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 $\mathsf{ht} \, \mathfrak{p}=1$?



Primes associated to a principal ideal
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).


