Let $R$ be a commutative ring.
(i) Prove that $R$ has ACCP if and only if every non-empty collection of principal ideals of $R$ has a maximal element.
(ii) Prove further that if $R$ is an integral domain and has ACCP, then $R[X]$ has ACCP.


(i) ($\Rightarrow$) Suppose that there exists a non-empty collection of ascending chain of principal ideals of $R$ that does not have a maximal element. Then, for every ideal $I_i$ in this collection we can always take an ideal $I_{i+1}$ such that $I_i \subseteq I_{i+1}$. If not, then $I_i$ is the maximal element in this collection which is not possible. Hence, $R$ does not have ACCP. Contradiction.

($\Leftarrow$) Suppose $R$ does not have ACCP. Then we can find a chain of principal ideals that do not terminate. This chain does not have a maximal element. Contradiction.

I don't really know how to prove it directly other than by contradiction. Can someone show me how?

(ii) I can't see how can I apply the first part.


This answer is for item ii) given in the question. The other answer is fine, with the corrections given in the comments, but there is not necessary at all to use the fact that $F[X]$ satisfies the ACCP.

As usual, let $$(P_1)\subseteq (P_2)\subseteq \ldots$$ be a chain of principal ideals of $R[X]$. Then $P_{i+1}\mid P_i$ for all $i\ge 1$, and this implies that $\deg(P_1)\ge \deg(P_2)\ge \ldots$ is a decreasing sequence of natural numbers, then it must stabilize and thus there is some $n\in \Bbb{N}$ such that $\deg(P_n)=\deg(P_{n+i})$ for all $i\ge 0$. As $P_{n+i+1}\mid P_{n+i}$ it follows that $P_{n+i}=r_iP_{n+i+1}$ for some $r_i\in R$.

Now, let's denote $a_i$ the leading coefficient of $P_{n+i}$, therefore $a_i=r_ia_{i+1}$ for all $i\ge 0$ and then $a_{i+1}\mid a_{i}$, which lead us to the following chain of principal ideals of $R$: $$(a_0)\subseteq (a_1)\subseteq \ldots$$ Since $R$ satisfies the ACCP, then there exists $k\in \Bbb{N}$ such that $(a_k)=(a_{k+j})$ for all $j\ge 0$. Let's set $t=n+k$, we claim that $(P_t)=(P_{t+j})$ for all $j\ge 0$. Indeed, since $P_{t+j}\mid P_t$, it's enough to prove that $P_t\mid P_{t+j}$. Let's write $P_t=dP_{t+j}$, with $d\in R$, then $da_{k+j}=a_k=ua_{k+j}$, with $u\in R^{\times}$ because $a_k\sim a_{k+j}$. As $a_{k+j}\neq 0$, we deduce that $d=u$. Therefore $d^{-1}P_t=(d^{-1}d)P_{t+j}=P_{t+j}$, so $P_t\mid P_{t+j}$. Hence, $R[X]$ satisfies the ACCP.


For the first, I think there is no direct proof other than contradiction, since this result is independent of ZF and to prove it one must at least assume axiom of dependent choice.

For the second, let $F$ denote the fraction field of $R$. For any ascending chain $(f_1)\subseteq(f_2)\subseteq(f_3)\subseteq\cdots$ of principal ideals in $R[X]$, note that since $F[X]$ is a PID, $(f_1)\subseteq(f_2)\subseteq(f_3)\subseteq\cdots$ stabilizes eventually in $F[X]$, that is to say, there exists $n\in\mathbb N$ such that for any $i\in\mathbb N$, we have $f_n\sim f_{n+i}$ in $F[X]$. Then since $(f_n)\subseteq(f_{n+i})$ in $R[X]$, $f_{n+i}=r_if_n$ for a certain $r_i\in R$, and since $R$ is an integral domain, $$(f_n)=(r_0f_n)\subseteq(f_{n+1})=(r_1f_n)\subseteq(f_{n+2})=(r_2f_n)\subseteq\cdots$$ induces an ascending chain $(r_0)\subseteq(r_1)\subseteq(r_2)\subseteq\cdots$, which must stabilize eventually since $R$ satisfies ACCP and thus in turn implies that the chain $$(f_n)=(r_0f_n)\subseteq(f_{n+1})=(r_1f_n)\subseteq(f_{n+2})=(r_2f_n)\subseteq\cdots$$ stabilizes eventually.

  • $\begingroup$ is my proof ok? $\endgroup$ – user10024395 Apr 25 '15 at 10:42
  • $\begingroup$ @Aha I think it's OK. $\endgroup$ – Censi LI Apr 25 '15 at 11:20
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    $\begingroup$ I think this proof is not correct because actually $f_n=r_if_{n+i}$ in the sixth line , not the other way round. $\endgroup$ – Georges Elencwajg Dec 13 '15 at 11:16
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    $\begingroup$ @GeorgesElencwajg This proof can be "saved" if write $f_n=r_if_{n+i}$ for $i\ge0$, and notice that this leads to $\text{LC}(f_n)=r_i\text{LC}(f_{n+i})$ for all $i\ge0$, where $\text{LC}(f)$ denotes the leading coefficient of $f$. Then we get an ascending chain of principal ideals generated by $\text{LC}(f_k)$ for $k≥n$, and now the proof is (almost) over. $\endgroup$ – user26857 Dec 13 '15 at 15:40
  • $\begingroup$ @user26857: yes, your rectification has saved the proof. Well done! $\endgroup$ – Georges Elencwajg Dec 13 '15 at 15:49

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