Every irreducible element is prime: always holds under surjective ring homomorphism?

Question

Let $f:R\rightarrow S$ be a surjective ring homomorphism. $R,S$ are both integral domains.

If every irreducible element in $R$ is prime, then is it true that every irreducible element in $S$ is prime?

How about the converse:

If every irreducible element in $S$ is prime, then is it true that every irreducible element in $R$ is prime?

Answer 1

Both are false. Consider the ring homomorphisms: $$\mathbb Z[x]\xrightarrow{\varphi}\mathbb Z[x]/(x^2+5)=\mathbb Z[\sqrt{-5}]\xrightarrow{\psi} \mathbb Z/(5)$$

The homomorphism $\varphi$ is the map given by taking the quotient by the indicated ideal. The homomorphism $\psi$ can be constructed as follows. Start with the mapping $\mathbb Z[x]\to\mathbb Z/(5)$ that maps any polynomial $f(x)$ to the congruence class of $f(0)$ mod. $5$. We note that $x^2+5$ maps to $0$, so that the ideal $(x^2+5)$ is a subset of the kernel of this mapping. Taking the quotient by this ideal, we obtain $\psi$.

Since every ring here is Noetherian (cf. Hilbert's Basis Theorem), they are UFDs if and only if every irreducible is prime. However, as in Example 1.8 here, $\mathbb Z[\sqrt{-5}]$ is not a UFD. Both $\varphi$ and $\psi$ are surjective, while $\mathbb Z/(5)$ and $\mathbb Z[x]$ are UFDs.