Are projective modules flat?

Are projective modules flat?

Every projective module is flat. The converse is in general not true: the abelian group Q is a Z-module which is flat, but not projective. Conversely, a finitely related flat module is projective.

Is Q flat Z-module?

Exercise 11.2: Show that Q is a flat Z-module but is not projective.

Are vector spaces flat modules?

Finite dimensional vector spaces are flat. Every vector space is the filtered colimit of its finite dimensional subspaces. Tensor products commute with colimits. Filtered colimits of exact sequences of vector spaces are exact.

Are fields flat modules?

Therefore, if your field is torsion-free, it is flat, and if it has torsion, it is not flat. A Z-module is flat if and only if it is torsion free, so it might depend on the characteristic of the field.

Is Z an Injective module?

The factor group Q/Z and the circle group are also injective Z-modules. The factor group Z/nZ for n > 1 is injective as a Z/nZ-module, but not injective as an abelian group.

How do you prove a module is flat?

Faithful flatness A module is faithfully flat if taking the tensor product with a sequence produces an exact sequence if and only if the original sequence is exact. Although the concept is defined for modules over a non-necessary commutative ring, it is used mainly for commutative algebras.

Is Q za projective 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. So Q cannot find a basis of Z-module.

What is Z module?

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 integers Z in a unique way. For n > 0, let n ⋅ x = x + x + + x (n summands), 0 ⋅ x = 0, and (−n) ⋅ x = −(n ⋅ x).

Is a ring Injective over itself?

Self-injective rings Every ring with unity is a free module and hence is a projective as a module over itself, but it is rarer for a ring to be injective as a module over itself, (Lam 1999, §3B).

What is a divisible module?

A module over a unit ring is called divisible if, for all which are not zero divisors, every element of can be “divided” by , in the sense that there is an element in such that . This condition can be reformulated by saying that the multiplication by defines a surjective map from to .

How can I see what modules are 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.

When do you call a module a flat module?

An -module is called flat if whenever is an exact sequence of -modules the sequence is exact as well. An -module is called faithfully flat if the complex of -modules is exact if and only if the sequence is exact.

Which is a flat module over a ring R?

Flat module. In homological algebra and algebraic geometry, a flat module over a ring R is an R-module M such that taking the tensor product over R with M preserves exact sequences. A module is faithfully flat if taking the tensor product with a sequence produces an exact sequence if and only if the original sequence is exact.

Why are direct limits of a flat module flat?

In general, arbitrary direct sums and direct limits of flat modules are flat, a consequence of the fact that the tensor product commutes with direct sums and direct limits (in fact with all colimits), and that both direct sums and direct limits are exact functors.

How is a flat module equivalent to a morphism?

The module M being flat is equivalent to being able always to do this. There is an alternative way to phrase this which is less element-centric. The elements mi correspond to a morphism into M from a free module, say m: F → M. The ai correspond to a morphism a: F → F, multiplying the i th term by ai.