= Solution
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 therefore
$$
\begin{pmatrix}A^2&A\cdot B&A\cdot D\\B\cdot A&B^2&B\cdot D\\D\cdot A&D\cdot B&D^2\end{pmatrix}
=\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix},\qquad\det=2.
$$
If 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 \b[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 $\operatorname{CH}^1(E)\otimes\operatorname{CH}^0(E)$ and $\operatorname{CH}^0(E)\otimes\operatorname{CH}^1(E)$. Here $\operatorname{CH}^0(E)=\mathbb Z[E]$, 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 \b[the natural map of Chow groups is not surjective].
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