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?
Best Answer
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 !