]> 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 $\left\{{M}_{i}\mid i\in I\right\}$ be a family of submodules of $M$ (for some indexing set $I$). Define their sum as for ideals: $\sum {M}_{i}$ consists of all finite sums $\sum {x}_{i}$ where ${x}_{i}\in {M}_{i}$ and almost all ${x}_{i}$ are zero. Note $\sum {M}_{i}=⟨\mathrm{union}{M}_{i}⟩$.

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 $L\supset M\supset N$ are $R$-modules then

$\left(L/N\right)/\left(L/M\right)\cong L/M$

2. If ${M}_{1},{M}_{2}$ are submodules of an $R$-module $M$ then

$\left({M}_{1}+{M}_{2}\right)/{M}_{1}\cong {M}_{2}/{M}_{1}\cap {M}_{2}$

Proof:

1. The map $\theta :L/N\to L/M$ given by $x+N↦x+M$ 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 $\theta :{M}_{2}\to \left({M}_{1}+{M}_{2}\right)/{M}_{1}$ given by $x\to x+{M}_{1}$.

Let $I$ be an ideal of $R$. Then define the product $IM$ to be the set of all finite sums $\sum {a}_{i}{x}_{i}$ where ${a}_{i}\in I,{x}_{i}\in M$. It is a submodule of $M$.

Let $N,P$ be submodules of $M$. Define

$\left(N:P\right)=\left\{a\in R\mid aP\subset N\right\}$

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

$\mathrm{Ann}\left(M\right)=\left(0:M\right)$

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

An $R$-module is faithful if $\mathrm{Ann}\left(M\right)=0$. Note $M$ is faithful as an $R/\mathrm{Ann}\left(M\right)$-module.

The following can be easily verified:

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

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

Let $x\in M$. Then define $Rx$ to be $⟨x⟩$, that is the set of all $ax$ where $a\in R$. A set $X\subset M$ 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.