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For a finitely generated $A$-module $M$ the support of the module $M$, $\operatorname{Supp}(M)$ is the same as the set of all prime ideals of $A$ containing the ideal $\operatorname{Ann}(M)$.

The equivalent definition of $\operatorname{Supp}(M)$ is all prime ideals such that $M_p$ is not zero, where $M_p$ is now an $A_p$-module.

I believe that these definitions are equal.

Here the case of $\mathbb{C}[x,y]/(xy)$ is discussed.

I want to find $\operatorname{Supp}(\mathbb{C}[x]/(x^2-1))$ as a $\mathbb{C}[x]$-module.

UPD: The primes in $\mathbb{C}[x]$ are $(x-a)$ and $(0)$. $\operatorname{Ann}(M) = (x^2-1)$. That means that two ideals $(x-1), (x+1)$ contain $\operatorname{Ann}(M)$. So the answer is $(x-1)$ and $(x+1)$.

Is it correct?

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No, it is not correct, because $\operatorname{Ann}(M) \neq (x-1,x+1)$ : $x-1$ doesn't kill $\overline{1} \in \mathbb C [x]/(x^2-1)$ so it is not in the annihilator.

In fact, you can prove that $\operatorname{Ann}(M) = (x^2-1)$, and this will let you compute the support.

With your edited computations, you are correct !

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  • $\begingroup$ Amended the mistake with annihilator in edited question $\endgroup$ Aug 27 '19 at 10:36
  • $\begingroup$ Edited my answer $\endgroup$ Aug 27 '19 at 10:52

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