- An element $ t \in M $ is called a torsion element of $ M $ if there exists a non-zero $ a \in A $ such that $ at = 0 $.
- The torsion elements of $ M $ form a submodule of $ M $ which is denoted by $ T(M) $ and is called the torsion submodule of $ M $.
- The module $ M $ is called torsion-free if there is no non-zero torsion element in $ M $ (or, what is the same, if $ T(M)=0 $).
First, prove that actually the torsion elements of $ M $ form a submodule of $ M $. Second, prove the following assertions:
- $ M / T(M) $ is torsion-free.
- If $ f \ \colon M \longrightarrow N $ is an $ A $-module homomorphism, then $ f(T(M)) \subseteq T(N) $.
- If \[ 0 \rightarrow M^{\prime} \rightarrow M \rightarrow M^{\prime \prime} \]is an exact sequence, then the sequence \[ 0 \rightarrow T(M^{\prime}) \rightarrow T(M) \rightarrow T(M^{\prime \prime}) \]is exact.
- $ T(M) $ is the kernel of the mapping \[ M \longrightarrow Q(A) \otimes_{A} M \ , \ x \mapsto 1 \otimes x \]
- Being torsion-free is a local property, i.e. for an $ A $-module $ M $ the following are equivalent:
- $ M $ is a torsion-free $ A $-module.
- $ M_{\mathfrak{p}} $ is a torsion-free $ A_{\mathfrak{p}} $-module for all prime ideals $ \mathfrak{p} $ of $ A $.
- $ M_{\mathfrak{m}} $ is a torsion-free $ A_{\mathfrak{m}} $-module for all maximal ideals $ \mathfrak{m} $ of $ A $.