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Q/Z

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Idea

The quotient group /\mathbb{Q}/\mathbb{Z} of the additive abelian group of rational numbers by its subgroup of integers.

Properties

Relation to cyclic groups

For each natural number nn \in \mathbb{N} the cyclic group /n\mathbb{Z}/n\mathbb{Z} is a subgroup

/n / [k] [k/n] \array{ \mathbb{Z}/n\mathbb{Z} &\overset{\;\;\;\;\;\;}{\hookrightarrow}& \mathbb{Q}/\mathbb{Z} \\ [k] &\mapsto& [k/n] }

In fact /\mathbb{Q}/\mathbb{Z} is the filtered colimit of a diagram of cyclic groups and inclusions between them. In more detail, whenever m|nm|n, multiplication by n/mn/m induces an inclusion

/m/n\mathbb{Z}/m\mathbb{Z} \hookrightarrow \mathbb{Z}/n\mathbb{Z}

and this gives a functor to abelian groups from the lattice of natural numbers ordered by divisibility. The colimit of this functor is the abelian group /\mathbb{Q}/\mathbb{Z}.

Relation to Prüfer groups

For each prime pp the pp-Prüfer group [1/p]/\mathbb{Z}[1/p]/\mathbb{Z}, similarly formed as the colimit of groups /p k\mathbb{Z}/p^k\mathbb{Z} and inclusions between them, embeds in /\mathbb{Q}/\mathbb{Z}. Furthermore, it follows from the Chinese remainder theorem? that the induced map

p[1/p]//\bigoplus_p \mathbb{Z}[1/p]/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z}

is an isomorphism.

Relation to the circle group

We have a canonical subgroup inclusion into the circle group: If the latter is identified as the canonical subgroup U(1) ×U(1) \;\subset\; \mathbb{C}^\times in the group of units of the complex numbers, this inclusion is given by

/ U(1) [q] exp(2πiq). \array{ \mathbb{Q}/\mathbb{Z} &\overset{\;\;\;\;\;\;}{\hookrightarrow}& U(1) \\ [q] &\mapsto& \exp\big( 2\pi\mathrm{i} q\big) \,. }

From this point of view, /\mathbb{Q}/\mathbb{Z} is the torsion subgroup of U(1), whose elements are precisely the roots of unity.

Further properties

  • The group /\mathbb{Q}/\mathbb{Z} is an injective object in the category Ab of abelian groups.

  • It is also a cogenerator in the category of abelian groups.

    Proof: let AA be an abelian group. It suffices to check that for every aAa \in A there exists f:A/f: A \to \mathbb{Q}/\mathbb{Z} such that f(a)0f(a) \neq 0. But if a\langle a \rangle is the cyclic subgroup generated by aa, then it is easy to find a map g:a/g: \langle a \rangle \to \mathbb{Q}/\mathbb{Z} such that g(a)0g(a) \neq 0, and then we can extend gg to a map f:A/f: A \to \mathbb{Q}/\mathbb{Z} using injectivity of /\mathbb{Q}/\mathbb{Z}.

This means every abelian group embeds into an injective abelian group,

A f:A//,A \to \prod_{f: A \to \mathbb{Q}/\mathbb{Z}} \mathbb{Q}/\mathbb{Z},

and into an algebraic double dual, Ahom(hom(A,/),/)A \hookrightarrow \hom(\hom(A, \mathbb{Q}/\mathbb{Z}), \mathbb{Q}/\mathbb{Z}). The algebraic double dual of \mathbb{Z} is its profinite completion

p p\prod_{p} \mathbb{Z}_p

where p\mathbb{Z}_p is the group of pp-adic integers (this is connected with the direct sum decomposition of /\mathbb{Q}/\mathbb{Z} into Prüfer pp-groups).

  • For any ring RR, its algebraic dual hom(R,/)\hom(R, \mathbb{Q}/\mathbb{Z}), equipped with the left RR-module structure defined by rf:sf(sr)r \cdot f: s \mapsto f(s r), is an injective cogenerator in the category of left RR-modules. This follows easily from the corresponding property for /\mathbb{Q}/\mathbb{Z}.

Last revised on December 18, 2020 at 10:57:51. See the history of this page for a list of all contributions to it.