I recently read the theorem that in a commutative ring R with unity and X,Y two unital cyclic left R-modules, X is isomorphic to Y as R-modules iff for every $x\in X$ such that <x>=X and for every $y\in Y$ such that <y>=Y, we have $Ann_R(x)=Ann_R(y)$. I am trying to prove this theorem. I believe I have shown the forward direction: If X and Y are isomorphic, then <x> $\cong$<y> and since X and Y are unital R-modules, we also have Rx=<x>=<y>=Ry. Then there exists some isomorphism $\phi$ from Rx to Ry. I believe that this works:
To show that Y=<$\phi(x)$>, let $x_0 \in$<x>. Then $\exists r\in R$ such that $x_0=rx$. Since $\phi$ is an R-module homomorphism, $\phi(x_0)=\phi(r_0x)=r_0\phi(x)$, and without loss of generality we can let $\phi(x)$=y.
Then if $r\in Ann_R(x)$, rx=0, so we have $0=\phi(0)=\phi(rx)=r\phi(x)=ry$, and thus $r\in Ann_R(y)$. Just use $\phi^{-1}$ for the other direction.
My main question is how to get started showing the other direction. My attempt so far has been to define a map $\phi:$<x> $\rightarrow$ <y> by $\phi(rx)=ry$ for any $r\in R$. I believe it is straightforward to show that this is an R-module homomorphism. To show that $\phi$ is 1-1, I aimed to show that $ker(\phi)={\{0}\}$. To this end, I supposed that $x_0\in ker(\phi)$. Let $x_0=r_0x$ for some $r_0\in R$. Then $0=\phi(x_0)=r_0\phi(x)=r_0y$, so $r_0\in Ann_R(y)=Ann_R(x)$ by assumption, so $r_0x=0$, thus $x_0=0$. To show that $\phi$ is onto, let $y'\in Y$ be arbitrary. Let $r\in R$ be such that $y'=ry$, since <y>=Ry=Y. Then $\phi(rx)=r\phi(x)=ry$, so $\phi$ is onto. Thus $\phi$ is a bijection and an R-module homomorphism, so $\phi$ is an isomorphism of R-modules and thus X=<x>$\cong$<y>=Y. Am I on the right track with this, or have I gotten something completely wrong?
Any help would be greatly appreciated!