----- > [!proposition] Proposition. ([[integral domain is a field iff its cyclic modules are torsion-free]]) > Let $R$ be an [[integral domain]]. We already know that if $R$ is a [[field]], then every [[cyclic module|cyclic]] $R$-[[module]] is necessarily [[torsion element of a module|torsion-free]] [^1]. > > Less obvious is that this condition is sufficient: If every [[cyclic module|cyclic]] $R$-[[module]] is [[torsion element of a module|torsion-free]], then $R$ is a [[field]]. ^proposition [^1]: Well, in this case *every* $R$-[[module]] is torsion-free because every [[vector space]] is torsion-free ([[torsion element of a module#^nonexample]][!note] Remark. > This lemma illustrates a more general fact that one can understand much about a [[ring]] $R$ by understanding the [[category]] $R$-$\mathsf{Mod}$ it entails. ^note > [!proof]- Proof. ([[integral domain is a field iff its cyclic modules are torsion-free]]) > Let $c \in R$ be nonzero; then $M=R / \langle c \rangle$ is a [[cyclic module|cyclic module]] by the equivalence (here, $\langle c \rangle$ is the [[principal ideal]] generated by $c$); indeed, the [[equivalence class|class]] of $1$ generates $R / \langle c \rangle$. Since the class of $1$ belongs to $\text{Tor}(M)$ (for $1 (cR)=cR=0_{M}$ and $c \neq 0$), we have that $\text{Tor}(M)=M$. But $M$ is torsion-free! Thus $M=R / \langle c \rangle$ is the zero $R$-module, hence $\langle c \rangle=\langle 1 \rangle=R$. Therefore, $c$ is a [[unit]], and since $c$ was arbitrary this proves $R$ is a [[field]]. ----- #### ---- #### 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 > ```