If $f:R\to S$ is a surjective homomorphism of integral domains, $p$ is irreducible in $R,$ and $f(p)\neq 0_R$ is $f(p)$ irreducible in $S?$

Question

Background:

Exercise 24: If $f:R\to S$ is a surjective homomorphism of integral domains, $p$ is irreducible in $R,$ and $f(p)\neq 0_R$ is $f(p)$ irreducible in $S?$

Questions:

My guess is no. If we let $p=6,$ and $f:\mathbb{Z}\to \mathbb{Z}_6,$ and $f(2\cdot 2)=[2]_6\cdot [2]_6=[4]_6.$ I am wondering what happens if $p$ being prime, would the result hold?

Thank you in advance

Answer 1

Take $\pi:K[X,Y,Z]\to K[X,Y,Z]/(YZ-X)$ the canonical projection.
Clearly $X$ is irreducible in $K[X,Y,Z]$ but $\overline{X}=\overline{Y}\overline{Z}$.

Answer 2

Each prime number $p$ is irreducible in $\mathbf Z[x]$. The evaluation homomorphism $\mathbf Z[x] \to \mathbf Z$ where $x \mapsto p$ is surjective and it maps the irreducible polynomial $x^n + p$ in $\mathbf Z[x]$, where $n \geq 1$, to the integer $p^n + p = p(p^{n-1} + p)$, which is composite.

Answer 3

$\mathbb{Z}[x]\mapsto \mathbb{Z}$, $x\mapsto 0$, surjective, $x+a$ irreducible, maps to $a$, which can be any element.