing that $nD \sim_{\text{alg}} 0$. $\square$

By Riemann–Roch, we have $$\chi(X,\mathcal O(D)) = \int_X \operatorname{ch}(\mathcal O(D)) \cdot \operatorname{td}(T_X),$$ and similarly for $\mathcal O$. The point is that the right hand side is defined purely in terms of intrinsic geometry of $X$ and the intersection behaviour of $D$. Indeed, for a line bundle $\mathcal L$, we have $$\operatorname{ch}(\mathcal L) = 1 + c_1(\mathcal L) + \frac{c_1(\mathcal L)^2}{2} + \ldots + \frac{c_1(\mathcal L)^n}{n!};$$ we cup it with some fixed thing $\operatorname{td}(T_X)$, and we integrate (i.e. take the degree of the component in dimension $0$). This only depends on the intersection behaviour of $c_1(\mathcal L)$. We conclude that $$\chi(X,\mathcal O(D)) = \chi(X,\mathcal O),$$ since the intersections of $D$ and $0$ with any curve agree.

Edit: An argument is missing here; I thought I fixed it but that created a different gap. In order to deploy the theory of Quot schemes, we have to write $\mathcal O$ and $\mathcal O(D)$ as quotient of the same vector bundle. For this, some boundedness argument is needed. This is carried out in characteristic $0$ in Lazarsfeld's Positivity in Algebraic Geometry I, Prop 1.4.37 (using Fujita's vanishing theorem), or in general in FGA Explained, Lemma 9.6.6 (using a variant of Mumford's argument for boundedness of the Hilbert scheme, cf. Lectures on curves on an algebraic surface, Lecture 14, Theorem).

Now by the theory of Quot schemes, when we fix the Hilbert polynomial, the Quot scheme is projective. In particular, it has finitely many components, so a multiple of $\mathcal O(D)$ has to land in the identity component $\operatorname{Pic}^0_{X/k}$.

Now write $D = D_1 - D_2$ with $D_1, D_2$ effective; then $D_1$ and $D_2$ lie in the same component of the Picard scheme. Choosing a curve connecting these two points in $\operatorname{Pic}^0_{X/k}$ gives a family of effective divisors from $nD_1$ to $nD_2$, showing that $nD \sim_{\text{alg}} 0$. $\square$

Note that \begin{align*} h(w,v) &= \frac{1}{4 \pi^2 i} w \cup \bar v \cup H^{n-1} \\ &= \frac{1}{4 \pi^2 i} \cdot (-1) \bar v \cup w \cup H^{n-1} \\ &= \overline{h(v,w)}, \end{align*} since the cup product is graded commutative. Thus $h$ is hermitian. Moreover, for $v, w \in H^1(X,\mathbb Z)$, we have \begin{align*} h(v,w) &= \frac{1}{4\pi^2 i} \cdot (2\pi i v) \cup (-2\pi i w) \cup H^{n-1}\\ &= -i \cdot v \cup w \cup H^{n-1}. \end{align*} This is a purely imaginary algebraic integer, since $v \cup w \cup H^{n-1}$ is an integer.

Construct the Picard scheme $\operatorname{Pic}_{C\mathbb C}$ (a lot of work), and prove that $\operatorname{Pic}^0_{X/\mathbb C}$ is proper. Hence, it is projective (some general result about abelian varieties), so it is a complex torus. Prove that it coincides with the complex torus $H^(X,\mathcal O_X)/H^1(X,\mathbb Z)$. $\suqare$

Note that \begin{align*} h(w,v) &= \frac{1}{4 \pi^2 i} \cdot w \cup \bar v \cup H^{n-1} \\ &= \frac{1}{4 \pi^2 i} \cdot (-1) \cdot \bar v \cup w \cup H^{n-1} = \overline{h(v,w)}, \end{align*} since the cup product is graded commutative. Thus $h$ is hermitian. Moreover, for $v, w \in H^1(X,\mathbb Z)$, we have \begin{align*} h(v,w) &= \frac{1}{4\pi^2 i} \cdot (2\pi i v) \cup (-2\pi i w) \cup H^{n-1}\\ &= -i \cdot v \cup w \cup H^{n-1}. \end{align*} This is a purely imaginary algebraic integer, since $v \cup w \cup H^{n-1}$ is an integer.

Construct the Picard scheme $\operatorname{Pic}_{X/\mathbb C}$ (a lot of work!), and prove that $\operatorname{Pic}^0_{X/\mathbb C}$ is projective, so it is a complex torus. Prove that it coincides with the complex torus $H^1(X,\mathcal O_X)/H^1(X,\mathbb Z)$. $\square$

Remark. In the first proof, we need that the complex torus $\operatorname{Pic}^0_{X/\mathbb C} = H^1(X,\mathcal O_X)/H^1(X,\mathbb Z)$ is actually algebraic when $X$ is smooth projective.

Proof 1 (analytic).

Recall that a complex torus $V/U$ is projective (hence algebraic) if and only if there is a positive definite hermitian metric $h$ on $V$ such that $\operatorname{im} h$ is integral on $U \times U$ (see for example Mumford's book on abelian varieties, p. 33).

To construct such a form for $V = H^1(X,\mathcal O_X)$ and $U = H^1(X,\mathbb Z)$, let $H$ be a hyperplane class, and set \begin{align*} h \colon V \times V &\to \mathbb C\\ (v,w) &\mapsto \frac{1}{4\pi^2 i} \cdot v \cup \bar w \cup H^{n-1}. \end{align*} Here, we view elements in $H^1(X,\mathcal O_X)$ as living in $H^1(X,\mathbb C) = H^1(X, \mathcal O_X) \oplus H^0(X,\Omega^1_{X/\mathbb C})$. Remember that $H^1(X,\mathbb Z)$ for our purposes maps to $H^1(X,\mathbb C)$ by multiplication by $2 \pi i$; this is different from the natural inclusion $H^1(X, \mathbb Z) \subseteq H^1(X, \mathbb C)$.

Note that \begin{align*} h(w,v) &= \frac{1}{4 \pi^2 i} w \cup \bar v \cup H^{n-1} \\ &= \frac{1}{4 \pi^2 i} \cdot (-1) \bar v \cup w \cup H^{n-1} \\ &= \overline{h(v,w)}, \end{align*} since the cup product is graded commutative. Thus $h$ is hermitian. Moreover, for $v, w \in H^1(X,\mathbb Z)$, we have \begin{align*} h(v,w) &= \frac{1}{4\pi^2 i} \cdot (2\pi i v) \cup (-2\pi i w) \cup H^{n-1}\\ &= -i \cdot v \cup w \cup H^{n-1}. \end{align*} This is a purely imaginary algebraic integer, since $v \cup w \cup H^{n-1}$ is an integer.

Finally, we need to check that $h$ is positive definite. This is an immediate consequence of the Hodge index theorem. $\square$

Proof 2 (algebraic, sketch).

Construct the Picard scheme $\operatorname{Pic}_{C\mathbb C}$ (a lot of work), and prove that $\operatorname{Pic}^0_{X/\mathbb C}$ is proper. Hence, it is projective (some general result about abelian varieties), so it is a complex torus. Prove that it coincides with the complex torus $H^(X,\mathcal O_X)/H^1(X,\mathbb Z)$. $\suqare$