---- > [!definition] Definition. ([[submodule]]) > A **submodule** $N$ of an $R$-[[module]] $M$ is a [[subgroup]] preserved by the [[linear map|action]] of $R$: for all $r \in R$ and $n \in R$ we have $rn \in N$. > > Put differently, $N$ is itself an $R$-[[module]], and the [[inclusion map|inclusion]] $N \subset M$ is a $R$-[[linear map]]. ^definition > [!basicexample] > - $R$ may itself be viewed as an (left-)$R$-[[module]] (take $S=R, \alpha=\id$ in [[module induced by a ring homomorphism]]). In this case the [[submodule|submodules]] are precisely the (left-)[[ideal|ideals]] of $R$. > - [[kernel iff submodule]] ([[linear map|images are submodules too]]) > - If $r\in$ [[center of a ring|center]]$(R)$ and $M$ is an $R$-[[module]], then $rM$ is a [[submodule]] of $M$. > - If $I$ is any (left-)[[ideal]] of $R$, then $IM=\left\{ \sum_{i}r_{i}m_{i}:r_{i} \in I, m_{i}\in M \right\}$ is a [[submodule]] of $M$. ^basic-example > [!proof] Proof of Examples. > The first two are done, let us quickly do the second two. > Let $r \in \text{center}(R)$. Let $s \in R$, $rm \in rM$ be arbitrary, then $s(rm)=(sr)m=r(sm) \in rM$ as desired. > Let $r \in R$. Then for any $r_{i} \in I$ we have $r r_{i} \in I$ by definition. $r\sum_{i}r_{i}m_{i}= \sum_{i} r(r_{i}m_{i})= \sum_{i}\overbrace{(r r_{i})}^{ \in I}m_{i}$ shows the result. ^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 > ```