In a Noetherian integral domain, a principal prime ideal can't have proper non-zero prime ideals

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.