Given $M$ an abelian group which is an R-module, where $R$ is a P.I.D., then, for $\alpha, \beta \in M$ and if
$Ann(\alpha) = (s),$ $Ann(\beta) = (t)$
Where both $s$ and $t$ are irreducible elements in $R$, is it true that $Ann(\alpha + \beta) = (st)$? It seems true for me, but I just cannot prove (or disprove) it.
Just proved it:
Let $\alpha, \beta$ as previously. If $r \in \(st)$ then $r\alpha + r\beta = 0$ clearly.
Conversely, if $r\alpha + r\beta = 0$ then $r\alpha = -r\beta$ therfore $0 = rs\alpha = -rs\beta$, thus $rs \in Ann(\beta)$ which means $t|rs$, but $(t,s)=1$ (they are distinct irreducibles), thus $t|r$, with the same argument $s|r$ and then $r \in (st)$.