Question

Prove that every ideal I of a ring R is the kernel of a ring homomorphism of R.

Answer 1

Let us prove that every ideal $I$ of a ring $R$ is the kernel of a ring homomorphism of $R$. Consider the quotient ring $R/I=\{a+I:a\in R\}.$ Let $\overline{a}$ denotes $a+I.$ Then $\overline{a}+\overline{b}=\overline{a+b},\ \overline{a}\cdot\overline{b}=\overline{a\cdot b}.$ Consider the canonical ring homomorphism $\pi:R\to R/I,\ \pi(a)=\overline{a}.$ It is indeed a homomorphism because of $\pi(a+b)=\overline{a+b}=\overline{a}+\overline{b}=\pi(a)+\pi(b),$ $\pi(a\cdot b)=\overline{a\cdot b}=\overline{a}\cdot \overline{b}=\pi(a)\cdot\pi(b).$ It follows that $\ker\pi=\{a\in R: \pi(a)=\overline{0}\}=\{a\in R: \overline{a}=\overline{0}\} =\{a\in R: a+I=0+I\}=\{a\in R: a\in I-I\}=\{a\in R: a\in I\}=I.$

We conclude that $I$ is a kernel of a ring homomorphism $\pi:R\to R/I.$