Super vast generalisation: for divisors on a smooth projective variety over an algebraically closed field of *any* characteristic, the notions of algebraic, homological (for any Weil cohomology theory), and numerical equivalence agree (up to torsion).

**Remark.** Recall: the group of cycles $\alpha \sim_{\text{alg}} 0$ is generated by cycles of the form
$$\pi_{1,*} (\pi_2^* ([t_1] - [t_2]))$$
for $C$ a curve, $t_1, t_2 \in C$, and $V \subseteq X \times C$ a subvariety flat over $C$, with projections $\pi_1 \colon V \to X$, $\pi_2 \colon V \to C$. To show that $\operatorname{cl}(\alpha) = 0$, one reduces to the case $[t_1] - [t_2]$ on a curve.

On the other hand, if $\operatorname{cl}(\alpha) = 0$, then compatibility between the intersection product and the cup product shows that $\alpha \sim_{\text{num}} 0$.

**Lemma.** If $D$ is a divisor with $D \sim_{\text{num}} 0$, then $nD \sim_{\text{alg}} 0$ for some $n \in \mathbb Z_{>0}$.

**Remark.** If $k = \mathbb C$ and we use singular cohomology, then we can actually say a little bit more:

 1. If $D \sim_{\text{num}} 0$, then $nD \sim_{\text{hom}} 0$ for some $n$,
 2. If $D \sim_{\text{hom}} 0$ (integrally), then $D \sim_{\text{alg}} 0$.

Fulton's Lemma 19.3.1 states both, but only proves the second assertion (it seems). I will prove both.

**Remark.** Since in general we don't have access to an integral Weil cohomology theory (the best we can do is $\mathbb Z_\ell$), there is no analogue of the second statement for other Weil cohomology theories.

*Proof 1.* (Over $k = \mathbb C$ for singular cohomology) 

 1. Let $H$ be an ample class, and recall that $\operatorname{NS}(X)_\mathbb Q$ is the intersection $H^{1,1}(X) \cap H^2(X,\mathbb Q)$, by the Lefschetz $(1,1)$-theorem. By hard Lefschetz and Poincaré duality (and since $- \cup H$ is compatible with both the Hodge decomposition and the $\mathbb Q$-structure), the pairing
\begin{align*}
\operatorname{NS}(X)_\mathbb Q \times \operatorname{NS}(X)_\mathbb Q &\to H^{2n}(X,\mathbb Q) \cong \mathbb Q\\
(\alpha, \beta) &\mapsto \alpha \cup H^{n-2} \cup \beta
\end{align*}
  is a perfect pairing. Hence, $\operatorname{cl}(D) = 0 \in H^2(X,\mathbb Q)$ if and only if $D \cdot C = 0$ for all curves $C$. (The summary is that hard Lefschetz and Lefschetz $(1,1)$ give us enough curves to cup with.)
 2. We have the exact sequence
$$0 \to H^1(X, \mathbb Z) \to H^1(X, \mathcal O_X) \to \operatorname{Pic}(X) \to H^2(X,\mathbb Z) \to H^2(X,\mathcal O_X) \to \ldots.$$
The map $\operatorname{Pic}(X) \to H^2(X,\mathbb Z)$ is the Chern class map (which is the same thing as the cycle class map in this case). Thus, any element in the kernel will come from $H^1(X, \mathcal O_X)/H^1(X, \mathbb Z)$. But that's exactly the identity component $\operatorname{Pic}^0_{X/\mathbb C}$.

 Choose a curve connecting $D$ with the identity element of $\operatorname{Pic}^0_{X/\mathbb C}$. The universal family restricted to this curve is now a family of cycles from $D$ to the trivial divisor, showing that $D \sim_{\text{alg}} 0$. $\square$

*Proof 2.* (Over any (algebraically closed?) field for any Weil cohomology theory)

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. remember only the degree of the codimension $n$ component). 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 intersection of $D$ and $0$ with any curve are the same.

Now by the theory of Hilbert schemes, when we fix the Hilbert polynomial, the Hilbert scheme is projective. In particular, it has finitely many components, so a multiple of $\mathcal O(D)$ has to land in the identity component $\operatorname{Pic}^0_{X/k}$. Choosing a curve connecting this point with the origin in $\operatorname{Pic}^0_{X/k}$ gives a family of cycles from $nD$ to the trivial line bundle, showing that $nD \sim_{\text{alg}} 0$. $\square$