## Modules

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

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

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

3. $(a b)x = a (b x)$

4. $1 x = x$

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

Example

1. Any ideal $I\triangleleft R$ 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 $f:M\rightarrow N$ 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 $f(x+y) = f(x) + f(y), f(a x ) = a f(x)$. Alternatively we may say $f$ is a homomorphism between abelian groups that respects the actions of the ring.

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

Let $u:M'\rightarrow M$ and $v:N\rightarrow N'$ be $R$-module homomorphisms. They induce $R$-module homomorphisms $\bar(u) = f u : Hom(M,N)\rightarrow Hom(M',N)$ and $\bar(v) = v f : Hom(M,N) \rightarrow Hom(M,N')$.

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 $Hom_R(R,M) \cong M$. This can be seen by considering the map given by $f \mapsto f(1)$ for all $f \in Hom_R(R,M)$.