Linear Algebra over a Ring/Modules and linear functions

Definition (module):

Let $R$ be a ring. A left $R$-module is an abelian group $(M,+)$ together with a function $R\times M\to M$, denoted by juxtaposition, that satisfies the following axioms for all $r,s\in R$ and $m,n\in M$:

1. $1m=m$
2. $(r+s)m=rm+sm$
3. $r(sm)=(rs)m$
4. $r(m+n)=rm+rn$

Definition (homogenous):

Let $M$, $N$ be left modules over a ring $R$. A function $f:M\to N$ is called homogenous if and only if for all $r\in R$ and $m\in N$ the identity

$f(rm)=rf(m)$

holds.

Definition (linear):

Let $M$, $N$ be left modules over a ring $R$. A function $f:M\to N$ is called linear if and only if it is both homogenous and a morphism of abelian groups from $M$ to $N$.

Theorem (first isomorphism theorem):

Let $M$ and $N$ be left modules over a ring $R$. Let $\varphi :M\to N$ be linear. Then

$M/\ker \varphi \cong \operatorname {im} \varphi$.

Proof:   $\Box$

Exercises

1. Prove that for a function $f:M\to N$ between left $R$-modules, the following are equivalent:
   1. $f$ is linear
   2. For all $m,n\in M$ and $r\in R$, we have $f(m+n)=f(m)+f(n)$ and $f(rm)=rf(m)$
   3. For all $m,n\in M$ and $r\in R$, we have $f(m+rn)=f(m)+rf(n)$
   4. For all $m,n\in M$ and $r,s\in R$, we have $f(rm+sn)=rf(m)+sf(n)$