Question

Show that any ideal (resp. subring) can be realized as the kernel (resp. the image) of a ring homomorphism.?!

Answer 1

Question 1.

Show that any ideal (resp. subring) can be realized as the kernel (resp. the image) of a ring homomorphism?

Solution.

Let $I$ be an ideal of a ring $R$. Consider the natural projection $j:R\to R/I$, which maps any $r\in R$ to $a+I\in R/I$. Since the zero of $R/I$ is the class $0+I=I$, we have

$r\in\ker j\Leftrightarrow j(r)=I\Leftrightarrow r+I=I\Leftrightarrow r\in I.$

Thus, $\ker j=I$.

Now let $S$ be a subring of $R$. Then there is an embedding $i$ of $S$ into $R$, namely, $i(s)=s\in S\subseteq R$ for all $s\in S$. Obviously, $\operatorname{im}i=S$. ∎