# Linear Algebra over a Ring/Homomorphism and dual modules

Proposition (multiple of module homomorphism by a ring element over a commutative ring is module homomorphism):

Let ${\displaystyle M,N}$ be modules over the commutative ring ${\displaystyle R}$. Let ${\displaystyle \phi :M\to N}$ be a homomorphism of ${\displaystyle R}$-modules, and let ${\displaystyle r\in R}$. Then the function

${\displaystyle rf:M\to N,(rf)(m):=rf(m)}$

is an ${\displaystyle R}$-module morphism too.

Proof: Let ${\displaystyle s\in R}$ and ${\displaystyle m,n\in M}$. Then ${\displaystyle (rf)(m+n)=(rf)(m)+(rf)(n)}$ and since ${\displaystyle R}$ is commutative, also ${\displaystyle (rf)(sm)=rsf(m)=s(rf)(m)}$. ${\displaystyle \Box }$

{{definition|homomorphism module|Let ${\displaystyle R}$ be a commutative ring

Definition (dual):

Let ${\displaystyle R}$ be a commutative ring, and let ${\displaystyle M}$ be an ${\displaystyle R}$-module. Then ${\displaystyle \operatorname {Hom} _{R}(M,R)}$ has a module structure given by addition and pointwise multipliction (since the category of modules over a ring is additive and the addition that is implied is compatible with pointwise multiplication, under which ${\displaystyle \operatorname {Hom} _{R}(M,R)}$ is closed), and this module, denoted more briefly by ${\displaystyle M^{*}}$, is called the dual of ${\displaystyle M}$.

Proposition (the category of modules is abelian):

Let ${\displaystyle R}$ be a ring. Then the category ${\displaystyle R-{\textbf {Mod}}}$ of left ${\displaystyle R}$-modules is abelian. Also, the category ${\displaystyle {\textbf {Mod}}-R}$ of right ${\displaystyle R}$-modules is abelian.

Proof: Existence of kernels and cokernels has been dealt with, and the same holds for existence of binary biproducts, since then product and sum coincide in this category. Also, the Noether isomorphism theorem holds. Then, the category is additive. Indeed, the summation on the homomorphism group indicated above is precisely the addition in the categorical sense. ${\displaystyle \Box }$