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== 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.)
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