I guess that the standard example of a prime power that is not a primary ideal is $$\mathfrak p^2 :=(x,z)^2\subset k[x,y,z]/(xy-z^2):=A.$$

Because $\mathfrak p^2 = (x^2,xz,xy)$, we see that $x\not \in \mathfrak p^2$ as well as $y\not\in r(\mathfrak p^2)$. But since $\mathfrak p \in\operatorname{Spec} A$ is also the standard example of a codimension 1 prime ideal that is not principal, I was wondering if there is the geometric intuition from the proof of the latter fact carries on to the former. In particular, how can I think geometrically to find examples of prime powers that are not primary? Should I look for codim 1 primes that are not principal? Should I think of varieties with singularities?

  • $\begingroup$ Why is this the standard example? What about $\mathfrak{p}=(y) \subset k[x,y]/(xy,y^2)$? Here $(y)^2=(0)$ is not primary, since $yx=0$, but $y\neq 0$, and $x$ is not nilpotent. In the words of Vakil, we are dealing with a line, with some "fuzz" around the origin. I can't quite figure out the geometric intuition here, but it is a bit clearer what's going on... $\endgroup$ Nov 12, 2015 at 13:43
  • 1
    $\begingroup$ Recall that $\mathfrak p^{(n)} = \mathfrak p^n$ iff $\mathfrak p^n$ is $\mathfrak p$-primary, where $\mathfrak p^{(n)} = (\mathfrak p^n)^{ec} \subset R \to R_{\mathfrak p}$ denotes the symbolic power of $\mathfrak p$. Then this gives a really good geometric interpretation of symbolic powers, in particular on page 9. $\endgroup$
    – Watson
    Feb 10, 2017 at 19:37

2 Answers 2


I don't really know how to think of powers of ideals geometrically. I believe this is fairly difficult answer how to conclude $p^n= p^{(n)}$, where the latter denote the $n$-th symbolic power of a prime ideal. At least in $\mathbb{A}^d$ powers of a complete intersection prime ideal is primary. In your example I believe that $p^{(2)} = (x)$ as expected since $CL(R) = \mathbb{Z}/(2)$.

There are a number of questions with symbolic powers of ideals in the direction of equality or containment. Eisenbud-Mazur conjecture says when $(R, m)$ is a regular local ring which contains a field of characteristic $0$ and $P$ is a prime ideal, then is $$ P^{(2)} \subseteq m P? $$

I hope somebody can provide a geometric explanation.


There is a geometric meaning to the difference between primary and prime as I will explain. Let me first recall some useful definitions:

Definition: Let $f: X \to Y$ be a morphism of schemes (you can think of affine scheme throughout and it will not make any difference), the scheme-theoretic image of $f$ is the universal (w.r.t. inclusion) closed subscheme of $Y$ though which $f$ factors.

Definition: A open subscheme $U \subset X$ is scheme-theoretically dense if for every $V \subset X$ open the scheme theoretic image of $U \cap V \subset V$ is $V$.

Definition: With notation as above $U$ is naively dense if $U \subset X$ is dense when we consider $U$ and $X$ as topological spaces (forgetting the structure sheaves). Equivalently we could ask for $U \subset X_{\text{red}}$ is scheme theoretically dense where $X_{\text{red}}$ is the underlying reduced scheme of $X$.

I hope you are convinced that these definitions are very natural and useful generalizations of concepts from point-set topology. We now have the following statements:

Let $I \subset R$ be an ideal of a ring $R$ (noetherian - not sure if this is needed here...). Then:

  1. The radical $\sqrt{I}$ is prime iff any open subset $U \subset \operatorname{Spec}R/I$ is naively dense.
  2. The ideal $I$ is primary iff any open subset $U \subset \operatorname{Spec}R/I$ is scheme-theoretically dense.

This statement is pretty straightforward to prove. The main thing one needs to use in order to prove this is the following (slightly non-trivial) statement:

Proposition: An open subset $U \subset X$ in a locally noetherian scheme is scheme theoretically dense iff it contains all associated points of $X$.

Recall: Associated points of $\operatorname{Spec}R$ are the associated primes of $R$.

So to summarize: Primary means geometrically that the scheme cut out by the ideal is not only irreducible but also has no embedded points. So an ideal with $\sqrt{I}$ prime and $I$ not primary cuts out a scheme which necessarily will have an embedded point. In your example $R= k[x,y,z] / (xy-z^2)$, $I=(x^2,xz,z^2)$ so $\operatorname{Spec}R/I$ is a nilpotent thickening of line (in 3-space) with an embedded point at the origin (this embedded point is in some sense equipped with an infinitesimal arrow pointing in the direction of the $x$-axis - reminiscent of the fact that intersecting $\operatorname{Spec}R$ with the $(x,y)$-plane ($\{z=0\}$) gives the union of the $x$ and the $y$ axes).


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