Commutative Algebra

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 $a(x+M\prime )=ax+M\prime$. 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 $f$

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

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

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

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 $\ker f$. Then $f$ induces a homomorphism $\bar{f}:M/M\prime \to N$ given by $x+M\prime \mapsto f(x)$. Note $\ker \bar{f}=\ker f/M\prime$. If $M\prime =\ker f$ we have the fundamental homomorphism theorem for modules:

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