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Theorem: Let $R$ be Noetherian and $P$ be a minimal prime ideal over $(a)$ for some nonunit $a$ of $R$. Then $\operatorname{ht}(P)\leq 1$.

My lecture notes prove this as follows.

  1. WLOG $R$ is local with unique maximal ideal $P$.
  2. Show $R/(a)$ is Artinian.
  3. Suppose for contradiction that there is a strict chain of primes $Q' \leq Q \leq P$. Let $I_m=\{x\in R : x/1 \in S^{-1}Q^m\}$ where $S=R\setminus Q$. Then $Q=I_1 \geq I_2 \geq I_3 \geq ...$. Use the fact that the chain $(R+I_m)/I_m$ must terminate to conclude that the chain $I_m$ terminates.
  4. Use Nakayama's lemma and the correspondence between primes in a ring and its localisation to get a contradiction.

I am fine with steps 1, 2 and 4. But I don't understand step 3. The proof in my notes (see below) appears to use the fact that $Q^m$ is prime, which it is not. (The bit I don't like is inside the red box) Proof of Krull's PIT Presumably I'm missing some obvious thing - could someone please set me straight?

user26857
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user3131035
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1 Answers1

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Let $f:R\to R_Q$ be the canonical homomorphism ($r\mapsto\frac r1$). Note that $I_m=f^{-1}(Q^mR_Q)$. In particular, $I_1=Q$ and $Q^m\subseteq I_m$.

Now let me recall that $I_m$ is called the $m$th symbolic power of $Q$ and it's denoted by $Q^{(m)}$. Since $Q^mR_Q$ is a $QR_Q$-primary ideal we get that $Q^{(m)}$ is $Q$-primary (why?), and now from $ax\in Q^{(m)}$, $a\notin Q$ we get $x\in Q^{(m)}$.

Edit. If don't want to use primary ideals, then the proof can go as follows:
One knows that $ax\in I_m\Leftrightarrow\frac{ax}{1}\in S^{-1}Q^m\Rightarrow\frac{ax}{1}=\frac{b}{s}$ with $b\in Q^m$ and $s\in S$ $\Rightarrow\frac{x}{1}=\frac{b}{as}\in S^{-1}Q^m\Rightarrow x\in I_m$. (I've used that $a\notin Q\Leftrightarrow a\in S$.)

user26857
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