Editing Module (mathematics) (section)

== Examples ==

*If ''K'' is a [[field (mathematics)|field]], then ''K''-modules are called ''K''-[[vector space]]s (vector spaces over ''K'').
*If ''K'' is a field, and ''K''[''x''] a univariate [[polynomial ring]], then a [[Polynomial ring#Modules|''K''[''x'']-module]] ''M'' is a ''K''-module with an additional action of ''x'' on ''M'' by a group homomorphism that commutes with the action of ''K'' on ''M''. In other words, a ''K''[''x'']-module is a ''K''-vector space ''M'' combined with a [[linear map]] from ''M'' to ''M''. Applying the [[structure theorem for finitely generated modules over a principal ideal domain]] to this example shows the existence of the [[Rational canonical form|rational]] and [[Jordan normal form|Jordan canonical]] forms.
*The concept of a '''Z'''-module agrees with the notion of an abelian group. That is, every [[abelian group]] is a module over the ring of [[integer]]s '''Z''' in a unique way. For {{nowrap|''n'' > 0}}, let {{nowrap|1=''n'' ⋅ ''x'' = ''x'' + ''x'' + ... + ''x''}} (''n'' summands), {{nowrap|1=0 ⋅ ''x'' = 0}}, and {{nowrap|1=(−''n'') ⋅ ''x'' = −(''n'' ⋅ ''x'')}}. Such a module need not have a [[basis (linear algebra)|basis]]—groups containing [[torsion element]]s do not. (For example, in the group of integers [[modular arithmetic|modulo]] 3, one cannot find even one element that satisfies the definition of a [[linearly independent]] set, since when an integer such as 3 or 6 multiplies an element, the result is 0. However, if a [[finite field]] is considered as a module over the same finite field taken as a ring, it is a vector space and does have a basis.)
*The [[decimal fractions]] (including negative ones) form a module over the integers. Only [[singleton (mathematics)|singletons]] are linearly independent sets, but there is no singleton that can serve as a basis, so the module has no basis and no [[rank of a free  module|rank]], in the usual sense of linear algebra. However this module has a [[torsion-free rank]] equal to 1.
*If ''R'' is any ring and ''n'' a [[natural number]], then the [[cartesian product]] ''R''<sup>''n''</sup> is both a left and right ''R''-module over ''R'' if we use the component-wise operations. Hence when {{nowrap|1=''n'' = 1}}, ''R'' is an ''R''-module, where the scalar multiplication is just ring multiplication. The case {{nowrap|1=''n'' = 0}} yields the trivial ''R''-module {0} consisting only of its identity element. Modules of this type are called [[free module|free]] and if ''R'' has [[invariant basis number]] (e.g. any commutative ring or field) the number ''n'' is then the rank of the free module.
*If M<sub>''n''</sub>(''R'') is the ring of {{nowrap|''n''&thinsp;×&thinsp;''n''}} [[matrix (mathematics)|matrices]] over a ring ''R'', ''M'' is an M<sub>''n''</sub>(''R'')-module, and ''e''<sub>''i''</sub> is the {{nowrap|''n'' × ''n''}} matrix with 1 in the {{nowrap|(''i'', ''i'')}}-entry (and zeros elsewhere), then ''e''<sub>''i''</sub>''M'' is an ''R''-module, since {{nowrap|1=''re''<sub>''i''</sub>''m'' = ''e''<sub>''i''</sub>''rm'' ∈ ''e''<sub>''i''</sub>''M''}}. So ''M'' breaks up as the [[direct sum]] of ''R''-modules, {{nowrap|1=''M'' = ''e''<sub>1</sub>''M'' ⊕ ... ⊕ ''e''<sub>''n''</sub>''M''}}. Conversely, given an ''R''-module ''M''<sub>0</sub>, then ''M''<sub>0</sub><sup>⊕''n''</sup> is an M<sub>''n''</sub>(''R'')-module. In fact, the [[category of modules|category of ''R''-modules]] and the [[category (mathematics)|category]] of M<sub>''n''</sub>(''R'')-modules are [[equivalence of categories|equivalent]]. The special case is that the module ''M'' is just ''R'' as a module over itself, then ''R''<sup>''n''</sup> is an M<sub>''n''</sub>(''R'')-module.
*If ''S'' is a [[empty set|nonempty]] [[Set (mathematics)|set]], ''M'' is a left ''R''-module, and ''M''<sup>''S''</sup> is the collection of all [[function (mathematics)|function]]s {{nowrap|''f'' : ''S'' → ''M''}}, then with addition and scalar multiplication in ''M''<sup>''S''</sup> defined pointwise by {{nowrap|1=(''f'' + ''g'')(''s'') = ''f''(''s'') + ''g''(''s'')}} and {{nowrap|1=(''rf'')(''s'') = ''rf''(''s'')}}, ''M''<sup>''S''</sup> is a left ''R''-module.  The right ''R''-module case is analogous.  In particular, if ''R'' is commutative then the collection of ''R-module homomorphisms'' {{nowrap|''h'' : ''M'' → ''N''}} (see below) is an ''R''-module (and in fact a ''submodule'' of ''N''<sup>''M''</sup>).
*If ''X'' is a [[smooth manifold]], then the [[smooth function]]s from ''X'' to the [[real number]]s form a ring ''C''<sup>∞</sup>(''X''). The set of all smooth [[vector field]]s defined on ''X'' forms a module over ''C''<sup>∞</sup>(''X''), and so do the [[tensor field]]s and the [[differential form]]s on ''X''.  More generally, the sections of any [[vector bundle]] form a [[projective module]] over ''C''<sup>∞</sup>(''X''), and by [[Swan's theorem]], every projective module is isomorphic to the module of sections of some vector bundle; the [[category (mathematics)|category]] of ''C''<sup>∞</sup>(''X'')-modules and the category of vector bundles over ''X'' are [[equivalence of categories|equivalent]].
*If ''R'' is any ring and ''I'' is any [[ring ideal|left ideal]] in ''R'', then ''I'' is a left ''R''-module, and analogously right ideals in ''R'' are right ''R''-modules.
*If ''R'' is a ring, we can define the [[opposite ring]] ''R''<sup>op</sup>, which has the same [[underlying set]] and the same addition operation, but the opposite multiplication:  if {{nowrap|1=''ab'' = ''c''}} in ''R'', then {{nowrap|1=''ba'' = ''c''}} in ''R''<sup>op</sup>.  Any ''left'' ''R''-module ''M'' can then be seen to be a ''right'' module over ''R''<sup>op</sup>, and any right module over ''R'' can be considered a left module over ''R''<sup>op</sup>.
* [[Glossary of Lie algebras#Representation theory|Modules over a Lie algebra]] are (associative algebra) modules over its [[universal enveloping algebra]].
*If ''R'' and ''S'' are rings with a [[ring homomorphism]] {{nowrap|''φ'' : ''R'' → ''S''}}, then every ''S''-module ''M'' is an ''R''-module by defining {{nowrap|1=''rm'' = ''φ''(''r'')''m''}}. In particular, ''S'' itself is such an ''R''-module.