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Can a finitely generated module $M$ over a commutative ring have $\operatorname{Ann}(x) \neq 0$ for all $x \in M$ while $\operatorname{Ann}(M) = 0$?

It's not difficult to show that there is no such module if the ring is a integral domain. For general, I guess the answer is yes. But I failed to find a desired example.

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3 Answers 3

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Let $R$ be a UFD which is not a PID, e.g. $R=\mathbb Z[X]$, and $M=\bigoplus_{p\text{ prime}} R/(p)$. Note that every non-invertible element of $R$ is a zero-divisor on $M$. Let $I=(p_1,p_2)$ with $p_1,p_2$ primes such that $I\ne R$. Since $I$ does not contain invertible elements, every element of $I$ is a zero-divisor on $M$. Moreover $(0:_MI)=0$.

Now consider the idealization $A=R(+)M$ of the $R$-module $M$. Let $J=IA$. We have that $J$ is finitely generated ideal and consists of zero-divisors, but no non-zero element of $A$ annihilates $J$.

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  • $\begingroup$ A very elegant example. Thank you very much. $\endgroup$
    – Censi LI
    Commented Mar 15, 2015 at 7:35
  • $\begingroup$ I remember seeing this when you first wrote it, and I forgot to ask a follow up. Is this something you had previously seen, or is it your own idea? I only ask because it seems these examples are fairly rare. If there was more literature on them, I would be interested in looking them up. Thanks $\endgroup$
    – rschwieb
    Commented Apr 15, 2019 at 13:53
  • $\begingroup$ This example is borrowed from Kaplansky, Commutative Ring Theory, Exercises 6, 7, Section 2-2. $\endgroup$
    – user26857
    Commented Apr 15, 2019 at 17:20
  • $\begingroup$ But the idealization is a well known trick for producing (counter)examples, and I used it in few answers on this site. $\endgroup$
    – user26857
    Commented Apr 15, 2019 at 17:21
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Here is a proof that no such examples exist over integral domains:

Let $M$ be finitely generated over an integral domain $R$. Let $x_1,\dots,x_n$ be generators of $M$. Take $0 \neq a_i \in \operatorname{ann}(x_i)$ for every $i=1,\dots,n$. Define $a=a_1 \cdots a_n$. Since $R$ is an integral domain, $a\neq 0$ and $a \in \operatorname{ann}(M)$.

Let me now answer the question negatively in the case where $R$ is Noetherian and $0 \neq M= J$ is an ideal of $R$. So let us assume that all elements of $J$ have zero annihilators. Then all elements of $J$ are zero-divisors of $R$. Let $p_1,\dots,p_s$ be the minimal prime ideals of $R$. We know that the set of all zero-divisors of $R$ is precisely $p_1 \cup \cdots \cup p_s$. Consequently, $ J \subset p_1 \cup \cdots \cup p_s$. Then by prime-avoidance we must have that $J \subset p_1$ without loss of generality. But the minimal prime ideals have the property that $p_i = \operatorname{ann}(x_i)$ for some $0 \neq x_i \in R, \, i=1,\dots,s$. This implies that $J \subset \operatorname{ann}(x_1)$, thus $x_1 \, J=0$ and so $0 \neq x_1 \in \operatorname{ann}(J)$.

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  • $\begingroup$ Actually my answer provides an example of a finitely generated ideal consisting of zero-divisors and having trivial annihilator. $\endgroup$
    – user26857
    Commented Mar 14, 2015 at 22:13
  • $\begingroup$ @user26857 This is old, but I should note that Manos proves no example exists when $R$ is Noetherian whereas your example uses a necessarily non-Noetherian ring. $\endgroup$
    – Anonymous
    Commented Sep 28, 2022 at 3:26
  • $\begingroup$ Although I believe there is a mistake here. The set of zero-divisors equals the union of the associated prime ideals, not minimal primes. The rest of the proof goes through. $\endgroup$
    – Anonymous
    Commented Sep 28, 2022 at 3:46
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What if you take $C=\{i^r | r\in \mathbb Q\}$, where $i\in\mathbb C$. $C$ is closed under multiplication. Let $\mathbb Q$ act on it by $r\cdot z = z^r$. Then it's finitely generated as a $\mathbb Q$ module (generated by $i$). Then $C$ is a $\mathbb Q$ module (I think) and every element has non-zero annihilator (I think), but no non-zero rational can annihilate everything (I think).

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    $\begingroup$ I believe you, but then where's the error in my example? $\endgroup$ Commented Mar 14, 2015 at 18:49
  • $\begingroup$ 1 is the identity element, so it is its own annihilator. Keep in mind this is a multiplicative abelian group. So the annihilator of an element $c\in C$ is an element $r\in\mathbb Q$ s.t. $r\cdot c = c^r=1$. Or am I confused about the definitions here? $\endgroup$ Commented Mar 14, 2015 at 22:13
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    $\begingroup$ I was trying to make this rigorous by defining $C=\{e^{ri\pi} | r\in\mathbb Q\}$. And action by $\mathbb Q$ by $s\cdot e^{ri\pi}=e^{sri\pi}$. It took me a while to see the flaw, but it turns out that is not a well-defined $\mathbb Q$ module. $\endgroup$ Commented Mar 14, 2015 at 23:05
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    $\begingroup$ It's easier to see if we translate into additive notation. As an abelian group it is $\mathbb Q/\mathbb Z$ and the action is $r\cdot \overline{a} = \overline{r\cdot a}$. But if we do that then $\overline{\frac12}=\overline{\frac32}$ and we'd have to have $\frac12 \cdot\overline{\frac12} = \frac12 \cdot\overline{\frac32}$. But the LHS is $\overline{\frac14}$ and the RHS is $\overline{\frac34}$ which are not equal in $\mathbb Q/\mathbb Z$. $\endgroup$ Commented Mar 14, 2015 at 23:15
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    $\begingroup$ Because it seemed to be an instructive mistake. I learned something from it and apparently others too who have read through it. $\endgroup$ Commented Mar 21, 2015 at 12:18

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