Ascending chain condition and ring homomorphism Let $f : R \to S$ be a surjective ring homomorphism between two integral domains. 
Could anyone advise me on how to prove/disprove the following statements:


*

*If $R$ satisfies the ascending chain condition for principal ideals, then so does $S$.

*If $S$ satisfies the ascending chain condition for principal ideals, then so does $R$.



I think both statements are false:
1.) There exists surjective ring homomorphism(a rather tedious/contrived construction) from $R=\mathbb{Z}[X]$ onto $S=\mathbb{Z}[\sqrt{5}].$ Since $R$ is $\text{UFD},$ every irreducible elements of $R$ is prime, so $R$ satisfies $\text{ACCP}.$  On the other hand, $2$ is irreducible but not prime in $S$ so $S$ does not satisfy $\text{ACCP}.$ 
2.) Define $\phi: R=\mathbb{Z} \to S=\mathbb{Z}_2$ by $\phi(a)= \overline{0},$ if $a$ is even and $\phi(a)=\overline{1},$ otherwise. $\phi$ is a surjective ring homomorphism. $R$ is not a field, so it doesn't satisfy $\text{ACCP}$ but $S$ is a field so it does. 
Thank you!
 A: For #1
It is natural to suspect 1) to be true since it is true when "for principal ideals" is replaced  with "for all ideals." This is the same as saying "if $R$ has it, then its quotient rings have it." 
Since $S\cong R/I$ for some ideal $I$, we would be considering an ascending chain of principal ideals in $R/I$. For each such ideal $(x+I)\lhd R/I$, the ideal $(x)\lhd R$ maps onto $(x+I)$, and this corresponds to and ascending chain of ideals in $R$. Work to show that if that chain stabilizes, then so does the one in the quotient ring.
For #2
You're right about 2, but the example you gave doesn't show this (we'll talk about that in the third section.)
An easy way to try to make a $S$ and ring $R$ such that $R\to S$ is onto and $S$ has some property that $R$ doesn't is to make $S$ a summand of $R$ and project onto $S$. Pick a ring $T$ that doesn't have ACCP, and a ring $S$ that does have ACCP. Let $R=S\times T$ and look at the projection $(s,t)\mapsto s$ from $R\to S$.
Your examples
In your first example, it looks like you are misusing some logic:


*

*"In a UFD, if irreducible elements are prime, then it satisfies the ACCP." in the second sentence and

*"If there is an irreducible element that's not prime in $S$, $S$ doesn't satisfy the ACCP." in the third sentence.
Both of these seem to be warped applications of this correct theorem:

"A commutative domain is a UFD iff it has ACCP and all irreducible elements are prime." 

You are apparently thinking there is some causal relationship between the ACC condition and irreducible-implies-prime conditions, but that's not what's going on. Rather, they are two halves of conditions that make a UFD. You can have ACCP without irreducibles being prime, and you can have irreducibles all prime without ACCP.
Now for the second argument. The claim "$R$ isn't a field, so it doesn't satisfy ACCP" is just false. You probably know that the integers satisfy the ACC on all ideals, and moreover that it is a principal ideal domain. But you see that ACC on all ideals and all ideals principal implies that the integers have ACCP right?
So the projection $\Bbb Z\to \Bbb Z/2\Bbb Z$ is not a counterexample to #2. Both rings have ACCP.
