]> Commutative Algebra - Operations on Submodules

Operations on Submodules

Let M be a module. Let X be a subset of M. Then define X to be the submodule of M generated by X, that is, the intersection of all submodules of M containing X. (Note that the intersection of modules is itself a module.)

Let {M iiI} be a family of submodules of M (for some indexing set I). Define their sum as for ideals: M i consists of all finite sums x i where x iM i and almost all x i are zero. Note M i=unionM i.

Thus the submodules of M form a complete lattice with respect to inclusion (the glb is interesction and lub is the sum).

Isomorphism Theorems

1. If LMN are R-modules then

(L/N)/(L/M)L/M

2. If M 1 ,M 2 are submodules of an R-module M then

(M 1 +M 2 )/M 1 M 2 /M 1 M 2

Proof:

  1. The map θ:L/NL/M given by x+Nx+M is a surjective R-module homomoprhism with kernel M/N, thus the result follows by the fundamental homomoprhism theorem.

  2. Apply the same reasoning to the map θ:M 2 (M 1 +M 2 )/M 1 given by xx+M 1 .

Let I be an ideal of R. Then define the product IM to be the set of all finite sums a ix i where a iI,x iM. It is a submodule of M.

Let N,P be submodules of M. Define

(N:P)={aRaPN}

(similar to ideal quotients). It is an ideal of R. Define the annihilator of M by

Ann(M)=(0 :M)

If I is an ideal of R contained in Ann(M) then M can be viewed as an R/I-module by defining (x+I)m=xm for all xR,mM. This map is well-defined since IM=0 .

An R-module is faithful if Ann(M)=0 . Note M is faithful as an R/Ann(M)-module.

The following can be easily verified:

1. Ann(M+N) = Ann(M) \cap Ann(N)

2. (N:P) = Ann((N+P)/N)

Let xM. Then define Rx to be x, that is the set of all ax where aR. A set XM is a set of generators of M if M=X (so every element of M can be written as a linear combination of elements of X). If M has a finite set of generators then M is said to be finitely-generated.