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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?

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  • $\begingroup$ 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. $\endgroup$
    – rschwieb
    Sep 16, 2019 at 17:02
  • $\begingroup$ 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? $\endgroup$
    – kevkev1695
    Sep 16, 2019 at 17:06
  • $\begingroup$ The answer has been given here: mathoverflow.net/questions/341760/… by Luc Guyot. $\endgroup$
    – kevkev1695
    Sep 18, 2019 at 12:57
  • $\begingroup$ 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.) $\endgroup$
    – rschwieb
    Sep 18, 2019 at 13:15
  • $\begingroup$ It posted it there because it was not answered here. Should i mark it next time or how is it done normaly? $\endgroup$
    – kevkev1695
    Sep 18, 2019 at 14:28

2 Answers 2

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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.

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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 $Ann_R(\langle a, b\rangle)\supseteq Ann_R(a-b)$.

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    $\begingroup$ 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. $\endgroup$
    – kevkev1695
    Sep 16, 2019 at 17:24

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