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Find an example of a unital commutative ring $R$ and an $R$-module $M$ such that $\text{Ass}(M)=\emptyset$.

Recall that

  • $\text{Ass}(M)=\{p\in\text{spec}(R):\exists 0\ne m\in M:\text{ann}(m)=p\}$.
  • $\text{ann}(m)=\{r\in R:rm=0\},\forall m\in M$.

I thought of $R=\mathbb Z$ and $M=\mathbb Z/n\mathbb Z$, which doesn't work.

J. Doe
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1 Answers1

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Take $A_n=K[[X^{2^{-n}}]]$, and $A=\bigcup_n{A_n}$. Let $M=R=A/I$, where $I$ is the ideal generated by the $X^{\alpha}$, $\alpha \geq 1/3$.

Edit: note that if $m \in M$ is nonzero with annihilating ideal is $I$, and if there is no $m’ \in m$ of which the annihilating ideal contains strictly $I$, then $I \in Spec(R)$. So we have to look for not-Noetherian rings, hence the complicated example.

Aphelli
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  • How did you come up with this example? I also do not quite see how to show that the annihilators are not prime. Could you give me a hint? – Severin Schraven Sep 22 '19 at 19:26
  • The original idea was to try with $\mathbb{Z}[T]/(T^2)$, but there was this element $T$ with a prime annihilator. So I tried to make it so that there wouldn’t be such a “maximally singular” annihilator. For your other points, note that it is enough to consider the annihilators of $X^{k/2^n}$ for all $3k < 2^n$. If the annihilator is prime and if $X^{s/2^l}$ is in the annihilator, it is the product of $s2^p$ $X^{2^{-l-p}}$ for all $p \geq 0$, so all of them are in the ideal, which entails $k/2^n \geq 1/3+o_p(1)$, a contradiction. – Aphelli Sep 22 '19 at 19:49
  • Thanks, I'll grab a pencil and try to fill in the details (my algebra is a bit rusty) :) – Severin Schraven Sep 22 '19 at 19:58