]> Commutative Algebra - Submodules and Quotient Modules

Submodules and Quotient Modules

A submodule $M\prime$ of a $R$-module $M$ is a subgroup of $M$ that is closed under scalar multiplication. The quotient group $M/M\prime$ becomes an $R$-module by defining $a\left(x+M\prime \right)=ax+M\prime$. The $R$-module $M/M\prime$ is the quotient of $M$ by $M\prime$.

The natural map $M\to M/M\prime$ given by $x\to x+M\prime$ is a surjective module homomorphism, and it induces a bijection between submodules of $M/M\prime$ and submodules of $M$ that contain $M\prime$.

Let $f:M\to N$ be a module homomorphism. The kernel of $f$

$\mathrm{ker}f=\left\{x\in M\mid f\left(x\right)=0\right\}$

is a submodule of $M$. The image of $f$ is

$\mathrm{im}f=f\left(M\right)=\left\{f\left(x\right)\mid x\in M\right\}$

is a submodule of $N$. The cokernel of $f$ is

$\mathrm{coker}f=N/\mathrm{im}f$

Let $M\prime$ be a submodule of $M$ contained in $\mathrm{ker}f$. Then $f$ induces a homomorphism $\stackrel{‾}{f}:M/M\prime \to N$ given by $x+M\prime ↦f\left(x\right)$. Note $\mathrm{ker}\stackrel{‾}{f}=\mathrm{ker}f/M\prime$. If $M\prime =\mathrm{ker}f$ we have the fundamental homomorphism theorem for modules:

$M/\mathrm{ker}f\cong \mathrm{im}f$