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:

  1. If $R$ satisfies the ascending chain condition for principal ideals, then so does $S$.
  2. 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!

  • $\begingroup$ Hmm, dunno what 'tedious/contrived' homomorphism you found from $\Bbb Z[x]\to \Bbb Z[\sqrt{5}]$, but you could replace it with the homomorphism that evaluates polynomials at $\sqrt{5}$ if that helps you in the future. It does not help you to find a counterexample, though. $\endgroup$ – rschwieb Mar 28 '14 at 13:19

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:

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

  2. "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.

  • $\begingroup$ THANKS FOR THE CORRECTION! $\endgroup$ – Alexy Vincenzo Mar 28 '14 at 13:25
  • $\begingroup$ Following your suggestion to (2), I would like to let $ \ T= \mathbb{Q}[X]$ and $S=\mathbb{Z}_2.$ $T$ has no ACCP because $(X) \subset (\frac{X}{2}) \subset ....$ is a strictly ascending chain of principal ideals. Also, there exists surjective homomorphism from $R=S \times T $ to $S.$ Since $\left((\overline{0},X)\right) \subset \left((\overline{0},\frac{X}{2})\right) \subset \left((\overline{0},\frac{X}{3})\right) \subset...$ is a strictly ascending chain of principal ideals, $R$ does not have ACCP. $\endgroup$ – Alexy Vincenzo Mar 28 '14 at 15:30
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    $\begingroup$ @AlexyVincenzo OK, nice try, but $\Bbb Q[x]$ is a principal ideal domain just like $\Bbb Z$, so it has the ACCP. We have talked about this point several times now, so I hope you are getting some impression about it! $(X)=(X/2)=(X/q)$ for whatever nonzero rational number $q$ you pick. However, the simplest example of a ring breaking the ACCP that I know looks like this. It is the ring $\Bbb Z + x\Bbb Q[x]$, with exactly the chain of ideals that you suggested. $\endgroup$ – rschwieb Mar 28 '14 at 15:42
  • $\begingroup$ Thank you. Can I let $T=\{f(X) \in \mathbb{Q}[X]: f(0) \in \mathbb{Z}\} \ ?$ $\endgroup$ – Alexy Vincenzo Mar 29 '14 at 7:53
  • $\begingroup$ @AlexyVincenzo that is exactly what I described as Z + xQ[x]. $\endgroup$ – rschwieb Mar 29 '14 at 11:26

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