Following [1], we say that a ring $R$ satisfies Property (A) if every finitely generated faithful ideal of $R$ is regular.
Claim 1. [1, Section 2] Let $R$ be a commutative unital ring.
- If the zero ideal of $R$ has a primary decomposition in $R$ then $R$ has Property (A). In particular, any Noetherian ring has Property (A).
- If $R$ is zero-dimensional, then $R$ has Property (A).
Thus a counter-example should be a non-Noetherian ring of positive Krull dimension, should this dimension be well-defined.
We present now an example, due to D. Anderson and J. Pascual, of a ring $R$ with a two-generated faithful ideal $I$ which is not regular.
This construction relies on Nagata's idealization method. Given a commutative unital ring $S$ and an $S$-module $A$, we define the commutative ring $R \Doteq S \oplus A$ with identity $(1, 0)$ by $$(s, a) \cdot (s', a') \Doteq (ss', sa' + s'a).$$
We can identify (and we will) $S$ with the subring $S \oplus \{0\}$ of $R$.
It is easily seen that $Z(S \oplus A) = \{(s, a) \, \vert \, s \in Z(S) \cup Z(A)\}$ where $Z(R)$ is the set of zero divisors of $R$ and $Z(A) \Doteq \{ s \in S \, \vert \, sa = 0 \text{ for some non-zero } a \in A \}$. If $S$ is an integral domain and $\mathcal{P}$ is a set of prime ideals of $S$ and $A = \bigoplus_{\mathfrak{p} \in \mathcal{P}} S/\mathfrak{p}$, then $Z(A) = \bigcup_{\mathfrak{p} \in \mathcal{P}} \mathfrak{p}$. Thus, if the set $S \setminus S^{\times}$ of non-units of $S$ is covered by the prime ideals in $\mathcal{P}$, then $(s, a) \in R = S \oplus A$ is regular if and only if $s$ is a unit of$S$. This covering property holds for instance if $\mathcal{P}$ is the set of height one prime ideals of $S$ for $S$ a Noetherian normal domain [2, Theorem 11.5].
Claim 2. [1, Example 2.1] Let $k$ be a field and let $S = k[X, Y]$ be the polynomial ring over two variables with coefficients in $k$. Let $A = \bigoplus_{\mathfrak{p}}S/\mathfrak{p}$ where $\mathfrak{p}$ ranges over the height one prime ideals of $S$. Let $R = S \oplus A$ be the idealization of $A$. Let $I$ be the ideal of $R$ generated by $(X, 0)$ and $(Y, 0)$. Then $I$ is faithful but no regular.
Showing that $I$ is faithful will be easy. In order to show that any element of $I$ is a zero divisor, the important fact to note is that $S \setminus S^{\times}$ is covered by the height one prime ideals of $S$.
Proof of Claim 2.
Let us show first that $I$ is faithful. Consider $r = (s, a) \in R$ such that $rI = \{0\}$. Since $S$ is a domain, it is trivial to check that $s = 0$. Let us write $a = (a_{\mathfrak{p}} + \mathfrak{p})_{\mathfrak{p}}$ with $a_{\mathfrak{p}} \in S$. Since $Xa = Ya = 0$, we have $a_{\mathfrak{p}} \in \mathfrak{p}$ for every $\mathfrak{p}$ distinct from $(X)$ and $(Y)$. As $Y a_{(X)} \in (X)$ and $Xa_{(Y)} \in (Y)$, we deduce that $a = 0$.
Let us prove now that every element of $I$ is a zero-divisor of $R$. Since $S$ is a Noetherian normal domain (it is actually a Noetherian UFD), every non-unit $s$ of $S$ is contained in some height one prime ideal of $S$ [2, Theorem 11.5]. As every element $r$ of $I$ is of the form $(s, a)$ with $s \in (X, Y)$, every such $r$ is a zero-divisor by the remark preceding the claim . Indeed, take $\mathfrak{q}$ a prime of height one that contains $s$ and let $\delta_{\mathfrak{q}} \in A$ be defined by $(\delta_{\mathfrak{q}})_{\mathfrak{p}} = 1 + \mathfrak{p}$ if $\mathfrak{p} = \mathfrak{q}$ and $(\delta_{\mathfrak{q}})_{\mathfrak{p}} = \mathfrak{p}$ otherwise. Then we have $(s, a) \cdot (0, \delta_{\mathfrak{q}}) = (0, 0).$
[1] D. Anderson and J. Pascual, "Regular ideals in commutative rings, sublattices of regular ideals, and Prüfer rings", 1987.
[2] D. Eisenbud, "Commutative algebra with a view towards geometric algebra", 1995.
