Lef $f$ be a module homomorphism , $f$ is one-one if and only if $\ker f={0}$

Question

I am following Herstein's Topics in Algebra and according to that book a module may be defined on a ring without 1 (https://i.stack.imgur.com/FMeIZ.png).

So if we consider a module $M$ over a ring $R$ without $1$, then will be the statement in the question true ? If yes , then how to prove it ?

Answer 1

You prove it exactly as in the case of modules over rings with unity.

In order to show the equivalence between injectivity of a map and triviality of the kernel you only use the fact that $$ f(x-y)=f(x)-f(y). $$ Mind that you still have $-y$ and $f(-y)=-f(y)$ even though you don't have $1$ (and so $-1$) because a module is nonetheless an additive group.

Answer 2

Suppose that$Ker f=0$, $f(x)=f(y)$ implies that $f(x-y)=0$ and $x-y\in Ker f$ we deduce that $x-y=0$ and $x=y$.

$f$ is one to one, $f(x)=0=f(0)$ implies $x=0$.

Answer 3

If $f$ is one-to-one, then for any $x,y$ (elements of module) $$ f(x)=f(y) \Leftrightarrow x=y $$ $$ 0 = f(x) - f(y) = f(x-y) \Leftrightarrow x = y $$ So $x-y$ is in the kernel of $f$ if and only if $x-y=0$. Hope you can work out the rest of the proof.