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 localization faithfully flat?
Any free A- module is faithfully flat over A. Any localization of A is flat over A. Any direct factor of A (as a ring) is flat over A. If M is finitely generated as an A-module, then M is flat iff M is free iff the natural map mA ⊗ M → M is injective.
How do you show a module is flat?
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.
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?
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.
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.
Is Z torsion-free?
Both these examples can be generalized as follows: if R is a commutative domain and Q is its field of fractions, then Q/R is a torsion R-module. The torsion subgroup of (R/Z, +) is (Q/Z, +) while the groups (R, +) and (Z, +) are torsion-free.
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.