When a module is called finitely generated?
An A-module M is said to be finitely generated if it has a finite set of generators. Examples: 1) Any finite dimensional vector space over a field k is a finitely generated k-module. 2) Any finitely generated abelian group is a finitely generated Z-module.
What does it mean for a ring to be finitely generated?
A ring is an associative algebra over the integers, hence a ℤ-ring. Accordingly a finitely generated ring is a finitely generated ℤ-algebra, and similarly for finitely presented ring. For rings every finitely generated ring is already also finitely presented.
Are finitely generated modules 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.
Are Noetherian rings finitely generated?
Since the domain is Noetherian, the codomain is Noetherian as well. This is a Noetherian ring, but it is not finitely generated, because there are infinitely many primes.
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 is the rank of a 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 submodule finitely generated?
In general, submodules of finitely generated modules need not be finitely generated. Since every polynomial contains only finitely many terms whose coefficients are non-zero, the R-module K is not finitely generated. In general, a module is said to be Noetherian if every submodule is finitely generated.
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 every Noetherian ring Artinian?
A ring A is noetherian, respectively artinian, if it is noetherian, respectively artinian, considered as an A-module. In other words, the ring A is noetherian, respectively artinian, if every chain a1 ⊆ a2 ⊆ ··· of ideal ai in A is stable, respectively if every chain a1 ⊇ a2 ⊇··· of ideals ai in A is stable.
Is a PID a Noetherian ring?
Thus every PID is a Noetherian integral domain. Proof. In a principal ideal ring R, every left or right ideal is generated by a single element and hence in particular, it is finitely generated. Thus R is a Noetherian ring by the Theorem 1.4.
Is the Z-module Q finitely generated?
Q is obviously a Z-module, however, it is not finitely generated.
Which of the following module is not free module?
The submodule 2Z/4Z is not free. For an explicit counterexample, consider R=k[X,Y] for any field k, which is free over itself. Then the ideal m=RX+RY is not free.
How are finitely generated modules over a division ring classified?
Finitely generated modules over the ring of integers Z coincide with the finitely generated abelian groups. These are completely classified by the structure theorem, taking Z as the principal ideal domain. Finitely generated (say left) modules over a division ring are precisely finite dimensional vector spaces (over the division ring).
What kind of module has a finite generating set?
In mathematics, a finitely generated module is a module that has a finite generating set. A finitely generated module over a ring R may also be called a finite R-module, finite over R, or a module of finite type .
When is a finitely generated module over a PID free?
A finitely generated module over a principal ideal domain is torsion-free if and only if it is free. This is a consequence of the structure theorem for finitely generated modules over a principal ideal domain, the basic form of which says a finitely generated module over a PID is a direct sum of a torsion module and a free module.
How is the left R-module M finitely generated?
The left R -module M is finitely generated if there exist a1, a2., an in M such that for any x in M, there exist r1, r2., rn in R with x = r1a1 + r2a2 + + rnan . The set { a1, a2., an } is referred to as a generating set of M in this case. A finite generating set need not be a basis, since it need not be linearly independent over R.