Why is every Noetherian zero-dimensional scheme finite discrete?

Question

In the book The geometry of schemes by Eisenbud and Harris, at page 27 we find the exercise asserting that

Exercise I.XXXVI.
The underlying space of a zero-dimensional scheme is discrete; if the scheme is Noetherian, it is finite.

Thoughts
I thought that a zero-dimensional scheme is one such that, in every local ring $\mathscr O_{X,p}$, the only prime ideal is its unique maximal ideal $\mathfrak m_{X,p}.$ Hence, supposing that $X=\operatorname{Spec} R$ is affine, we deduce that every prime ideal is maximal, so each ponit in $X$ is closed. But how does this imply the discreteness? It only implies the Hausdorff-ness of $X$, right?
Furthermore, I cannot perceive what the "Noetherianity" has to do with the finiteness claimed here. In the affine case, it seems to be claiming that there are only a finite number of prime ideals in a Noetherian ring, if its scheme is zero-dimensional? For example, $\mathbb Z$ is a Dedekind domain, hence its scheme is zero-dimensional, but it has infinitely many primes: does this contradict the statement?

Thanks in advance for any reference or hint, and point out any inappropriate point if any is presented.
Edit I see why my example cannot work now: it is one-dimensinoal: don't forget the pirme $(0)$. But I am still wondering why the statement is true...

Answer 1

a) A noetherian zero-dimensional scheme is covered by finitely many open affines which are spectra of zero-dimensional noetherian rings.
If these spectra are discrete finite, then the scheme is discrete finite too.
This answers the second part of your question.

b) Here is why the spectrum $X=\operatorname{Spec}(A)$ of a zero-dimensional ring $A$ is Hausdorff:
Consider two distinct points of $X$, i.e. two maximal ideals $\mathfrak m,\mathfrak n\subset A$.
There exist idempotent elements $m\in \mathfrak m, n\in \mathfrak n$ with $m+n=1$.
The disjoint open subsets $ D(n), D(m)$ are then separating neighbourhoods of $\mathfrak m$ and $\mathfrak n$.
The details (and more) are to be found in the Stacks Project here.

c) If moreover $A$ is noetherian, then $X=\operatorname{Spec}(A)$ is finite and thus discrete.
Inded a noetherian ring has only finitely many minimal prime ideals and all primes are minimal in a zero-dimensional ring.

d) It is false that the spectrum of a zero-dimensional ring is discrete if the ring is not assumed noetherian:
If $A=K^\mathbb N$ is the denumerable power of a field $K$ then $A$ is zero-dimensional and $\operatorname{Spec}(A)$ is homeomorphic to the Stone-Čech compactification of $\mathbb N$, a dreadful beast which is certainly not discrete since it is compact and infinite.

Answer 2

As a complement to Georges's excellent answer, I think it is worth noting a systematic construction of non-discrete zero-dimensional schemes.

Start with an arbitrary scheme $S$, say noetherian to simply what follows. There is a finer topology on $S$ than the initial one, called the constructible topology on $S$. The open subsets for this topology consists in constructible subsets of $S$ (i.e. finite union of locally closed subsets). Denote this topological space by $S^{\rm cons}$. It turns out that $S^{\rm cons}$ is always totally disconnected and compact, see EGA IV.1.9.15 ($S$ is noetherian hence quasi-compact and quasi-separated). Edit So $S^{\rm cons}$ is discrete if and only if $S$ is finite (set-theoretically).

To finish, and this was really surprising for me, such a "dreadful beast" is the underlying space of a scheme ! See EGA IV.1.9.16. So

$S^{\rm cons}$ is non-discrete and zero-dimensional for any noetherian scheme such that the underlying space is infinite.

If $S_1, S_2$ are two integral algebraic varieties not birational to each other, then $S_1^{\rm cons}$ is not isomorphic to $S_2^{\rm cons}$ by looking at residue fields.