How do you prove a module is free?

In mathematics, a free module is a module that has a basis – that is, a generating set consisting of linearly independent elements. Every vector space is a free module, 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.

Do all modules have a basis?

It is well known that a vector space V is having a basis, i.e. a subset of linearly independent vectors that spans V. Unlike for a vector space, a module doesn’t always have a basis.

Is 0 a free module?

We will adopt the standard convention that the zero module is free with the empty set as basis. Any two bases for a vector space over a field have the same cardinality.

Are submodules of free modules free?

Submodules of free modules every submodule of a free R-module is itself free; every ideal in R is a free R-module; R is a principal ideal domain.

Is 2Z a free module?

Thus Z/2Z and Z/3Z are non-free modules isomorphic to direct summands of the free module Z/6Z. 43.2 Definition.

What is rank of a free module?

The rank of a free module M over an arbitrary ring R( cf. Free module) is defined as the number of its free generators. For rings that can be imbedded into skew-fields this definition coincides with that in 1). In general, the rank of a free module is not uniquely defined.

Is z’n free module?

n Z= (n) is free as Z-module. Since X = {n} spans n Z as a Z-module. Also nm = 0 for m Z implies m = 0 as .

Is every module free?

Let R be a unital ring. Then R is a division ring if and only if every left R -module is free. Λ={P⊆M∣∣P is linearly independent}. ⁢…rings whose every module is free.

Is Zn a free module?

A finitely generated abelian group Zn is a free Z-module. Ex.

Is Z nZ a free module?

A basic theorem of linear algebra says that any vector space (over a field k) has a basis, i. e. it is free as a module over k. This does not hold for general rings: the Z-module Z/nZ, for instance, cannot be free because it is finite, and any nontrivial free Z-module is infinite.

Is Z 6Z a free module?

2) Z/2Z and Z/3Z are non-free projective Z/6Z-modules. be a sequence of R-modules and R-module homomorphisms.

Is QA free Z-module?

Because Z is a PID, Q is also a free Z-module But It’s not. Because for all submodules of Q \ {0}, they are not linearly independent over Z. And thus the only independent submodule of Q is {0}, which cannot span the whole Q.