Let $R$ be a commutative ring, and let $\mathfrak{a}\subseteq R$ be an ideal. For an $R$-module $M$, we set $\Gamma_{\mathfrak{a}}(M)=\{x\in M\mid\exists n\in\mathbb{N}:\mathfrak{a}^nx=0\}$, and we denote by ${\rm Ass}_R^{\rm f}(M)$ the set of weakly associated primes of $M$. In lots of situations there are interesting relations between $\Gamma_{\mathfrak{a}}(M)$ and ${\rm Ass}_R^{\rm f}(M)$, and this question concerns the following two of them:
$(*)$ For any $R$-module $M$ with $\Gamma_{\mathfrak{a}}(M)=0$ we have ${\rm Ass}_R^{\rm f}(M)\cap{\rm Var}(\mathfrak{a})=\emptyset$.
$(**)$ For any $R$-module $M$ with ${\rm Ass}_R^{\rm f}(M)\subseteq{\rm Var}(\mathfrak{a})$ we have $\Gamma_{\mathfrak{a}}(M)=M$.
If $\mathfrak{a}$ is of finite type, then both statements are true, but in general neither of them need be so. Moreover, there are examples showing that $(**)$ need not imply $(*)$.
If the functor $\Gamma_{\mathfrak{a}}$ is a radical, i.e., if $\Gamma_{\mathfrak{a}}(M/\Gamma_{\mathfrak{a}}(M))=0$ for every $R$-module $M$, then $(*)$ implies $(**)$. This (and, of course, my inability to find a counterexample) leads to the following question:
Does $(*)$ imply $(**)$ in general?
