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Let $R$ be a Noetherian local domain which is not a UFD and let $P$ be a height one prime ideal of $R.$ Can we find an element $x\in P$ such that $P$ is the only minimal prime ideal containing $x$?

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  • $\begingroup$ If we use Krull's height Theorem, then we can say that $P$ is minimal over an element $y\in R$. Since $R$ is Noetherian, we know that there are only finitely many minimal primes $P_i$ over (y). We may find a new element $x$ that avoids all these $P_i$, but there might be other primes that might be minimal over this new element $x$. So now i feel that the above statement might be false, but i have no counterexample. $\endgroup$
    – messi
    Jun 21, 2012 at 12:06
  • $\begingroup$ Related to Regular local ring and a prime ideal generated by a regular sequence up to radical. $\endgroup$
    – user26857
    Apr 28, 2013 at 14:23

3 Answers 3

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I'll prove @messi's assertion that $R_\mathfrak{m}$, where $R=k[X,Y,U,V]/(XV-YU)$ and $\mathfrak{m}=(X,Y,U,V)/(XV-YU)$, is a counterexample to the question, that is, there exists $P$ a height one prime ideal of $R_\mathfrak{m}$ which is not generated up to radical by a single element. In other words, the arithmetical rank of $P$ is greater than $1$.

Let's write $R=k[x,y,u,v]$ with $xv=yu$. Then $R$ is a noetherian graded $k$-algebra with $\dim R=3$. Moreover, Proposition 14.5 from R. Fossum, The Divisor Class Group of a Krull Domain, tells us that $R$ is a Krull domain and its divisor class group $\operatorname{Cl}(R)$ is isomorphic to $\mathbb{Z}.$ Take $\mathfrak{p}=(x,y)$. It's easy to check that $\mathfrak{p}$ is a height one prime ideal of $R$ and from Fossum we also know that $[\mathfrak{p}]$, the class of $\mathfrak{p}$, generates $\operatorname{Cl}(R)$. Suppose that $\mathfrak{p}=\sqrt{aR}$. Since $aR$ is a divisorial ideal it follows that $aR=\mathfrak{p}^{(t)}$, $t\ge 1$. This implies that $[\mathfrak{p}]$ is a torsion element of $\operatorname{Cl}(R)$, a contradiction. Now we can pass from $R$ to $R_\mathfrak{m}$ by using Corollary 10.3 from the same book.

Remark. In the view of curious' answer form this related topic, another way to solve this problem is to show that the local cohomology module $H_\mathfrak{p}^2(R)\neq 0$.

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Suppose $R$ is a Dedekind domain. Then height one prime ideals are the same as non-zero prime ideals, and are also the same as maximal ideals. If $x \in R$ is such that $P$ (some given non-zero prime ideal) is the only prime ideal containing $x$, then the factorization of $(x)$ into non-zero prime ideals contains just copies of $P$, and so $(x) = P^n$ for some $n$. Thus another way to phrase your question in this case is to ask whether or not $P$ is of finite order in the class group of $R$.

This is not true in general, because the class group of a Dedekind domain need not be finite in general. E.g. $\mathbb C[x,y]/(y^2 - x^3 + x)$ has infinite class group.

Added: I overlooked the fact that the OP asked for a local ring. I hope to add a counterexample satisfying this condition.

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  • $\begingroup$ I asked the question with the hypothesis being local Noetherian domain which is not a UFD. $\endgroup$
    – messi
    Jun 21, 2012 at 20:52
  • $\begingroup$ @messi: Dear messi, Ah, I didn't notice that you wanted a local example. Regards, $\endgroup$
    – Matt E
    Jun 22, 2012 at 7:01
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@Matt E A counterexample should be $S=R_m$, where $R=k[x,y,u,v]/(xv-yu)$ and $m=(x,y,u,v)/(xv-yu)$. But i cant check(or prove) in an elementary way that for every $b\in (x,y)$(which is a height 1 minimal prime), there is another minimal prime $P$ containing $b$.

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