---- $R$ is a [[ring]]. > [!definition] Definition. ([[torsion element of a module]]) > Let $M$ be an $R$-[[module]]. We call $m \in M$ a **torsion element** if $\{ m \}$ is [[linearly independent|linearly dependent]], that is, if there exists nonzero $r \in R$ such that $rm=0$. > > The subset ([[submodule]] if $R$ is an [[integral domain]]) of torsion elements of $M$ is denoted $\text{Tor}_{R}(M)$. If $\text{Tor}_{R}(M)=(0)$ then we call $M$ **torsion-free**; if $\text{Tor}_{R}(M)=M$ we call $M$ a **torsion module**. ^definition > [!specialization] > In the special case where $M$ is an [[abelian group]] ([[every abelian group is a Z-module, in exactly one way|a]] $\mathbb{Z}$-[[module]]), this definition of torsion is precisely that given in [[order of an element in a group]]. ^specialization > [!note] Remark. > 'Having torsion' is the 'most spectacular way in which a module can fail to be [[free module|free]]'. > [!basicnonexample] > - [[submodule|Submodules]] and [[direct sum of modules|direct sums]] of torsion-free [[module|modules]] are torsion-free. > - [[free module|Free modules]] over an [[integral domain]] are torsion-free (follows from the above, since an [[integral domain]] is a torsion-free module over itself and a free module is — [[terminal objects are unique up to a unique isomorphism|up to isomorphism]] — a direct sum of $R$). > - [[every vector space has a basis|In particular]], every [[vector space]] is torsion-free > - [[ideal|Ideals]] in an [[integral domain]] $R$ are torsion-free ([[submodule#^basic-example|for they are]] [[submodule|submodules]] of the [[free module]] $R^{1}$). ^nonexample > [!justification] > We must show that if $R$ is an [[integral domain]], then $\text{Tor}_{R}(M)$ is a [[submodule]]. > > Let $m_{1},m_{2} \in \text{Tor}_{R}(M)$ and nonzero $r_{1},r_{2} \in R$ with $r_{1}m_{1}=0$ and $r_{2}m_{2}=0$. Clearly $\text{Tor}_{R}(M)$ is a [[subgroup]], since $r_{1}r_{2}(m_{1}-m_{2})=r_{2}r_{1}m_{1} - r_{1}r_{2}m_{2}=r_{2}0-r_{1}0=0$ (we needed [[commutative ring|commutativity]] here). And the action of $R$ preserves $\text{Tor}_{R}(M)$: for any $r \in R$, $r_{1}(rm)=r(r_{1}m)=r0=0,$ so $r m \in \text{Tor}_{R}(M)$. ^justification ---- #### ---- #### 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 > ```