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Let $R$ be a Noetherian ring. I want to prove that $\operatorname{height}(X) = 1$.

Here is how i do it: It is not hard to show that $\operatorname{height}(X) = \operatorname{height}(p,X)$, where $p$ is a minimal prime of $R$. Since the ring extension $R \rightarrow R[X]$ is flat, we have that $\operatorname{height}(p,X) = \operatorname{height}(p) + \dim (R[X])_{(p,X)}/ p (R[X])_{(p,X)} = \dim (R[X])_{(p,X)}/ p (R[X])_{(p,X)}$ since $p$ is minimal. Now $(R[X])_{(p,X)}/ p (R[X])_{(p,X)} \cong (R/p[X])_{(X)}$. Let $Q \neq (X) (R/p[X])_{(X)}$ be a prime ideal of $(R/p[X])_{(X)}$. Then $Q = (X) Q$ and by NAK $Q=0$ and so $\dim (R[X])_{(p,X)}/ p (R[X])_{(p,X)} = 1$.

Question: Is there an alternative proof?

Manos
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    Doesn't this follow immediately from Krull's principal ideal theorem? – Seth Mar 19 '14 at 20:27
  • Thanks, I made it an answer. – Seth Mar 19 '14 at 20:45
  • By the way, any comment on the proof that i give? – Manos Mar 19 '14 at 20:46
  • Sorry it looks complicated and I'm not that comfortable with commutative algebra yet so I can't really comment. I just figured the result should follow from some basic theorem on dimension so I opened up Atiyah and Macdonald. I've covered chapters 1-9 and part of 10 so far (and I just know a few random facts about dimension). – Seth Mar 19 '14 at 20:50
  • No problem, good job. – Manos Mar 19 '14 at 20:51

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This follows directly from Krull's principal ideal theorem.

Seth
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