nLab torsion-free module

Contents

Contents

Idea

A module over a ring whose underlying abelian group has trivial torsion subgroup is called torsion-free.

Definition

Torsion-free \mathbb{Z}-modules

In classical mathematics, a torsion-free \mathbb{Z}-module or torsion free abelian group MM could be defined using a variant of the zero-divisor property characteristic of integral domains: for all rr in \mathbb{Z} and mm in MM, if rm=0r m = 0, then r=0r = 0 or m=0m = 0, or the contrapositive, if r0r \neq 0 and m0m \neq 0, then rm0r m \neq 0.

There is also an equivalent definition: a torsion-free \mathbb{Z}-module MM or torsion free abelian group is such that right multiplication by mm is injective if m0m \neq 0 and left multiplication by rr is injective if r0r \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 rr in \mathbb{Z} and mm in MM, if rm=0r m = 0, then r=0r = 0 and m=0m = 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 rr in \mathbb{Z} and mm in MM, if r0r \neq 0 and m#0m \# 0, then rm#0r 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 MM is such that right multiplication by mm is injective if m#0m \# 0 and left multiplication by rr is injective if r0r \neq 0.

A torsion-free ring is a monoid object in torsion-free \mathbb{Z}-modules.

Torsion-free RR-modules

In classical mathematics, given a commutative ring RR, a torsion-free RR-module is a module MM such that for all rr in Can(R)Can(R), where Can(R)Can(R) is the multiplicative submonoid of cancellative elements in RR and mm in MM, if rm=0r m = 0, then r=0r = 0 or m=0m = 0. Equivalently, the contrapositive, if m0m \neq 0, then rm0r m \neq 0. Some authors require RR to be an integral domain, where Can(R)Can(R) is the monoid of nonzero elements in RR.

In constructive mathematics, given a ring RR, there are multiple inequivalent ways of defining a torsion-free RR-module. One could define a torsion-free module as a module such that for all rr in Can(R)Can(R) and mm in MM, if rm=0r m = 0, then m=0m = 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 MM has a tight apartness relations, then one could define a torsion-free module as a module such that for all rr in Can(R)Can(R) and mm in MM, if m#0m \# 0, then rm#0r m \# 0. This is valid in \mathbb{R}-modules, but is no longer capable of being defined in coherent logic.

A torsion-free RR-algebra is a monoid object in torsion-free RR-modules.

Properties

Proposition

Every divisible torsion-free \mathbb{Z}-module is a rational vector space.

Proposition

Every integral domain RR is a torsion-free RR-module.

Locality

Being torsion-free is not a local property in general.

Proposition

Let RR be a commutative and assume its total ring of fractions? Q(R)\mathrm{Q}(R) be absolutely flat?. Then for an RR-module MM the following are equivalent :

  1. MM is torsion-free over RR ;
  2. M 𝔭M_\mathfrak{p} is torsion-free over R 𝔭R_\mathfrak{p} for every prime ideal 𝔭R\mathfrak{p} \subset R ;
  3. M 𝔪M_\mathfrak{m} is torsion-free over R 𝔪R_\mathfrak{m} for every maximal ideal 𝔪R\mathfrak{m} \subset R.

Proof

121 \Rightarrow 2. Suppose that MM is a torsion-free RR-module and that α¯x¯=0\overline{\alpha}\overline{x} = 0 for a regular α¯R 𝔭\overline{\alpha} \in R_\mathfrak{p} and x¯M 𝔭\overline{x} \in M_\mathfrak{p}. Set 𝔞 𝔭={a;as=0 for some sR𝔭,aR}\mathfrak{a}_\mathfrak{p} = \{a ; a s = 0 \quad\text{ for some }\quad s \in R - \mathfrak{p}, a \in R\} and M={m;sm=0 for some sR𝔭,mM}M' = \{m ; s m = 0 \quad\text{ for some }\quad s \in R - \mathfrak{p}, m \in M\}. Then we may assume that α¯R/𝔞 𝔭R 𝔭\overline{\alpha} \in R/\mathfrak{a}_\mathfrak{p} \hookrightarrow R_\mathfrak{p} and x¯M/MM 𝔭\overline{x} \in M/M' \hookrightarrow M_\mathfrak{p} by multiplying suitably elements of R/𝔞 𝔭𝔭/𝔞 𝔭R/\mathfrak{a}_\mathfrak{p} - \mathfrak{p}/\mathfrak{a}_\mathfrak{p} to α¯,x¯\overline{\alpha}, \overline{x}.

Denote by αR\alpha \in R a representative of α¯\overline{\alpha} and xMx \in M a representative of x¯\overline{x}. Since Q(R)\mathrm{Q}(R) is absolutely flat, there exists a regular rRr \in R and βR\beta \in R such that α(αβr)=0\alpha ( \alpha \beta - r) = 0. Because α¯\overline{\alpha} is regular, r¯=α¯β¯\overline{r} = \overline{\alpha} \overline{\beta}. Then one has r¯x¯=0\overline{r} \overline{x} = 0 which means that there exists sR𝔭s \in R- \mathfrak{p} such that srx=0s r x = 0 but since MM is torsion-free and rr is regular, one has sx=0s x = 0, so x¯=0\overline{x} = 0.

232 \Rightarrow 3. Obvious.

313 \Rightarrow 1. Let rRr \in R be a regular element and xMx \in M such that rx=0r x = 0. For every maximal ideal 𝔪R\mathfrak{m} \subset R, the image r¯R 𝔪\overline{r} \in R_\mathfrak{m} is regular so that r¯x¯=0\overline{r} \overline{x} = 0 in M 𝔪M_\mathfrak{m} means that x¯=0\overline{x} = 0, for every 𝔪R\mathfrak{m} \subset R. As a consequence x=0x = 0 in MM.

References

  • Shizuo Endo, On semi-hereditary rings, J. Math. Soc. Japan Vol. 13, No. 2, 1961

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.