ag.algebraic geometry - difference between equivalence relations on algebraic cycles

I will focus on complex projective varieties.

Codimension one

The situation in codimension one is considerably simpler than in higher codimensions.
Codimension one rational equivalence classes are parametrized by $Pic(X)= H^1(X,mathcal O_X^{ast})$ while algebraic equivalence classes are parametrized by the Neron-Severi group of $X$, which can be defined as the image of the Chern class map from $Pic(X)$ to $H^2(X,mathbb Z)$. It follows that in codimension one

• the group of rational equivalence classes is a countable union of abelian varieties;


• the groups of algebraic equivalence classes and homological equivalence classes coincide, and are equal to $NS(X)$ a subgroup of $H^2(X,mathbb Z)$;


• the group of numerical equivalence classes is the quotient of $NS(X)$ by its torsion subgroup.

Higher codimension

The higher codimension case, as pointed out by Tony Pantev, is considerably more complicate and algebraic and homological equivalence no longer coincide.

Concerning rational equivalence, Mumford proved that the Chow group of zero cycles of surfaces admitting non-zero holomorphic $2$-forms are infinite dimensional, contradicting a conjecture by Severi. The paper is Mumford, D. Rational equivalence of $0$-cycles on surfaces. J. Math. Kyoto Univ. 9 1968.

Warning

The definitions of rational and algebraic equivalence at wikipedia are not correct.
I will commment below on the algebraic equivalence.

There one can find the following definition.

$Z ∼_{alg} Z'$ if there exists a curve $C$ and a
cycle $V$ on $X × C$ flat over C, such
that $$V cap left( X timeslbrace crbrace right) = Z quad text{ and }
quad V cap left( X timeslbrace crbrace right) = Z' $$
for two points $c$ and $d$ on the
curve.

This is not correct. The correct definition is

$Z ∼_{alg} Z'$ if there exists a curve $C$ and a
cycle $V$ on $X × C$ flat over C, such
that $$V cap left( X timeslbrace crbrace right) - V cap left( X timeslbrace drbrace right) = Z - Z' $$
for two points $c$ and $d$ on the
curve.

To construct an example of two algebraically equivalent divisors which do not satisfy the wikipedia definition let $X$ be a projective variety with $H^1(X,mathcal O_X) neq 0$ and
take a non-trivial line-bundle $mathcal L$ over $X$ with zero Chern class.
If $Y = mathbb P ( mathcal O_X oplus mathcal L)$ then $Y$ contains two copies $X_0$ and $X_{infty}$ of $X$ ( one for each factor of $mathcal O_X oplus mathcal L$ ) which are algebraically equivalent but can't be deformed because their normal bundles are $mathcal L$ and $mathcal L^{ast}$. This does not contradict the second definition because for sufficiently ample divisors $H$ it is clear $X_0 + H$ can be deformed into $X_{infty} + H$.