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.
Which modules are not flat?
The short answer is that a “typical” ideal in an integral domain R is torsionfree but not flat. First observe that if R is a domain, then any ideal I of R is a submodule of the torsionfree R-module R and is thus torsionfree.
Are free modules flat?
Flatness is related to various other module properties, such as being free, projective, or torsion-free. In particular, every flat module is torsion-free, every projective module is flat, and every free module is projective.
Why flat module is called flat?
A module over a ring R is called flat if its satisfies one of many equivalent conditions, the simplest to state of which is maybe: forming the tensor product of modules with N preserves submodules.
Are free modules projective?
Theorem 1.4. Every free module is projective.
Is QA 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.
Is QA projective Z-module?
Is Q flat over Z?
Z modules are flat iff they are torsion-free, and QZ is torsion-free.
Are submodules of free modules free?
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.
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.
Why projective modules are important?
2) Projective modules are important for at least the following reasons. a) Geometric: A finitely generated module over a ring R is projective iff it is locally free (in the stronger sense of an open cover of SpecR). In other words, projective modules are the way to express vector bundles in algebraic language.
Is Q Z projective?
Hence Q is not projective as an Z-module.
How to prove that a projective module is a free module?
Proof. Assume is projective. Choose a surjection where is a free -module. As is projective there exists a such that . In other words and we see that is a direct summand of . Conversely, assume that is a free -module. Note that the free module is projective as and the functor is exact. Then as functors, hence both and are projective.
Is it possible to prove that every free module is flat?
But the result that every free module is flat comes very handy in the exercises. Is it possible to prove that every free module is flat just by definitions and without appealing to projective modules?
Are there any modules that are not torsion free?
In particular, every flat module is torsion-free, every projective module is free, and every free module is projective. There are finitely generated modules that are flat and not projective.
Are there any flat modules over commutative rings?
Flat modules over commutative rings are always torsion-free (that is, the multiplication by any regular element of the ring in injective in the module). Projective modules (and thus free modules) are always flat.