# Exercise

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.

Outline:

• 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!