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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?

David C. Ullrich
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Subham Jaiswal
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1 Answers1

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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.

cqfd
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  • Thanks! I was trying to show that each element of $\mathcal{P}$ becomes nilpotent in the localisation. So, i was trying to show that for each $x \in \mathcal{P}$ there is a $n$ and $s$ in the $\mathcal{P}^c$ such that $x^n \cdot s = 0$. This is equivalent to $\mathcal{P}$ being minimal but infact it turns out as you have shown that something stronger is true. However, I am a bit surprised(though your proof seems correct)it because it seems like $A_{p}$ is a field. Is there a way to explain that? how would one expect it here? because $\mathcal{P}$ seems to be pretty general miniimal prime. – Subham Jaiswal Jul 05 '21 at 02:42
  • @SubhamJaiswal That was my initial plan. But then I noticed the fact I wrote in the first paragraph and was able to conclude quickly. However, right now, I can't think of a satisfactory answer to why $A_p$ is a field. It might be because $\mathcal P$ is inside a principal ideal, but I am not sure. I will think about it. – cqfd Jul 05 '21 at 03:47
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    By the way, I just realised that there is a slick way to conclude if you already know the result holds for Noetherian domains. Assume $\mathcal P$ is not minimal. Then you can find a prime ideal $\mathcal Q$ such that $\mathcal Q\subset \mathcal P$. Now work over the integral domain $A/\mathcal Q$ to conclude. – cqfd Jul 05 '21 at 03:51
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    See also the lemma given here. – cqfd Jul 05 '21 at 04:09
  • @Subham If you look at the proof there, Atticus finds an element $1-xy$, which kills off $\mathcal P$. We are doing a similar thing here. However, I was content with finding some $s$ for each $x\in\mathcal P$ satisfying $sx=0$. Anyways, in light of Attcus's proof, it is now clear why $A_p$ is a field. – cqfd Jul 05 '21 at 05:49