Torsion submodules of non-noetherian modules Let $R$ be a commutative ring, let $\mathfrak{a}\subseteq R$ be an ideal, and let $M$ be an $R$-module. The $\mathfrak{a}$-torsion submodule of $M$ is defined as $$\Gamma_{\mathfrak{a}}(M)=\{x\in M\mid\mathfrak{a}\subseteq\sqrt{(0:_Rx)}\}.$$ If $R$ or $M$ is noetherian, then this submodule has lots of nice properties, some of which are lost in a non-noetherian setting. While trying to understand what precisely is lost, I got stuck with the following question:

Is there an example of $R$, $\mathfrak{a}$ and $M$ such that:
(1) There exists $n\in\mathbb{N}$ with $\mathfrak{a}^n\Gamma_{\mathfrak{a}}(M)=0$;
(2) For every $m\in\mathbb{N}$ we have $\mathfrak{a}^mM\cap\Gamma_{\mathfrak{a}}(M)\neq 0$.

One should note that (1) implies $$\Gamma_{\mathfrak{a}}(M)=\{x\in M\mid\exists n\in\mathbb{N}:\mathfrak{a}^nx=0\}$$ (which is not true in general).
 A: Take the $k[x,y]$-module with generators $a_n, n\in \mathbb N$ and relations
$$x a_1= ya_1=0$$ $$x a_{2n}+ ya_{2n+1} =a_n$$
Let $\mathfrak a=(x,y)$, then clearly $\mathfrak a M = M$. The $\mathfrak a$-torsion submodule contains $a_1$, so is nontrivial, and thus your second condition is clearly satisfies.
To check the first condition, it suffices to check that the $\mathfrak a$-torsion module of $M$ is generated by $a_1$.
So I guess let $M_n$ be the module generated by $a_1, \dots ,a_{2n-1}$ with all the relations from my list that are defined, so that $M_1=k[x,y]/(x,y)$. There is a natural map $M_n \to M_{n+1}$, and $M$ is the forward limit of the $M_n$.
The cokernel of this map is isomorphic to a module with two generators $a,b$ and one relation $xa+yb=0$. This modulee is torsion-free, for instance because it is isomorphic to the ideal $(x,y)$.
The kernel is trivial because if $f \in M_n$ is sent to $0$ under this map, then $f$ must be sent to a multiple of $xa_{2n+}+ya_{2n+1}-a_n$ in the module where we add $a_{2n}, a_{2n+1}$ and no relations. But every nonzero multiple of $xa_{2n}+ya_{2n+1}-a_n$ involves the new variables with nonzero coefficient and so can't be an element of $M_n$.
So we have a short exact sequence $0 \to M_n \to M_{n+1} \to (x,y) \to 0$, which induces an exact sequence $0 \to \Gamma_{\mathfrak a}(M_n) \to \Gamma_{\mathfrak a}(M_{n+1}) \to \Gamma_{\mathfrak a}((x,y))=0$, so $\Gamma_{\mathfrak a}(M_n) =\Gamma_{\mathfrak a}(M_{n+1})$, so by induction all the $M_n$ have torsion module generated by $a_1$ and thus the torsion module of $M$ is generated by $a_1$ as well.
