Reminder of definitions: For \(M\) an abelian group and \(R\) a ring, an \(R{\hbox{-}}\)module structure on \(M\) is a choice of a morphism \begin{align*} (\cdot)\in \hom_{\text{set}}(R\times M, M) \text{ or } \Phi\in \hom_{\text{ring}}(R, M^M) .\end{align*}

In the first instance, we think of \(\cdot\) as a binary operation \((r, m) \mapsto r\cdot m\) and in the second case, a choice of morphism \(r\mapsto \phi_r\).

For any group element \(m\in M\), we define \begin{align*} \operatorname{Ann}(m) \coloneqq\left\{{r\in R {~\mathrel{\Big\vert}~}r\curvearrowright_f m = 0_m}\right\} \leq R .\end{align*}

This is an ideal, because \(R\) is an \(R{\hbox{-}}\)module over itself, so we can define a morphism of \(R{\hbox{-}}\)modules: \begin{align*} g_m: R &\to M \\ r &\mapsto r\curvearrowright_f m .\end{align*}

Then clearly \(\operatorname{Ann}(m) = \ker g_m\), which is a \(R{\hbox{-}}\)submodule of \(R\), which correspond precisely to ideals of \(R\).

Let \(R\) be a ring and \(M\) a cyclic \(R{\hbox{-}}\)module. Then \(M \cong R/\operatorname{Ann}(M)\) as \(R{\hbox{-}}\)modules.


  • Does the statement make sense categorically? I.e. are both sides actually \(R{\hbox{-}}\)modules?
  • Construct an element of \(f \in \hom_{R{\hbox{-}}\text{mod}}(R, M)\)
  • Show that \(f\) is surjective and \(\ker f = \operatorname{Ann}(M)\), finish by 1st isomorphism theorem.

Lemma: For any \(I{~\trianglelefteq~}R\), the quotient \(R/I\) does in fact have an \(R{\hbox{-}}\)module structure.

Suppose the Lemma holds. To see that this finishes the proof, let \(M\) have an \(R{\hbox{-}}\)module structure \((\cdot)\in \hom_{{\mathsf{Set} }}(R\times M, M)\).

Since \(M\) is cyclic, we can write \(M = Rm = \left\{{r\cdot m {~\mathrel{\Big\vert}~}r\in R}\right\}\) for some group element \(m\in M\).

I.e. the \(R{\hbox{-}}\)orbit of \(m\) is transitive.

We then define the usual map: \begin{align*} g_m: R &\to M \\ r &\mapsto r\cdot m .\end{align*}

  • It’s clear that this is a morphism of \(R{\hbox{-}}\)modules: we have \((rx+_Ry)\cdot m = r(x\cdot m) +_M (y\cdot m)\) which just follows because \(\cdot\) is a well-defined action.
  • \(g_m\) is surjective: obvious from \(M = Rm\).
  • \(\ker g_m = \left\{{r\in R {~\mathrel{\Big\vert}~}r\cdot m =0}\right\} \coloneqq\operatorname{Ann}(m)\).

So \(R/\operatorname{Ann}(m) = R/\ker g_m \cong M\).

Note that \(R\) is an \(R{\hbox{-}}\)module with action given by \(f: R \to \endo(R)\) given by \(f(r) = (x\to rx)\) using the ring multiplication.

Then we are supplied with a map \(\Phi:R \to \endo(R)\) where, say, \(m \mapsto (\phi_r: R\to R)\). Let \(\pi: R \to R/I\) be the canonical projection. We want to build a map \(\Psi: R \to \endo(R/I)\), so fix an \(r\) and examine the obvious thing:

\begin{align*} \tilde \phi_r: R &\to R/I \\ x &\mapsto \pi\circ\phi_r(x) = \phi_r(x) + I .\end{align*}

Note the alternative, i.e. we could try defining this on the quotient directly: \begin{align*} R/I &\to R/I \\ x + I &\mapsto \phi_r(x) + I .\end{align*}

but we’d need to check if this was well-defined.

We can instead appeal to the universal property: any morphism \(f:R\to S\) such that \(I\subset \ker f\) factors uniquely through \(R/I\). So here we can just take \(S = R/I\), and check that \(I \subset \ker \tilde\phi_m\).

If \(i\in I\), then \(\tilde\phi_r(i) = \phi_r(i) + I\). Here, we actually need to use the fact that \(\phi_r(i) = ri\) and \(I\) is an ideal, so \(ri \in I\), and this \(\phi_r(i) + I = (ri) + I = 0 + I\), which is indeed the zero element of \(R/I\).

Note: it’s not clear that the analogous theorem “If \(R\) is an \(M{\hbox{-}}\)module then \(R/I\) is an \(M{\hbox{-}}\)module” goes through with this approach!