Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

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ADVANCED ABSTRACT ALGEBRA
Example 2
Any homomorphic image of a module M is isomorphic to a quotient module of M.
Proof
Let
––––→ M 1
be a module epimorphism and let Ker ψ = N. Now we define a mapping
M
f : ––––→ M
N
1
defined by f ( x + N ) = ψ ( x).
∀ x + N ∈ M N
f ( x + y + N ) = f ( x + y + N ) = ψ ( x + y)
=
From above f is a module homomorphism. Further f is injective since Ker
f is surjective also since ψ is.
, the zero element of
Hence
Im f = M1 = ψ ( M ) i.e. M N
≅ ψ ( M)
Q. 1. Let I be an ideal in a commutative ring R with 1. If M is an R-module, show that the set
S = { xm x ∈ I , m ∈ M}
is not in general an R- module. When is S an R- module
Q. 2. If M is an R-module and if r ∈ R
, prove that the set
is an R-module.
Q. 3. Let M be a right R-module. Show that is an ideal of R. It is called
ψf∀ rM (0
mS
Mf
ψ
annihilator of M.
K
Example 3
If M is a finitely generated R-module, it does not follow that each submodule of N & M is also finitely
generated.
Let M be a cyclic right R-module, i.e. M = mR for some of M. The right R-submodules of M is if
the form mS, where S is a right ideal in the ring R. Suppose S is a finitely generated right ideal,
say
. Now the submodule mS is generated by the elements ma 1 , ma 2 ,............, ma k ,
i.e. ms = ma1, ma2
,...... ma k and so is a finitely generated R-module. Actually, ms is a cyclic
S-module.
If R is a Noetherian ring (i.e. R has the ascending chain condition on right ideals. I1 < I
2
< I3