Closed Subset of Spec$(\mathbb{C}[X])$ is a union of closed points in the Zariski topology.

Question

I am trying to show that any proper closed subset of Spec$(\mathbb{C}[X])$ is a union of closed points in the Zariski topology. I know that the closed points in Spec$(\mathbb{C}[X])$ are exactly the ideals of the form $(X-\alpha)$ for all $\alpha \in \mathbb{C}$ and I feel like this should be enough to prove the claim but I am not sure how to argue this. Also, does this mean that Spec$(\mathbb{C}[X])$ is Hausdorff? I haven't seen much topology in a while and I am not sure what the best way to see this is.

Answer 1

The point of $Spec(\mathbb{C}[X])$ are irreducible ideals of $\mathbb{C}[X])$. Since $\mathbb{C}[X])$ is principal, every non zero prime ideal is generated by an irreducible polynomial $(X-\alpha)$ which is maximal and represents a closed point. The zero ideal which is dense (the generic point). So it is not separated.