[Math] “Closure” and “neighborhoods” in Spec(A)

abstract-algebraalgebraic-geometrycommutative-algebra

While trying to work through the sequence of problems in Atiyah-Macdonald's first chapter regarding the prime spectrum of a ring, I've run across a small point of confusion. Namely:

In the point spectrum Spec(A), what exactly is a limit point or a neighborhood of a set?

Any definition I can find for those two things, and the related concept of the closure of a set, seems to appeal to some kind of metric on the space, but I'm not at all clear on what sort of metric Spec(A) has, since it appears as if the work I've done so far only handles topological properties of the space. But problem 18 in particular asks for a proof that, if x denotes the prime ideal $P_x$ of A, the closure of the single-point set {x} is $V(P_x)$. It's clear that that set is closed and contains {x}, but I don't quite understand how to distinguish a closed set containing a set from the closure of a set in this context. Any help would be appreciated.

Best Answer

We've gotten a little sidetracked into broader matters and I will hopefully remember to update this to reflect that, but I wanted to push you in the direction of solving 1.18.

There are many ways to characterize the closure of a subset $S$ in a topological space $X$, so the first thing to do is pick your favorite and try to work with that. I think the one best suited to the problem is the following:

$\overline{S}$ is a closed subset of $X$ containing $S$, and if $T$ is another such set then $\overline{S} \subset T$.

That this uniquely defines a set $\overline{S}$ and is equivalent to the other definitions is an exercise in general topology worth doing.

As you say, $V(\mathfrak{p}_x)$ is closed and contains $\{x\}$, so our candidate satisfies half of the requirements. Now take any other closed set containing $\{\mathfrak{p}_x\}$ and show that it contains $V(\mathfrak{p}_x)$ too. It may help to recall the form that all closed subsets of $\operatorname{Spec} A$ take, as well as the fact that $V$ reverses inclusions.