---- Let $R$ be a [[ring]]. > [!definition] Definition. ([[cyclic module]]) > An $R$-[[module]] $M$ is **cyclic** if it is [[submodule generated by a subset|generated by]] a singleton: $M=\langle m \rangle$ for some $m \in M$. > That is, the [[cyclic module|cyclic]] modules are those which admit an [[epimorphism]] from $R$, viewing the latter as a [[free module|free]] [[rank of a free module|rank]]-1 $R$-[[module]]: the [[chain complex of modules|sequence]] $R^{1} \to M \to 0$ is [[exact sequence|exact]] [^1]. ^definition > [!equivalence]+ > $M$ is cyclic if and only if $M \cong R / I$ for some [[ideal]] $I$ of $R$. ^equivalence [^1]: See examples in [[exact sequence]]; also recall [[module homomorphism is surjective iff cokernel is trivial iff is an epimorphism]]. The extra formalism offered here is meant to demonstrate that cyclic modules are a special case of finite-rank free modules $R^{m} \xrightarrow{\pi} M \to 0$. > [!basicexample] > - For $R=k$ a [[field]], a [[cyclic module]] is just a 1-[[dimension|dimensional]] [[vector space]] — that is, a 'copy of $k. ^basic-example > [!proof]+ Proof of Equivalence. > > $\to$. Suppose $M$ is [[cyclic module|cyclic]]; write $M=\langle m \rangle$. Then a [[linear map]] $\varphi:R \to M$ is determined by setting $\varphi(1):=m$ > and extending homomorphically, so that in general $\varphi (r)=r\varphi(1)=rm$. $\varphi$ is a [[surjection]], since the elements of $M$ are precisely of the form $rm$ for $r \in R$. Thus the [[first isomorphism theorem for modules]] > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRACUQBfU9TXfIRQBGclVqMWbALLdeIDNjwEiZYePrNWiDgAIA9LoA6RnDAAeOYAGsYAJ11cT9O2gAWWXQF5d7fQElucRgoAHN4IlAAMzsIAFskMhAcCCRREDcYOig2HAB3CEzshGpNKR1nOlcPOWjYhMQklKQAJh46+NbqZsR0nDosBjY3CAhrEGoinJ18wqyoBC4KLiA > \begin{tikzcd} > R \arrow[r, "\varphi", two heads] \arrow[d] & M \\ > R / \text{ker }\varphi = R/I \arrow[ru, two heads, hook] & > \end{tikzcd} > \end{document} > ``` > yields a [[module isomorphism]] $M \cong R / I$, where $I=\text{ker }\varphi$. > > $\leftarrow$. Conversely suppose $M \cong_{\psi} R / I$ for some [[ideal]] $I$ of $R$. A [[surjection]] $R^{1} \twoheadrightarrow M$ is given by $\psi \circ \pi$, where $\pi$ projects $R$ onto $R / I$. Then we're done by [[submodule generated by a subset]]. > ^proof ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```