Let \(A=Ra\) be a cyclic module over a commutative ring \(R\). Define the order ideal of \(a\) by \[ \mathcal O_a=\{r\in R:ra=0\}. \] Show that \[ A\cong R/\mathcal O_a. \] In particular, if \(\mathcal O_a=(r)\) is principal, show that \(ra=0\) and \(A\cong R/(r)\).
Solution
Consider the surjective \(R\)-module homomorphism \[ \phi:R\longrightarrow A,\qquad s\longmapsto sa. \] Its kernel is exactly \[ \ker\phi=\{s\in R:sa=0\}=\mathcal O_a. \] The first isomorphism theorem therefore gives \[ A\cong R/\mathcal O_a. \]
If \(\mathcal O_a=(r)\), then \(r\in\mathcal O_a\), hence \(ra=0\), and the preceding isomorphism becomes \[ A\cong R/(r). \] Thus the principal generator of the order ideal records the annihilator of the cyclic generator and therefore the cyclic module itself, up to associates.