Commutative Algebra

Let $R$ be a ring. An $R$-module or module over $R$ is an abelian group $(M,+)$ and a map $\mu :R\times M\to M$ (scalar mulitplication) such that for all $a,b\in R$ and $x,y\in M$ we have

1. $a(x+y)=ax+ay$

2. $(a+b)x=ax+bx$

3. $(ab)x=a(bx)$

4. $1x=x$

Alternatively, we may say that an $R$-module $M$ is an abelian group with a ring homomophism $R\to \mathrm{End}(M)$ where $\mathrm{End}(M)$ is the ring of endomorphisms of $M$.

Example

1. Any ideal $I◃R$ is an $R$-module. (Scalar multiplication is ring multiplication.) In particular $R$ is an $R$-module

2. If $R$ is a field then $R$-modules are precisely the vector spaces over $R$.

3. All abelian groups are $\mathbb{Z}$-modules.

4. Let $R=K[x]$ for some field $K$. Then an $R$-module is a $K$-vector space together with some linear transformation.

5. Let $K$ be a field, $G$ be a group and consider the group ring $R=K[G]$. Then $R$ modules are precisely $K$-representations of $G$.

A mapping $f:M\to N$ between $R$-modules $M,N$ is an $R$-module homomorphism or $R$-linear if addition and scalar multiplication are preserved, that is for all $x,y\in M$ and $a\in R$ we have $f(x+y)=f(x)+f(y),f(ax)=af(x)$. Alternatively we may say $f$ is a homomorphism between abelian groups that respects the actions of the ring.

Let ${\mathrm{Hom}}_R(M,N)$ be the set of all $R$-module homomorphisms from $M$ to $N$. For all $f,g\in \mathrm{Hom}(M,N),a\in R$, define $f+g$ and $af$ by $(f+g)(x)=f(x)+g(x),(af)(x)=af(x)$. Then it can be easily verified that $\mathrm{Hom}(M,N)$ is an $R$-module.

Let $u:M\prime \to M$ and $v:N\to N\prime$ be $R$-module homomorphisms. They induce $R$-module homomorphisms $\overline{(}u)=fu:\mathrm{Hom}(M,N)\to \mathrm{Hom}(M\prime ,N)$ and $\overline{(}v)=vf:\mathrm{Hom}(M,N)\to \mathrm{Hom}(M,N\prime )$.

Example: If $R$ is a field then $R$-module homomorphisms are linear transformations which may be written as matrices. The induced homomorphisms can be computed via matrix multiplications.

For any $R$-module $M$ we have ${\mathrm{Hom}}_R(R,M)\cong M$. This can be seen by considering the map given by $f\mapsto f(1)$ for all $f\in {\mathrm{Hom}}_R(R,M)$.