Can someone please give me an example of a Noetherian normal local domain of dimension two such that there exists a prime ideal $P$ of height one having the property $P^{(n)}$ is not a principal ideal for any $n \geq 1$. Here $P^{(n)}$ is the symbolic $n$-power.
Symbolic Powers – Prime Ideal of Height One
ac.commutative-algebraag.algebraic-geometryra.rings-and-algebrassymbolic computation
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I cannot fully answer the question but some keywords:
If you drop 1. what you get is called a weakly Krull domain.
If you just weaken 1. to valuation ring what you get is called a generalized Krull domain.
There are a bunch of related notions. Thus, yes, you get something new and these types of rings got studied.
Anon's answer gives a beautiful geometric proof when $A$ is a variety. Below I am trying to give some geometric interpretation of the usual algebraic proof.
First a disclaimer: I'm not an algebraist, so the explanation below will be a learner's perspective, probably from an analytic perspective, and thus may seem idiosyncratic to experts.
I am going to start by interpreting two key ingredients used in the proof.
1. Symbolic power
Let $\frak p$ be a prime ideal of $A$. I think of the localization $A_{\frak p}$ as capturing the behavior of functions on a neighborhood of the generic point of $V({\frak p})$. To see what I mean, take for example $A=k[x,y]/(xy,y^2)$ and ${\frak p}=(y)$. Geometrically $Spec A$ is the $x$-axis plus some fuzz of order 2 at the origin, and $V({\frak p})$ is just the $x$-axis. Now look at $y\in A$. We have $y$ is nonzero in $A$ but becomes zero in $A_{\frak p}$ (where $y=xy/x=0$.) The geometric explanation is that $y$ is indeed zero on a neighborhood of $(x_0,0)$ for any $x_0\neq0$, because there is no fuzz around that point. The only reason for $y\neq0$ in $A$ is that it vanishes only to order 1 near the origin, which is captured by the fuzz (of order 2) there. But this happens at a single point in $V({\frak p})$ (a so called embedded prime), so that behavior is not generic (in the colloquial sense of the word), so we can still say that $y$ vanishes on a neighborhood of the generic point of $V({\frak p})$, which explains why it vanishes in $A_{\frak p}$.
Generalizing this example a bit, if we take $A=k[x,y]/(xy^n,y^{n+1})$, we see that $Spec A$ is the $x$-axis with multiplicity $n$ (in other words, with fuzz of order $n$ in the $y$ direction), plus some fuzz of order $n+1$ in the $y$ direction at the origin. Again let ${\frak p}=(y)$. Then ${\frak p}^{(m)}$ (the symbolic power) consists of functions that vanish to order $m$ at the generic point of $V({\frak p})$. Thus for $m<n$, ${\frak p}^{(m)}=(y^m)$, and for all $m\ge n$, ${\frak p}^{(m)}=(y^n)$. The fact that $y^n$ only vanishes to order $n$ near the origin does not matter, because again the origin is only a single point, not generic enough for $V({\frak p})$.
2. Nakayama's Lemma
The OP doesn't ask about this but since this lemma is used often in the proof I will also try to interpret it geometrically. Let $(A,{\frak m})$ be a local ring and $M$ be a finitely generated $R$ module. I think of $A$ as the germ of holomorphic functions on a neighborhood of the origin and $M$ as some sort of holomorphic vector bundle over that neighborhood. The condition of Nakayama's Lemma says that $M={\frak m}M$. Iterating this we get $M={\frak m}^nM$ for any $n\in\mathbb N$. This means that all sections of $M$ vanishes to arbitrarily high orders near the origin. By the holomorphic heuristics, all sections of $M$ vanish identically, so $M=0$.
Now we turn to the actual proof of Krull's principal ideal theorem, which can found, for example, here.
By standard reduction, we can assume that $(A,\frak m)$ is a local domain, $f\neq0$ with a minimal prime ideal $\frak m$. Assume that there is a prime ideal $\frak p$ properly contained in $\frak m$, and our aim is to show that ${\frak p}=(0)$.
