Abstract Algebra: integral domain and principal ideal domain

Question

I am studying by myself and I needed help for few question which I am confused how give proof of that. Let $\varphi : J \to K$ be a ring epimorphism with $\varphi(1) = 1$, where $J$ and $K$ are commutative rings with $1$. Prove the following or give a choice of $J$, $K$, and $\varphi$ where the claim fails.

1. If $J$ is an integral domain, then $\varphi(J)$ is an integral domain.
2. If $(k) \unlhd K$ is a principal ideal, i. e., generated by a single element, then the preimage $\varphi^{-1}((k))$ is a principal ideal in $J$.

Help me to understand to how to solve this question?

Answer 1

I will assume that $\varphi: J\rightarrow K$ because of the mixed notation.

For the first question, suppose $J=\mathbb{Z}$ and $K=\mathbb{Z}/m\mathbb{Z}$, where $m$ is not prime, and $\varphi: J\rightarrow K$ is the canonical quotient transformation (which is surjective and thus epic). $\mathbb{Z}/m\mathbb{Z}$ is not an integral domain; if $m=ab$, then $ab\equiv 0 \mod m$ but $a,b$ are non-zero modulo $m$.

For the second question, take $J=\mathbb{F}[x,y]$ and $K=\mathbb{F}$. Then take $\varphi:\mathbb{F}[x,y]\rightarrow \mathbb{F}$ such that $\varphi$ is constant on constant polynomials and sends $x$ and $y$ to 0. The kernel of this is the (non-principal ideal) $(x,y)$. Thus, we take $k=0$, so that $(0)=\{0\}=\varphi(\ker \varphi)$ and thus $\varphi^{-1}[(0)]=\ker\varphi$ is not a principal ideal.