It's very well known that if $A$ is a Noetherian normal domain then every prime ideal $\mathfrak{p}\in \mathsf{Spec} \, A$, associated to a principal ideal $aA\subset A$, has height equal to $1$. Moreover, in this setup one can prove that $\mathfrak{p}A_\mathfrak{p}$ is a principal ideal.

My question is the following.

QUESTION: If $A$ is a Noetherian domain, not necessarily normal, and $\mathfrak{p}\subset A$ is a prime associated to a principal ideal $aA$, then is it true that $\mathsf{ht} \, \mathfrak{p}=1$?


1 Answer 1


Geometrically, the question you ask is the following:

Question. Let $X$ be an integral Noetherian scheme, and let $D \subseteq X$ be a Cartier divisor. Then can $D$ have embedded points?

Indeed, your question is the case $X = \operatorname{Spec}(A)$ and $D = V(a)$ for some element $a \in A$.

Answer. The answer is no in general, but it's a bit complicated to write down an actual example. We will use Example III.9.8.4 of Hartshorne's Algebraic geometry, which we recall below. (Note that Hartshorne doesn't explain his notation, but it works out to what we compute below.)

In the example, we start with the twisted cubic curve $X_1 \subseteq \mathbb P^3$ given by the map \begin{align*} \mathbb P^1 &\to \mathbb P^3\\ [x:y] &\mapsto [x^3:x^2y:xy^2:y^3]. \end{align*} Explicitly, this is given by the equations $V(x_0x_3-x_1x_2,x_1^2-x_0x_2,x_2^2-x_1x_3) \subseteq \mathbb P^3$. However, Hartshorne uses a linear change of variables, and instead uses \begin{align*} \mathbb P^1 &\to \mathbb P^3\\ [x:y] &\mapsto [y^3:x^2y-y^3:x^3-xy^2:xy^2]. \end{align*} Next, we look at the constant family $\mathbb P^1 \times U \to U$, where $U = \mathbb A^1 \setminus \{0\} \subseteq \mathbb A^1$. We map this to a family of subschemes of $\mathbb P^3$ by \begin{align*} \mathbb P^1 \times U &\to \mathbb P^3 \times U\\ ([x:y],a) &\mapsto ([y^3:x^2y-y^3:x^3-xy^2:axy^2],a). \end{align*} Then Proposition III.9.8 shows that this $U$-family of subschemes of $\mathbb P^3$ extends uniquely to a flat family $X$ over $\mathbb A^1$. Moreover, this is done by taking the scheme-theoretic closure, hence $X$ is integral since $X|_U = \mathbb P^1 \times U$ is.

It is computed in Example III.8.9.4 that on the affine patches in $\mathbb P^1$ and $\mathbb P^3$ with coordinates $[t:1]$ and $[1:x:y:z] = [1:t^2-1:t^3-t:at]$, the equations for $X$ can be given as $$X = V(a^2(x+1)-z^2,ax(x+1)-yz,xz-ay,y^2-x²(x+1)) \subseteq \mathbb A^3 \times \mathbb A^1.$$ Restricting to the divisor $a = 0$, we get the ideal $$I_0 = (z^2,yz,xz,y^2-x^2(x+1)) \subseteq k[x,y,z].$$ We see that this has an embedded prime $(x,y,z) = \operatorname{Ann}(z)$. Since $\dim X = 2$, we see that the closed point $(x,y,z)$ has height $2$. $\square$

Remark. If $A$ is Cohen-Macaulay, then so is $A/(a)$, and hence it has no embedded primes. This shows that we cannot make counterexamples that are e.g. complete intersections. Thus, the above may well be a 'minimal example' of some sort (at least in the sense that the twisted cubic is one of the easiest examples of a subscheme that is not a complete intersection).


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