symmetric monoidal (∞,1)-category of spectra
A module over a ring whose underlying abelian group has trivial torsion subgroup is called torsion-free.
In classical mathematics, a torsion-free $\mathbb{Z}$-module or torsion free abelian group $M$ could be defined using a variant of the zero-divisor property characteristic of integral domains: for all $r$ in $\mathbb{Z}$ and $m$ in $M$, if $r m = 0$, then $r = 0$ or $m = 0$, or the contrapositive, if $r \neq 0$ and $m \neq 0$, then $r m \neq 0$.
There is also an equivalent definition: a torsion-free $\mathbb{Z}$-module $M$ or torsion free abelian group is such that right multiplication by $m$ is injective if $m \neq 0$ and left multiplication by $r$ is injective if $r \neq 0$, where “multiplication” refers to the $\mathbb{Z}$-action.
In constructive mathematics, there are multiple inequivalent ways of defining a torsion-free $\mathbb{Z}$-module. One could define a torsion-free module as a module such that for all $r$ in $\mathbb{Z}$ and $m$ in $M$, if $r m = 0$, then $r = 0$ and $m = 0$. The first definition is valid in all modules with decidable equality, and could be defined using coherent logic, but is not valid for $\mathbb{R}$-modules.
If the module has a tight apartness relation, then one could define a torsion-free $\mathbb{Z}$-module as a module such that for all $r$ in $\mathbb{Z}$ and $m$ in $M$, if $r \neq 0$ and $m \# 0$, then $r m \# 0$. This is valid in $\mathbb{R}$, but is no longer capable of being defined in coherent logic. Similarly, one could define a torsion-free $\mathbb{Z}$-module $M$ is such that right multiplication by $m$ is injective if $m \# 0$ and left multiplication by $r$ is injective if $r \neq 0$.
A torsion-free ring is a monoid object in torsion-free $\mathbb{Z}$-modules.
In classical mathematics, given a commutative ring $R$, a torsion-free $R$-module is a module $M$ such that for all $r$ in $Can(R)$, where $Can(R)$ is the multiplicative submonoid of cancellative elements in $R$ and $m$ in $M$, if $r m = 0$, then $r = 0$ or $m = 0$. Equivalently, the contrapositive, if $m \neq 0$, then $r m \neq 0$. Some authors require $R$ to be an integral domain, where $Can(R)$ is the monoid of nonzero elements in $R$.
In constructive mathematics, given a ring $R$, there are multiple inequivalent ways of defining a torsion-free $R$-module. One could define a torsion-free module as a module such that for all $r$ in $Can(R)$ and $m$ in $M$, if $r m = 0$, then $m = 0$. The first definition is valid in all modules with decidable equality, and could be defined using coherent logic, but is not valid for $\mathbb{R}$-modules.
If $M$ has a tight apartness relations, then one could define a torsion-free module as a module such that for all $r$ in $Can(R)$ and $m$ in $M$, if $m \# 0$, then $r m \# 0$. This is valid in $\mathbb{R}$-modules, but is no longer capable of being defined in coherent logic.
A torsion-free $R$-algebra is a monoid object in torsion-free $R$-modules.
Every divisible torsion-free $\mathbb{Z}$-module is a rational vector space.
Every integral domain $R$ is a torsion-free $R$-module.
Being torsion-free is not a local property in general.
Let $R$ be a commutative and assume its total ring of fractions? $\mathrm{Q}(R)$ be absolutely flat?. Then for an $R$-module $M$ the following are equivalent :
$1 \Rightarrow 2$. Suppose that $M$ is a torsion-free $R$-module and that $\overline{\alpha}\overline{x} = 0$ for a regular $\overline{\alpha} \in R_\mathfrak{p}$ and $\overline{x} \in M_\mathfrak{p}$. Set $\mathfrak{a}_\mathfrak{p} = \{a ; a s = 0 \quad\text{ for some }\quad s \in R - \mathfrak{p}, a \in R\}$ and $M' = \{m ; s m = 0 \quad\text{ for some }\quad s \in R - \mathfrak{p}, m \in M\}$. Then we may assume that $\overline{\alpha} \in R/\mathfrak{a}_\mathfrak{p} \hookrightarrow R_\mathfrak{p}$ and $\overline{x} \in M/M' \hookrightarrow M_\mathfrak{p}$ by multiplying suitably elements of $R/\mathfrak{a}_\mathfrak{p} - \mathfrak{p}/\mathfrak{a}_\mathfrak{p}$ to $\overline{\alpha}, \overline{x}$.
Denote by $\alpha \in R$ a representative of $\overline{\alpha}$ and $x \in M$ a representative of $\overline{x}$. Since $\mathrm{Q}(R)$ is absolutely flat, there exists a regular $r \in R$ and $\beta \in R$ such that $\alpha ( \alpha \beta - r) = 0$. Because $\overline{\alpha}$ is regular, $\overline{r} = \overline{\alpha} \overline{\beta}$. Then one has $\overline{r} \overline{x} = 0$ which means that there exists $s \in R- \mathfrak{p}$ such that $s r x = 0$ but since $M$ is torsion-free and $r$ is regular, one has $s x = 0$, so $\overline{x} = 0$.
$2 \Rightarrow 3$. Obvious.
$3 \Rightarrow 1$. Let $r \in R$ be a regular element and $x \in M$ such that $r x = 0$. For every maximal ideal $\mathfrak{m} \subset R$, the image $\overline{r} \in R_\mathfrak{m}$ is regular so that $\overline{r} \overline{x} = 0$ in $M_\mathfrak{m}$ means that $\overline{x} = 0$, for every $\mathfrak{m} \subset R$. As a consequence $x = 0$ in $M$.
free module$\Rightarrow$ projective module $\Rightarrow$ flat module $\Rightarrow$ torsion-free module
See also
Last revised on July 23, 2023 at 16:53:51. See the history of this page for a list of all contributions to it.