Generating set of a module: Difference between revisions

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: <math>x = r_1 g_1 + \cdots + r_m g_m.</math>

Put in another way, there is an ''R''-lineara [[surjection]]

: <math>\bigoplus_{g \in \Gamma} R \to M., \, r_g \mapsto r_g g,</math>

where we wrote ''r''_''g'' for an element in the ''g''-th component of the direct sum. (Coincidentally, since a generating set always exists, e.g. ''M'' itself, this shows that a module is a [[quotient module|quotient]] of a [[free module]], a useful fact.)

A generating set of a module is said to be '''minimal''' if no [[proper subset]] of the set generates the module. If ''R'' is a [[field (mathematics)|field]], then a minimal generating set is the same thing as a [[basis (linear algebra)|basis]]. Unless the module is [[finitely- generated module|finitely- generated]], there may exist no minimal generating set.<ref>{{cite web|url=https://mathoverflow.net/q/33540 |title=ac.commutative algebra – Existence of a minimal generating set of a module – MathOverflow|work=mathoverflow.net}}</ref>

The [[cardinality]] of a minimal generating set need not be an invariant of the module; '''Z''' is generated as a principal ideal by 1, but it is also generated by, say, a minimal generating set {{nowrap|{2, 3}}}. What ''is'' uniquely determined by a module is the [[infimum]] of the numbers of the generators of the module.

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Let ''R'' be a [[local ring]] with [[maximal ideal]] ''m'' and [[residue field]] ''k'' and ''M'' finitely generated module. Then [[Nakayama's lemma]] says that ''M'' has a minimal generating set whose cardinality is <math>\dim_k M / mM = \dim_k M \otimes_R k</math>. If ''M'' is [[flat module|flat]], then this minimal generating set is [[linearly independent]] (so ''M'' is free). See also: [[minimal resolution (algebra)|Minimal resolution]].

A more refined information is obtained if one considers the relations between the generators; cf.see [[Free presentation of a module]].

== See also ==

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[[Category:Abstract algebra]]

{{algebra-stub}}