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Let $X$ be a $n$-dimensional ($n>2$) smooth projective variety over $k=\bar k$ of positive characteristic. Take a divisor $D\in Pic(X).$ Suppose we know that $\frac{[D]^2}{2}\in H^4(X,\mathbb Z_l(2))$ is algebraic (i.e. is the cohomology class of an element of $\mathrm{CH}^2(X)\otimes\mathbb Z_l$) for two relatively prime $l,l'$ (each being distinct from $char(k)$).
It must be obvious but I cannot find a convincing proof of the following statement: there is a $Z\in \mathrm{CH}^2(X)$ such that $[Z]= \frac{[D]^2}{2}\in H^4(X,\mathbb Z_l(2))\ \forall l\neq char(k).$

Edit: $X$ is an abelian variety so that $[D]^2/2$ does make sense ($H^4 = \wedge^4 H^1$) and $l=2$ and $l'$ odd.

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  • $\begingroup$ Do you really want to divide by $2$? I think that makes the statement false already in $X=\mathbb P^2$, $D$ a line. Then $Z$ would be half the class of a point. Without the division by $2$, use intersection theory. $\endgroup$
    – Will Sawin
    Sep 19, 2015 at 14:42
  • $\begingroup$ If I am not mistaken, even in your example, $H^4$ being generated by the class of a point, the point $D^2$ is divisible by $2$ in cohomology since $[D]^2/2$ is in $H^4$ (maybe I should put $X$ abelian variety). $\endgroup$
    – user3001
    Sep 19, 2015 at 15:02
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    $\begingroup$ For any odd $l$, $2$ is invertible in $\mathbb{Z}_{l}$, so your hypothesis is empty. $\endgroup$
    – abx
    Sep 19, 2015 at 16:15
  • $\begingroup$ yes, my example shows it's false without the abelian variety assumption. $\endgroup$
    – Will Sawin
    Sep 20, 2015 at 0:10

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