# Support of $\mathbb{C}[x]/(x^2-1)$ as $\mathbb C[x]$-module

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?

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.