Is intersection with the power of a polarization injective on the Chow Ring?

Let $A$ be an abelian variety of dimension g, and $x\in A$. Let $\theta$ be a polarization. Let $\theta_x$ denote its translate by $x$. Let $0\leq k\leq g$. If we have $$\theta^{g-k}(\theta_x-\theta)^k=0$$ in $\text{Ch}(A)$, where the exponents are intersection products, then is it true that $(\theta_x-\theta)^k=0$?

For context, a formula of Beauville gives us that $$\frac{\theta^{g-k}}{(g-k)!}(\theta_x-\theta)^k=\frac{\theta^g}{g!}*\gamma(x)^{*k}$$ where $*$ is the Pontryagin product and $\gamma(x)=\sum_{v=1}^{g}\frac{1}{v}(-[x])^{*v}$. I want to show that if $\gamma(x)^{*k}=0$, then $(\theta_x-\theta)^k=0$. This comes up in the following preprint of Voisin in lemma 0.7: https://arxiv.org/pdf/1802.07153.pdf

• Which Chow group are you considering? If it is the usual one (cycles modulo rational equivalence), I suggest you consider the case of $g=1$. – Mohan May 23 '18 at 1:50
• Yes it is the usual one. For $g=1$, we have the two cases $k=0,1$. For $k=0$, the equation is just $\theta=0$, which doesn't happen because $\theta$ is ample. The case $k=1$ gives us the equation we want. I'm not sure how to generalize. – Samir Canning May 23 '18 at 4:26
• When $k=1$, you get $\theta_x-\theta=0$, which is not true in the Chow group. So, what exactly do you mean? – Mohan May 23 '18 at 12:46
• I want to check: if $\theta^{g-k}(\theta_x-\theta)^k=0$, then $(\theta_x-\theta)^k=0$. So in the case $g=k=1$, the assumption doesn't occur, so the implication is vacuously true. – Samir Canning May 23 '18 at 13:59
• I am unclear what you are saying. If $g=k=1$, your equation says $\theta_x-\theta=0$, which is untrue. – Mohan May 23 '18 at 14:52

Let $$A$$ be an abelian variety of dimension g, and $$x\in A$$. Let $$\theta$$ be a polarization. Let $$\theta_x$$ denote its translate by $$x$$. Let $$0\leq k\leq g$$. If we have $$\theta^{g-k}(\theta_x-\theta)^k=0$$ in $$\text{Ch}(A)$$, where the exponents are intersection products, then is it true that $$(\theta_x-\theta)^k=0$$?
You can prove it directly by dominating your abelian variety $$A$$ by a product $$C^g$$---mapping to $$A$$ by the sum map. Then the two conditions are equivalent to the fact that $$(\theta_x - \theta)^k$$ pullback to $$0$$ in $$\text{CH}_0(C^k)$$, where $$C^k$$ maps to $$A$$ by the sum map.