proving that the height of $(X)$ in $R[X]$ is equal to 1

Question

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?

Answer 1

This follows directly from Krull's principal ideal theorem.