Content deleted Content added
Herpesklaus (talk | contribs) Remove a false statement. Decimal fractions as a Z-module have rank one. |
Undid revision 1263214016 by Herpesklaus (talk)Instead of removing ambiguous statements, you would habe better to clarify them |
||
Line 14:
In a vector space, the set of [[scalar (mathematics)|scalars]] is a [[field (mathematics)|field]] and acts on the vectors by scalar multiplication, subject to certain axioms such as the [[distributive law]]. In a module, the scalars need only be a [[ring (mathematics)|ring]], so the module concept represents a significant generalization. In commutative algebra, both [[ideal (ring theory)|ideals]] and [[quotient ring]]s are modules, so that many arguments about ideals or quotient rings can be combined into a single argument about modules. In non-commutative algebra, the distinction between left ideals, ideals, and modules becomes more pronounced, though some ring-theoretic conditions can be expressed either about left ideals or left modules.<!-- (semi)perfect rings for instance have a litany of "Foo is true for all left ideals iff foo is true for all finitely generated left ideals iff foo is true for all cyclic modules iff foo is true for all modules" -->
Much of the theory of modules consists of extending as many of the desirable properties of vector spaces as possible to the realm of modules over a "[[well-behaved]]" ring, such as a [[principal ideal ___domain]]. However, modules can be quite a bit more complicated than vector spaces; for instance, not all modules have a [[basis (linear algebra)|basis]], and, even for those that do ([[free module]]s), the number of elements in a basis need not be the same for all bases (that is to say that they may not have a unique [[Free_module#Definition|rank]]) if the underlying ring does not satisfy the [[invariant basis number]] condition, unlike vector spaces, which always have a (possibly infinite) basis whose [[cardinality]] is then unique. (These last two assertions require the [[axiom of choice]] in general, but not in the case of [[finite-dimensional]] vector spaces, or certain well-behaved infinite-dimensional vector spaces such as [[Lp space|L<sup>''p''</sup> space]]s.)
=== Formal definition ===
Line 37:
*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'' × ''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.
| |||