Is a ring a module over itself?

Note that a ring R is automatically finitely generated. In fact it is cyclic, considered as a module over itself, generated by 1, that is R = 〈1〉. This is clear, since if r ∈ R, then r = r · 1 ∈ 〈1〉.

Which is a module over itself?

Thus the submodules of A are simply the subgroups. If R is a ring, hence a module over itself, the submodules are those subsets closed under addition and also under multiplication by any r ∈ R, in other words, the left ideals. (If we take R to be a right R-module, then the submodules are the right ideals.)

Does every finitely generated module have a basis?

I know that a free module is a module with a basis, and that a finitely generated module has a finite set of generating elements (ie any element of the ring can be expressed as a linear combination of those generators). …

Does every module 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 R an R-module?

A module taking its scalars from a ring R is called an R-module. Like a vector space, a module is an additive abelian group, and scalar multiplication is distributive over the operation of addition between elements of the ring or module and is compatible with the ring multiplication.

Is every abelian group AZ module?

Any abelian group is a Z-module, where the action of Z is defined by na := a + ··· + a (n-fold sum). Every abelian group is a Z module in a unique way, and every homomorphism of abelian groups is a Z-module homomorphism in a unique way. Example. Just as any field F is vector space over F, any ring R is an R-module.

Is finitely generated module free?

A finitely generated torsion-free module of a commutative PID is free. A finitely generated Z-module is free if and only if it is flat. See local ring, perfect ring and Dedekind ring.

How do you prove a module is finitely generated?

A module M is finitely generated if and only if any increasing chain Mi of submodules with union M stabilizes: i.e., there is some i such that Mi = M. This fact with Zorn’s lemma implies that every nonzero finitely generated module admits maximal submodules.

What’s the difference between ring and field?

A RING is a set equipped with two operations, called addition and multiplication. A RING is a GROUP under addition and satisfies some of the properties of a group for multiplication. A FIELD is a GROUP under both addition and multiplication.

When is a left your module flat?

A left R – module is flat precisely if the tensoring functor from right R modules to abelian groups is an exact functor. By def. 0.3, or its immediate consequence, remark 0.5. N ∈ RMod is flat if for every injection i: A ↪ also i ⊗RN: A ⊗RN → B ⊗RN is an injection.

Are semi-simple rings direct product of matrix rings over division rings?

Here is a different answer, but it is much less elementary. If every module is free, then every module is projective, so every module is semi-simple, and the ring is an Artinian semi-simple ring, so a direct product of matrix rings over division rings.

How do you know if a module is flat?

Theorem 0.6. A module is flat if and only if it is a filtered colimit of free modules. This observation (Wraith, Blass) can be put into the more general context of modelling geometric theories by geometric morphisms from their classifying toposes, or equivalently, certain flat functors from sites for such topoi.

Is every module semisimple or projective?

If everymodule is projective, then every module is semisimple ( A ≤ B implies B/A is projective implies A is a direct summand of B). The key thing about the ring direct factors is they aretwo-sided ideals, so the earlier argument works: they have a nontrivial annihilator.