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 $\operatorname{Pic}^0_{X/\mathbb C}$.

 Write $D = D_1 - D_2$ as a difference of effective divisors. Then $D_1$ and $D_2$ live in the same component of $\operatorname{Pic}_{X/\mathbb C}$. Choose a curve connecting them. The universal family restricted to this curve is now a family of effective divisors from $D_1$ to $D_2$, showing that $D \sim_{\text{alg}} 0$. $\square$

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

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 $nD \sim_{\text{alg}} 0$. $\square$

**Remark.** If you want to see this Riemann–Roch thing in action: just look at the formula of the Euler characteristic of a line bundle on a surface. The point is that the formula only depends on *certain* intersection numbers; we don't even need to know *which* intersection numbers.