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In a module, the scalars need only be a ring, so the module concept represents a significant generalization. In commutative algebra, both ideals and quotient rings 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 ...
The isomorphism theorems for vector spaces (modules over a field) and abelian groups (modules over ) are special cases of these. For finite-dimensional vector spaces, all of these theorems follow from the rank–nullity theorem. In the following, "module" will mean "R-module" for some fixed ring R.
If the ring R is considered as a left module over itself, then its cyclic submodules are exactly its left principal ideals as a ring. The same holds for R as a right R -module, mutatis mutandis . If R is F [ x ], the ring of polynomials over a field F , and V is an R -module which is also a finite-dimensional vector space over F , then the ...
Every module over a division ring is a free module (has a basis); consequently, much of linear algebra can be carried out over a division ring instead of a field. The study of conjugacy classes figures prominently in the classical theory of division rings; see, for example, the Cartan–Brauer–Hua theorem.
Noncommutative rings and associative algebras (rings that are also vector spaces) are often studied via their categories of modules. A module over a ring is an abelian group that the ring acts on as a ring of endomorphisms , very much akin to the way fields (integral domains in which every non-zero element is invertible) act on vector spaces.
For example, a polynomial ring R[x] is finitely generated by {1, x} as a ring, but not as a module. If A is a commutative algebra (with unity) over R, then the following two statements are equivalent: [5] A is a finitely generated R module. A is both a finitely generated ring over R and an integral extension of R.