Exercise A0.III.6.16

Definition

A cyclic $R$-module $M$ is a module generated by a single element $m \in M$.

Proposition (C1)

Simple $R$-modules are cyclic.

Proof

Suppose $M$ is a simple module. Since a simple module is (by definition) non-zero, there’s a map $\varphi : R \to M$ mapping $1$ to some non-zero element of $M.$ The image of this map is a submodule of $M$ so it must be either $\{0\}$ or $M$. Since $\varphi(1) \neq 0,$ the image has to be the whole $M,$ that is, $\varphi$ must be surjective. Since $R$ is a free module on one generator, $M$ must be generated by one element.

$\blacksquare$

Proposition (C2)

An $R$-module $M$ is cyclic if and only if $M \cong R/I$ for some ideal $I.$

Proof

Suppose $M$ is cyclic. Then there’s a epimorphism $\pi : R \to M$ and by 6.12.C4 we know that this identifies $M$ with the cokernel of the inclusion $i : \ker \pi \to R.$ Thus $M = \coker i = R/\ker\pi.$

Suppose instead that $M \cong R/I.$ In this case we already have a surjection $\pi : R \to R/I$ and this shows that $M$ is cyclic.

$\blacksquare$

Proposition (C3)

Every quotient of a cyclic module is cyclic.

Proof

A cyclic $R$-module $M$ has a surjection $\tau : R \to M.$ A quotient map from $M$ to $M/I$ is also a surjection. So there is a surjection $\tau' : R \to M/I.$ This makes $M/I$ cyclic.

$\blacksquare$