Algebraic Geometry – Direct Proof: Spec(A) is T? Implies Spec(A) is Hausdorff

algebraic-geometrycommutative-algebra

In the book of Atiyah and MacDonald, I was doing exercise 3.11. One has to show that for a ring $A$, the following are equivalent:

  1. $A/\mathfrak{N}$ is absolute flat, where $\mathfrak{N}$ is the nilradical of $A$.
  2. Every prime ideal of $A$ is maximal.
  3. $\operatorname{Spec}(A)$ is $T_1$.
  4. $\operatorname{Spec}(A)$ is Hausdorff.

I already proved this by doing 1 $\Leftrightarrow$ 2, 2 $\Leftrightarrow$ 3, 4 $\Rightarrow$ 3, and 1 $\Rightarrow$ 4. Then I wondered: is there a direct way of showing that $X=\operatorname{Spec}(A)$ has the Hausdorff property when it already is $T_1$? I don't see any "canonical" disjoint open subsets of $X$ that I could construct, given $x,y\in X$ distinct points.

Googling for a solution which takes this approach led me here (again), where he shows 2 $\Rightarrow$ 4, which should be nearly the same as 3 $\Rightarrow$ 4. But I don't know the notation $X_{\mathfrak{p}}$ he uses. I guess it means $X\smallsetminus V(\mathfrak{p})$, analogous to the notation $X_f$ for basic open subsets of the Zariski topology. But why are $X_{\mathfrak{p}}$ and $X_{\mathfrak{q}}$ disjoint there? Assuming 2 (resp., 3), we have $V(\mathfrak{p})=\{\mathfrak{p}\}$, thus $X_{\mathfrak{p}}=X\smallsetminus\{\mathfrak{p}\}$, $X_{\mathfrak{q}}=X\smallsetminus\{\mathfrak{q}\}$, and hence the intersection would contain all points different from $\mathfrak{p}$ and $\mathfrak{q}$. Did I misinterpret the notation, or am I maybe just not getting what the author is doing here?

Thanks in advance!

Best Answer

Here is a quite simple, though not obvious proof, inspired by Lemma 10.24.4 and Lemma 10.24.5 of the Stacks Project:

Suppose every prime ideal of $A$ is maximal and $\mathfrak p \neq \mathfrak q$ are two distinct prime ideals of $A$, say there exists $f \in \mathfrak p$ with $f \not\in \mathfrak q$. The ring $A_{\mathfrak p}$ has exactly one prime ideal $\mathfrak pA_{\mathfrak p}$, which coincides with its nilradical. So $f/1$ is nilpotent, i.e. we have $sf^n = 0$ for some $s \in A\setminus \mathfrak p$ and $n \geq 0$. Using the notation $D(a) := \operatorname{Spec}(A) \setminus V(a)$, we have $$\mathfrak q \in D(f), \; \\\mathfrak p \in D(s), \;\\ D(f) \cap D(s) = D(sf^n) = \emptyset.$$ Thus, $\operatorname{Spec}(A)$ is Hausdorff.

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