# 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 $$f \in Hom_R(R,M)$$.

Ben Lynn blynn@cs.stanford.edu 💡