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Let $R$ be an integral domain and Noetherian. Let $P \subset R$ be a non zero prime ideal. Prove that if $P$ is principal then there is no prime ideal $Q$ such that $0 \subsetneq Q \subsetneq P$.

I couldn't do much. A Noetherian ring can be defined as (1) a ring in which every ideal is finitely generated or, equivalently, (2) a ring which satisfies the ascending chain condition.

Since $P$ is principal and nonzero then $P=\langle p \rangle$ with $p \in R \setminus \{0\}$. Suppose there is $Q$ prime ideal such that $0 \subsetneq Q \subsetneq P$. Using definition (1), we have $Q=\langle q_1,...,q_r\rangle$. Since $Q \subset P$ then $q_i=pp_i$ for all $1\leq i \leq r$. By hypothesis $Q \neq P$, from this and the fact that $Q$ is prime, it follows $p_i \in Q$ for all $i$.

Here I got completely stuck, I would appreciate hints to finish the problem.

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  • $\begingroup$ @user26857 I'll study the theorem and see how this proposition derives from it but is there an alternative approach without having to use this theorem? I mean, in a way that the result follows directly from the definitions of Noetherian, principal, integral and prime. $\endgroup$
    – user16924
    Dec 2, 2014 at 18:13
  • $\begingroup$ In fact the claim holds for integral domains satisfying ACC on principal ideals (ACCP for short). $\endgroup$
    – user26857
    Jan 25, 2015 at 18:56
  • $\begingroup$ @user16924 : let $x \in Q \setminus \{0\}$, write $x=x_1a$ where $P = (a)$. Since $Q$ is prime, we get either $a \in Q$ and we're done ($Q=P$), or $x_1 \in Q \subset P=(a) \implies x_1=x_2a$. You get a sequence $$(x_1) \subset (x_2) \subset \cdots$$ which must stabilize (ACC implies ACCP), so that $x_{n+1}=rx_n=r(ax_{n+1}) \implies ar=1$ since $R$ is a domain and $x_{n+1} \neq 0$ since $x \neq 0$. Then $P=(a)=R$, contradiction. $\endgroup$
    – Watson
    Jan 25, 2017 at 16:10

3 Answers 3

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Unless I'm missing something here, this is not too complicated. I'll leave it as a nice exercise in using definitions that in a Noetherian domain, every nonzero prime ideal contains an irreducible element. So there exists $q\in Q$ irreducible. Then $q\in P$, so $p\mid q$; since $q$ is irreducible and $p$ is not a unit, $q$ is an associate of $p$, so $(q)\subseteq Q \subseteq P = (p) = (q)$ and all the containments are equalities.

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  • $\begingroup$ Thanks, this is much simpler than the accepted answer. @John Brevik $\endgroup$
    – Babai
    Oct 23, 2019 at 15:01
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Assume the hypothesis. Let $q$ be a non-zero element of $Q$. Suppose there exists an integer $k$ such that $q$ may not be written as $p^{k+1}u$ for any element $u$. Let $n$ be the least such integer. Then $q$ may be written as $p^nu$ for some $u$. But since $p$ is not in $Q$, $u$ must be in $Q$ and on the form $pu'$, a contradiction.

Thus we conclude that for any integer $k$, there is an element $u_k$ such that $q = p^ku_k$. Consider the ideal $I$ generated by $u_k$ for each integer $k$. By the noetherian hypothesis, $I$ is generated by $u_1,...,u_n$ for some $n$. Thus $u_{n+1} = a_1u_1+...+a_nu_n$ for elements $a_i$ and

$$q = p^{n+1}u_{n+1} = a_1p^{n+1} u_1 + \cdots + a_np^{n+1}u_n = qa_1p^n + \cdots + qa_np.$$

Hence $q(1-p(a_1p^{n-1}+\cdots+a_n)) = 0$. $R$ being an integral domain, this implies that $1 = pA$ for some element $A$, hence $P$ is not a proper ideal. A contradiction.

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  • $\begingroup$ In the second part of your answer you actually proved that $\bigcap_{n\ge 1}p^nR=(0)$ for any non-invertible element $p∈R$. $\endgroup$
    – user26857
    Dec 1, 2014 at 22:28
  • $\begingroup$ @Bubbles I have some doubts: why there has to be a least integer $k$? I know the statement is true for positive integers $k$, but not for all integers. Then, you've picked the first $n$ $u_k$ as the generators but those aren't necessarily the generators. And finally, were we assuming that $P$ is a proper ideal? For $P$ we were only assuming it is principal, non zero and prime so I don't see why we've arrived to a contradiction in the last sentence, why can't $P=R$? $\endgroup$
    – user16924
    Dec 2, 2014 at 16:49
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    $\begingroup$ @user16924 1. Of course, I mean positive integers only. 2. I am using a common equivalent definition of the noetherian property. The noetherian property ensures that the ideal I is finitely generated. Say the generators are v_1,...,v_m. Then each of these v_i's may be written as finite linear combinations of the u_k's. Since we use only finitely many u_k's to write any v_i, choose the largest index n which is used. Then we may write each v_i as a linear combination of the u_k's for k <= n, and thus u_1,...,u_n generate the ideal. 3. Every prime ideal is by definition a proper ideal. $\endgroup$
    – Mellon
    Dec 2, 2014 at 20:11
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Use the following corollary to Krull's principal ideal theorem: Prime ideals in a Noetherian satisfy the descending chain condition, with length of primes descending from a prime $P$ bounded by the number of generators of $P$.

Thus any principal prime must contain at most one other prime. Since the ring is an integral domain this must be zero ideal.

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