Do the zeros of a prime ideal on closed points of the Zariski topology uniquely determine it?

Question

That is, if two prime ideals share the exact same zeros on maximal ideals, are they the same ideal? Or at least is there a result with other assumptions that shows this?

Learning algebraic geometry at the moment and not that deep into it, but this keeps popping into my mind as I read. It seems like it should be true (at least for nice conditions like Noetherian rings), but I'm struggling to see a proof.

Answer 1

It seems like the question is whether $V(\mathfrak{p})$ and $V(\mathfrak{q})$ having the same closed points would imply $\mathfrak{p} = \mathfrak{q}$. As you suspect, it's not true in general: if $A$ is a local ring then you can't distinguish any prime ideals in this way.

On the other hand, if $A$ is a finitely generated algebra over a field then closed points are dense as a consequence of the Nullstellensatz (great exercise!), and I think you should be able to use that to verify your conjecture in that case.

Answer 2

Just to add to Hoot's excellent answer, a ring $R$ is called Jacobson if every prime ideal $\mathfrak{p}$ of $R$ is equal to the intersection of the maximal ideals in $V(\mathfrak{p})$. It follows immediately that if $R$ is a Jacobson ring and $\mathfrak{p},\mathfrak{q}\in \mathrm{Spec}(R)$ satisfy $V(\mathfrak{p})\cap\mathrm{mSpec}(R)=V(\mathfrak{q})\cap\mathrm{mSpec}(R)$, then $\mathfrak{p}=\mathfrak{q}$.

One (general) version of the Nullstellensatz says that if $R$ is a Jacobson ring and $S$ is a finitely generated $R$-algebra, then $S$ is Jacobson as well. Examples of Jacobson rings include $\mathbf{Z}$ and any field. So finitely generated algebras over either of these rings will have the property you ask about.

This will not be true in general for non-Jacobson rings. For example, if $R$ is a local ring with at least two distinct prime ideals, then $R$ will not be Jacobson and it will not have the property you want. Indeed, for such a ring, if $\mathfrak{p}$ is a prime different from the maximal ideal $\mathfrak{m}$, then $V(\mathfrak{p})\cap\mathrm{mSpec}(R)=V(\mathfrak{p})\cap\{\mathfrak{m}\}=\{\mathfrak{m}\}=V(\mathfrak{m})$.