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The cycle class of an algebraic cycle is its image under the cycle class map. For a codimension-$r$ cycle on a smooth complex projective variety, it belongs to $H^{2r}(X,\mathbb{Z})$ and is the Poincare dual of its fundamental homology class. The map factors through the Chow group, but a Chow class retains finer information than its cohomological cycle class.

For a smooth complex projective algebraic variety, the cycle class map sends an algebraic cycle of codimension $r$ to the Poincare dual of its homology class. Rational equivalence gives the same cohomology class, so the map descends to the Chow group. It respects intersection products, pullback and proper pushforward; a cohomological obstruction therefore obstructs the corresponding proposed identity of Chow classes.

For smooth projective varieties $X,Y$, pull back Chow classes from the two factors and take their intersection product on $X\times Y$. This gives the external product and a graded map ${\mathrm{CH}}^{\ast }(X)\otimes {\mathrm{CH}}^{\ast }(Y)\to {\mathrm{CH}}^{\ast }(X\times Y)$. Unlike the cohomological Künneth theorem, it need not generate every class of the product.

Write $A=E\times \{p\}$, $B=\{p\}\times E$ and $D={\Delta }_E$. Their self-intersection numbers vanish: each factor curve has trivial normal bundle, and the self-intersection of the diagonal of a curve is $2-2g(E)=0$. Each pair meets transversely once. Their intersection pairing is thereforeIf a rational linear combination of their cycle classes vanished in $H^2(E\times E,\mathbb{Q})$, intersecting it with $A,B,D$ would multiply its coefficients by this invertible matrix and give zero. All coefficients must vanish, proving the three classes are linearly independent.

The cycle class map detects the Chow Künneth failure for an elliptic self-product. The codimension-one image of the external product of Chow classes consists of sums from ${\mathrm{CH}}^1(E)\otimes {\mathrm{CH}}^0(E)$ and ${\mathrm{CH}}^0(E)\otimes {\mathrm{CH}}^1(E)$. Here ${\mathrm{CH}}^0(E)=\mathbb{Z}[E]$, and every divisor's cohomology class is its degree of a divisor times the class of a point. Thus the image's cohomology is contained in the span of $[A],[B]$. The diagonal's independent class lies outside that span, so its Chow class lies outside the external-product image. Therefore the natural map of Chow groups is not surjective.