MHB Division Rings and RIng Homomorphisms .... A&W Corollary 2.4 ....

[Math Amateur]

I am reading "Algebra: An Approach via Module Theory" by William A. Adkins and Steven H. Weintraub ...

I am currently focused on Chapter 2: Rings ...

I need help with an aspect of the proof of Corollary 2.4 ... ...

Corollary 2.4 and its proof read as follows:View attachment 7933
In the above proof of Corollary 2.4 we read the following:

" ... ... If $$\text{Ker} (f) = \{ 0 \}$$ then $$f$$ is injective ... ... "
Can someone please explain exactly how/why $$\text{Ker} (f) = \{ 0 \}$$ implies that $$f$$ is injective ... ?
Help will be appreciated ...

Peter

 

[castor28]

Hi Peter,

If $f$ is not injective, there are distinct elements $a$ and $b$ such that $f(a)=f(b)$. Because $f$ is a homomorphism, this implies $f(a-b)=0$, and, by definition, $a-b\in\ker f$. If $a\ne b$, this gives you a non-zero element of $\ker f$.

Note that this is more generally true for any group homomorphism $f$: $\ker f$ is trivial if and only if $f$ is injective.

 

[Math Amateur]

Hi Peter,

If $f$ is not injective, there are distinct elements $a$ and $b$ such that $f(a)=f(b)$. Because $f$ is a homomorphism, this implies $f(a-b)=0$, and, by definition, $a-b\in\ker f$. If $a\ne b$, this gives you a non-zero element of $\ker f$.

Note that this is more generally true for any group homomorphism $f$: $\ker f$ is trivial if and only if $f$ is injective.

Thanks castor28 ...

You seem to have proved the contrapositive ... but then that proves it!

Appreciate your help ...

Peter

 

[castor28]

Thanks castor28 ...

You seem to have proved the contrapositive ... but then that proves it!

Appreciate your help ...

Peter

Hi Peter,

In fact, the converse is also true. If $f$ is injective, the equation $f(x)=0$ has no other solution than $0$, and the solution set is $\ker f$ by definition.