Consider $V(f)$. Since $\frak m$ is the maximal ideal of $A$ while at the same time a minimal prime ideal of $f$, $V(f)$ contains a single (scheme-theoretic) point, namely $\frak m$ itself (in algebraic terms, $A/(f)$ is an Artinian local ring), plus a finite order of fuzz around that point. Consider the symbolic power ${\frak p}^{(n)}$, that is, the ideal of functions that vanish to at least of order $n$ at a generic point of $V({\frak p})$. Then ${\frak p}^{(n)}|_{V(f)}$ (algebraically this is the ideal ${\frak p}^{(n)}+(f)/(f)$ in $A/(f)$) will include a finite order of fuzz near the unique point $\frak m$ of $V(f)$. The larger $n$ is, the more fuzz it can possibly include. Since the total order of fuzz around $\frak m$ is finite, for $n\gg1$ the order of fuzz included in ${\frak p}^{(n)}|_{V(f)}$ will not change (this is the DCC property for Artinian rings.) Written out algebraically, this amounts to ${\frak p}^{(n)}+(f)={\frak p}^{(n+1)}+(f)=\cdots$.
Now take $x\in{\frak p}^{(n)}$. Then $x$ vanishes at least to order $n$ generically on $V({\frak p})$, but we can write $x=y+fr$, where $y$ vanishes at least to order $n+1$ generically on $V({\frak p})$, so $fr$ vanishes at least to order $n$ generically on $V({\frak p})$. But $\frak p$ is not a point in $V(f)$ (whose only point is $\frak m$), so $f|_{V({\frak p})}\neq0$. Since $V({\frak p})$ is irreducible, $f$ does not vanish to any order generically on $V({\frak p})$. Hence $r$ vanishes at least to order $n$ generically on $V({\frak p})$. Translating back to algebra, we have ${\frak p}^{(n)}={\frak p}^{(n+1)}+f{\frak p}^{(n)}$.
Now we consider the module ${\frak p}^{(n)}/{\frak p}^{(n+1)}$. The above identity shows that every element in this module is a multiple of $f$. Iterating this we know that every element is a multiple of $f^m$ for all $m\in\mathbb N$. Since $f$ vanishes at $\frak m$, every element in ${\frak p}^{(n)}/{\frak p}^{(n+1)}$ vanishes to arbitrarily high order at $\frak m$. By Nakayama's Lemma, the module vanishes identically, so ${\frak p}^{(n)}={\frak p}^{(n+1)}$.
Now pass to the localization $A_{\frak p}$, that is, we forget about the behavior of functions at specific points of $V({\frak p})$, and only considers its behavior on a neighborhood of the generic point of $V({\frak p})$. Since the coordinate ring takes the generic behavior into account, it's unnecessary to restate it for the localization of the symbolic power. Thus ${\frak p}^{(n)}A_{\frak p}={\frak p}^nA_{\frak p}$, and it's stationary for $n\gg1$. Then any element in ${\frak p}^nA_{\frak p}$ vanishes to arbitrarily high order near $\frak p$. By Nakayama's Lemma again, ${\frak p}^nA_{\frak p}=(0)$.
To wrap up, I have to use some algebra (my previous geometric argument somewhat contradicts my earlier points.) Since $A$ is a domain, the localization $A\to A_{\frak p}$ is injective, so ${\frak p}^n=(0)$ in $A$. Again because $A$ is a domain, ${\frak p}=(0)$.
Best Answer
Take $E$ an elliptic curve $zy^2 - x(x-z)(x-tz)$ say over $\mathbb{C}$ and choose a point $Q$ of infinite order (or so for instance the divisor $Q - O$ has infinite order in the divisor class group, here $O$ is the point at infinity).
It follows that in the graded ring of dimension $2$, $$\mathbb{C}[x,y,z]/(zy^2 - x(x-z)(x-tz))$$ that the homogeneous ideal corresponding to $Q$, call it $P$, has the property that $P^{(n)}$ is never principal. Indeed, if $P^{(n)} = (f)$, then $(f)$ is homogeneous and the corresponding divisor $\mathrm{Div}_E(f/z^{\deg f})$ is linearly equivalent to 0. This contradicts the infinite order of $P$.
One should point out that for any rational double point, the divisor class group is finite by a result of Lipman (if and only if under some hypotheses), so one has to leave the setting of rational singularities. The cone over an elliptic curve is the probably the simplest singularity that is not rational.