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Every vector space is a free module, [1] but, if the ring of the coefficients is not a division ring (not a field in the commutative case), then there exist non-free modules. Given any set S and ring R, there is a free R-module with basis S, which is called the free module on S or module of formal R-linear combinations of the elements of S.
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 n is both a left and right R-module over R if we use the component-wise operations. Hence when n = 1, R is an R-module, where the scalar multiplication is just ring
A free module is a module that can be represented as a direct sum over its base ring, so free abelian groups and free -modules are equivalent concepts: each free abelian group is (with the multiplication operation above) a free -module, and each free -module comes from a free abelian group in this way. [21]
Jean-Pierre Serre, in his 1955 paper Faisceaux algébriques cohérents, remarked that the corresponding question was not known for algebraic vector bundles: "It is not known whether there exist projective A-modules of finite type which are not free." [1] Here is a polynomial ring over a field, that is, = [, …,].
In abstract algebra, Kaplansky's theorem on projective modules, first proven by Irving Kaplansky, states that a projective module over a local ring is free; [1] where a not-necessarily-commutative ring is called local if for each element x, either x or 1 − x is a unit element. [2]
For every category C, the free strict monoidal category Σ(C) can be constructed as follows: its objects are lists (finite sequences) A 1, ..., A n of objects of C; there are arrows between two objects A 1, ..., A m and B 1, ..., B n only if m = n, and then the arrows are lists (finite sequences) of arrows f 1: A 1 → B 1, ..., f n: A n → B ...
To formally construct a basis for the dual space, we shall now restrict our view to the case where is a finite-dimensional free (left) -module, where is a ring with unity. Then, we assume that the set X {\displaystyle X} is a basis for F {\displaystyle F} .
The module is injective if and only if it is a direct summand of a character module of a free module. [ 2 ] If N {\displaystyle N} is a submodule of M {\displaystyle M} , then ( M / N ) ∗ {\displaystyle (M/N)^{*}} is isomorphic to the submodule of M ∗ {\displaystyle M^{*}} which consists of all elements which annihilate N {\displaystyle N} .