# When is a faithful ideal of a unital commutative ring a regular ideal?

Let $$R$$ be a commutative ring with identity and let $$I$$ be an ideal of $$R$$. We denote by $$\mathrm{Ann}(I)$$ the annihilator of $$I$$, that is $$\mathrm{Ann}(I) \Doteq \{ r \in R \, \vert \, rI = \{0\}\}$$.

We say that $$I$$ is a faithful ideal of $$R$$ if $$\mathrm{Ann}(I) = \{0\}$$. We call $$r \in R$$ a regular element of $$R$$, if it is not a zero divisor, i.e., if $$\mathrm{Ann}(Rr) = \{0\}$$.

We say that $$I$$ is regular ideal of $$R$$ if $$I$$ contains a regular element. A regular ideal is clearly a faithful ideal.

Question 1. Do you know of an example of a two-generated ideal which is faithful but not regular?

Question 2. If such an example does exist, then under which conditions is a finitely generated faithful ideal also regular?

• My dissertation asked a more general question than this (concerning modules.) In short, I never found an example or proof either way either for the special case of ideals. If a counterexample exists, it would be extremely interesting to me. Sep 16, 2019 at 17:02
• I am particulary interested in the case of $a = 2$ and $b$ can be any ring element. Does that change anything? And is it more likely that the claim holds? Sep 16, 2019 at 17:06
• The answer has been given here: mathoverflow.net/questions/341760/… by Luc Guyot. Sep 18, 2019 at 12:57
• Thank you... I will consult that resource. By the way, it is considered bad form to crosspost to both here and mathoverflow without reason (for example, a question could turn out to be too hard here, and so it might be worthwhile posting there.) Sep 18, 2019 at 13:15
• It posted it there because it was not answered here. Should i mark it next time or how is it done normaly? Sep 18, 2019 at 14:28

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.

Your statement is not true in general. See ''F. Azarpanah, O. A. S. Karamzadeh and A. Rezai Aliabad, On ideals consisting entirely of zero divisors, $$\textit{Comm. Algebra} \textbf{28}$$ (2000) 1061--1073.'' for more information. The following lemma may be useful.

Lemma: Let $$R$$ be a reduced ring and $$a, b\in R$$ such that $$ab=0$$. Then $$Ann_R(\langle a, b\rangle)= Ann_R(a-b)$$.

Proof: Clearly $$Ann_R(\langle a, b\rangle)\subseteq Ann_R(a-b)$$. Now let $$r\in Ann_R(a-b)$$. Thus, $$ra=rb$$, and so $$(ra)^2=0$$. Hence, $$rb=ra=0$$, and so $$Ann_R(\langle a, b\rangle)\supseteq Ann_R(a-b)$$.

• In your example we always have $r\in \text{Ann}(a) \cap \text{Ann}(b)$ so in particular $\text{Ann}(a) \cap \text{Ann}(b) \neq \{0\}$ if $r\neq 0$. I do not understand how one can construct a counterexample like this. Sep 16, 2019 at 17:24