In my previous question I showed that the height of an irreducible polynomial $f \in \mathbb{C}[x_1, \cdots x_n]$ is $1$.
This I was able to generalize to any Noetherian domain. Basically, the idea of a smallest degree corresponds to stopping of an increasing chain. Here are the details,
If $\mathcal{P} \subset (f)$ is a prime ideal then like before we have $\mathcal{P}\times f = \mathcal{P}$. If $ x \in \mathcal{P}$ is a non zero element then we have $x = x_1 = x_2 f$ for some $x_2$ and in this way we get $x_3, x_4, \cdots$ and so on but this is where the noetherian condition comes in. IT is clear clear that these $x_i$s form an increasing ideal chain,
$$(x_1) \subset (x_2) \cdots$$ so this must terminate.
So, at some point $n$ we get $x_n = x_{n+1}f$ and $x_{n+1} = x_n r$
From here $x_n = x_n r f$ and BECAUSE we are in a domain $f$ becomes invertible which is a contradicion.
So, we must have that $\mathcal{P} = 0$. Done.
But My Question is how do I generalise this to the case we are potentially have zero divisors?
I want to do this without invoking the Krull's Principle Ideal theorem which I got to know about in the answer to my previous post.
I want to basically prove it myself and I think this will be a step towards it.
In the case of general Noetherian ring I need to show that $\mathcal{P}$ is a minimal prime. I have made an observation that all elements of $\mathcal{P}$ are zero divisors. Because we got $x_n(1 - rf) = 0$. If we assume that $1 - rf \neq 0$ then we get $x_n$ is a zero divisor. We can uniterate and show that $x_1$ is a zero divisor.
So, all elements of $\mathcal{P}$ are zero divisors. How can I now show that $\mathcal{P}$ is a minimal prime?
We have $x_n(1-rf)=0$ with $1-rf\neq 0$. Notice that $(1-rf)\notin \mathcal P$. Similarly, since $x_{n-1}=x_nf$, we have $x_{n-1}(1-rf)=0$. Repeating the argument, we can show that for any $x\in\mathcal P$, there exists some $s\notin \mathcal P$ such that $sx=0$.
Now localise at $\mathcal P$ and show that $\mathcal P A_{\mathcal P}=0$. Since $\mathcal P A_{\mathcal P}$ is the unique maximal ideal of $A_{\mathcal P}$, we see that the only prime ideal of $A_{\mathcal P}$ is $\mathcal P A_{\mathcal P}$. Since the prime ideals of $A_{\mathcal P}$ are in one-to-one correspondence with the prime ideals of $A$ contained in $\mathcal P$, we conclude that $\mathcal P$ is a minimal prime ideal.