Inverse image of a PID is a PID

Question

Let $f : R \to S$ be a ring homomorphism from $R$ onto $S$. If $S$ is a PID, is $R$ then a PID? If this is not possbile, is there an example to contradict it?

Answer 1

This is not true. Consider the evaluation map $\varphi : \mathbb{Z}[x] \to \mathbb{Z}$, $p(x) \mapsto p(0)$. Note that $\varphi$ is a surjective ring homomorphism so $\varphi^{-1}(\mathbb{Z}) = \mathbb{Z}[x]$. Furthermore, $\mathbb{Z}$ is a principal ideal domain (it is a Euclidean domain which is stronger), but $\mathbb{Z}[x]$ is not a principal ideal domain (for example, the ideal $(2, x)$ is not principal).

Answer 2

$R$ need not be principal, and it need not be a domain. It may not even be commutative.

Consider this quotient ring of polynomials in noncommuting indeterminates: $R=\Bbb Z\langle x,y\rangle/(x^2)$

Evaluation at $x=y=0$ yields a homomorphism from $R$ onto $\Bbb Z=S$. But as you can see $R$ isn't a domain (since $x^2=0$ and $x\neq 0$), $R$ isn't a principal ring (you can't generate $(x,y)\lhd R$ with a single element) and $R$ isn't commutative ($xy\neq yx$, since the indeterminates don't commute.)