For the definitions of the equivalence relations on algebraic cycles see <http://en.wikipedia.org/wiki/Adequate_equivalence_relation>.


I want to know how far away from each other the equivalence relations on algebraic cycles are and what the intuition is for them.


My impression is that rational equivalence gives much bigger Chow groups than algebraic equivalence, and that algebraic equivalence, homological equivalence and numerical equivalence are quite tight together.


Take for example an elliptic curve. We have $CH^1(E) = \mathbb{Z} \times E(K)$, algebraic equivalence (take $C = E$) $\mathbb{Z}$ = numerical equivalence.



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](http://en.wikipedia.org/wiki/Adequate_equivalence_relation) 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 $$V \cap \left( X \times\lbrace c\rbrace \right) = Z \quad \text{ and } 
> \quad V \cap \left( X \times\lbrace c\rbrace \right) = Z' $$
>  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 $$V \cap \left( X \times\lbrace c\rbrace \right) - V \cap \left( X \times\lbrace d\rbrace \right) = Z - Z' $$
>  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 $X\_{\infty} + H$.


