5
$\begingroup$

I am trying to construct an example of a ring satisfying the followings.

A commutative noetherian ring with $1$ which is neither domain nor local and has a principal prime ideal of height $1.$

I know that a local noetherian ring having a height $1$ principal prime ideal is a domain. Actually I wanted to prove this without the local condition. I couldn't prove this hence I am looking for a counterexample. I need some help. Thanks.

2 Answers 2

3

Let $k$ be a field and $A=k[x,y]/(xy)$. Then the ideal $(y-1)\subset A$ is a principal height $1$ prime. (Note that this ring is not local, and the non-localness is essential to the example in that if you localize at $(y-1)$ then the ring becomes a domain.)

As a hint to what's going on here that can't happen in the local case, notice that $(y-1)x=-x$, so $x$ is divisible by $y-1$ arbitrarily many times. In a local Noetherian ring this would imply $x=0$ by the Krull intersection theorem.

3

Let $R$ be a noetherian ring with principal prime ideal $P$ of height $1$.

Then $R\times R$ is Noetherian, non local, not a domain, and still has a principal prime ideal of height $1$: $P\times R$.

  • 0
    So, like $\mathbb Z\times \mathbb Z$. – 2017-02-25
  • 0
    $P$ has height $1$ in $R$ by choice of $P$, but why does $P\times R$ have height $1$ in $R\times R\,?$ – 2017-02-25
  • 1
    @quasi The prime ideals of the product have to be of the form $P\times R$ or $R\times P$ for some prime ideal $P$ of $R$. Therefore the primes contained in $P\times R$ look like $P'\times R$ for a prime ideal $P'$ contained in $P$. – 2017-02-25
  • 0
    But if the prime ideal $P \times R$ has height $1$, what is the prime ideal of $R \times R$ below $P \times R\,?$ Ok, so you're saying $0\times R$ is the one below it? That makes sense. But wait; $0 \times R$ is not height $0$, is it? – 2017-02-25
  • 0
    Ok, I got it. If $R$ is a domain, $0 \times R$ _does_ have height $0$ in $R \times R$. If $R$ is not a domain, then whatever prime ideal $Q$ in $R$ of height $0$ was below $P$, the prime ideal $Q \times R$ is a height $0$ prime ideal of $R \times R$ which is below $P \times R$. So I withdraw my question, and thanks. – 2017-02-25
  • 0
    @quasi well it doesn't really matter if $R$ is a domain. I have just explained why the primes contained in $R\times P$ correspond with primes in $P$. – 2017-02-25
  • 0
    Yes, I observed that in the last part of my last comment. That comment was kind of written in "real-time" (by the end of the comment, I saw that R being a domain wasn't needed). – 2017-02-25
  • 0
    @quasi ok, I wasn't sure. – 2017-02-25