Notes - Rings and Modules HT24, Torsion

\[\text{Ann}_R(m) = \\{r \in R \mid rm = 0\\} \trianglelefteq R\]

• $m$ is a torsion element if $\text{Ann} _ R(m) = \{r \in R \mid rm = 0\}$ is nonzero.
• $M$ is torsion if all elements of $M$ are torsion.
• $M$ is torsion-free if $M$ has no non-zero torsion elements.
• $R$ is an integral domain iff $R$ is torsion-free viewed as a module over itself.

Consider the $R$-module homomorphism $\phi : R \to M$ given by $r \mapsto rm$. Then $\text{Im } \phi = \langle m \rangle$ and $\ker \phi = \text{Ann} _ R(m)$, we have by the first isomorphsim theorem

\[\langle m \rangle \cong R/\text{Ann}_R(M)\]

as required.

\[M^\text{tor} = \\{m \in M \mid \text{Ann}_R(m) \ne \\{0\\}\\}\]

Suppose $x, y$ are elements of $M^\text{tor}$, then $\exists s, t \ne 0$ such that $sx = 0$ and $ty = 0$. Note $st \ne 0$ since $R$ is an integral domain. Then we have

\[st(x + y) = stx + sty = t(sx) + s(ty) = t\cdot 0 + s \cdot 0 = 0\]

and for arbitrary $r \in R$,

\[s(rx) = r(sx ) = 0\]

Hence $x + y \in M^\text{tor}$ and $rx \in M^\text{tor}$ so $M^\text{tor}$ is a submodule.

Now suppose that $x + M^\text{tor} \in M / M^\text{tor}$ is a torsion element. We aim to show that actually $x + M^\text{tor} = 0 + M^\text{tor}$ for a contradiction (since $M/M^\text{tor}$ being torsion requires there to be at least one non-zero torsion element).

Since $x + M^\text{tor}$ is a torsion element, then $\exists r \in R, r \ne 0$ such that $r(x + M^\text{tor}) = 0 + M^\text{tor}$, i.e. $rx \in M^\text{tor}$. Hence $rx$ is a torsion element, which means $\exists s \in R, s \ne 0$ such that $srx = 0$. But since $s, r$ are non-zero and $R$ is an integral domain, $sr \ne 0$. This means $x \in M^\text{tor}$ since $sr \in \text{Ann} _ R(x)$, but then $x + M^\text{tor} = 0 + M^\text{tor}$, as required.