Division Rings & 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:

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

 

[fresh_42]

Injective means $f(x)=f(y) \Longrightarrow x=y$ which is equivalent to $f(x-y)=0 \Longrightarrow x-y=0$ which means $\operatorname{ker}f = \{0\}$.

 

[member 587159]

Suppose $f(x) = f(y)$ for some $x,y \in R$. Then, since $f$ is a ring homomorphism, it follows that $0 = f(x) - f(y) = f(x-y)$, which means that $x-y \in \ker(f) = \{0\}$, and thus $x-y = 0$ or equivalently $x = y$. So, we may conclude that $f$ is injective

The converse is also true: indeed, suppose that $f$ is injective. Let $x \in \ker(f)$. Then $f(x) = 0$. But, because $f$ is a homomorphism, it follows that $f(0) = 0 = f(x)$, and because of injectivity it follows that $0 = x$, so that $\ker(f) \subseteq \{0\}$ and equality follows.

 

[Math Amateur]

Suppose $f(x) = f(y)$ for some $x,y \in R$. Then, since $f$ is a ring homomorphism, it follows that $0 = f(x) - f(y) = f(x-y)$, which means that $x-y \in \ker(f) = \{0\}$, and thus $x-y = 0$ or equivalently $x = y$. So, we may conclude that $f$ is injective

The converse is also true: indeed, suppose that $f$ is injective. Let $x \in \ker(f)$. Then $f(x) = 0$. But, because $f$ is a homomorphism, it follows that $f(0) = 0 = f(x)$, and because of injectivity it follows that $0 = x$, so that $\ker(f) \subseteq \{0\}$ and equality follows.

Thanks fresh_42, MAth_QED ...

Very clear on that matter now!

Thanks again,

Peter