Module (mathematics): Difference between revisions

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m Changed wording to avoid possible confusion
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Remove a false statement. Decimal fractions as a Z-module have rank one.
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*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.
*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.