Direct way to show: $\operatorname{Spec}(A)$ is $T_1$ $\Rightarrow$ $\operatorname{Spec}(A)$ is Hausdorff

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!

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Dear Rand: +1 for your nice question. I'm unable to answer it, but here are two links that might interest you (if you don't know them already): pitt.edu/~yimuyin/papers%20and%20notes.html, www-users.math.umd.edu/~karpuk/commalg.html –  Pierre-Yves Gaillard Feb 13 '12 at 15:07
Hello Pierre! I did not know these documents yet. Unfortunately, they both show the same implication that I did myself. But I'm sure they will be a great help for when I'm stuck again, which happens quite often ;) So thank you very much! –  InvisiblePanda Feb 13 '12 at 15:15
Yes, he is wrong, $X_p$ and $X_q$ don't partition $X$, but $X_{ax}$ and $X_{ax-1}$ do, for $x \in p-q$ and an appropriate $a \in A$. But the existence of this element $a$ relies on (i). –  Parsa Feb 14 '12 at 0:45
Hello and thanks for your answers. I'll work my way through them as soon as I am home again! –  InvisiblePanda Feb 14 '12 at 7:51

3 Answers

First of all, here is a meta-argument that such a proof exists:

The spaces that can arise as Spec of a ring were characterized by their abstract topological properties in Hochster's thesis; they are known as spectral spaces. Since $T_1 \implies$ Hausdorff is a purely topological statement about such spaces, it must admit a proof that is purely topological (without making reference to any ring of which the space is the spectrum).

Of course this is not an actual argument, but since in the end the statement in question (that $T_1 \implies$ Hausdorff) is not so deep, it is reasonable.

Okay, now here is an actual argument. First, one proves the following:

The quasi-compact open sets in a spectral space satisfy the finite intersection property (i.e. given a collection of such open sets, if any finite collection has non-empty intersection then the whole collection does).

The proof is in Hochster's thesis; see the Transactions referenced in the linked Wikipedia page. Here I will just prove a special case (which is in fact not hard to check is equivalent, and which will suffice for our application), and I will cheat, by using ring-based concepts. Here is the statement and proof:

Suppose $S$ is a set of elements of $A$ such that $D(a_i)$ (the distinguished open corresponding to elements $a_i$) have non-empty intersection for each finite subset $\{a_1, \ldots, a_n\} \subset S$; then the $D(a)$ (for $a \in S$) have a non-empty common intersection.

Proof: Consider the localization $A_S$. The assumption that finite intersections are non-empty is enough to prove that $A_S$ is non-zero. Then Spec $A_S$ is non-empty, and its image in Spec $A$ is contained in the intersection of all the $D(a)$. QED

[Aside: Hochster's topological proof replaces the application of Zorn's lemma that is used to prove that Spec $A_S$ is non-empty by a different application of Zorn's lemma, working directly with certain subsets of the topological space Spec $A$.]

Now we deduce the following corollary:

If $x$ and $y$ are two elements of Spec $A$ that cannot be separated by open sets, then there is a point $z$ containing both $x$ and $y$ in its closure.

Proof: Consider all the quasi-compact open sets of Spec $A$ each of which contains either $x$ or $y$. (If you want, you can retrict to distinguished open sets, and then apply the more restricted version of the preceding topological result.) By assumption, any finite collection of elements in this set of opens has non-empty intersection, and so the whole set of elements has non-empty intersection. If we choose $z$ to be an element of this intersection, then it contains both $x$ and $y$ in its closure. QED

Now if Spec $A$ is $T_1$, then we deduce from the preceding result that if $x$ and $y$ cannot be separated by open sets, then they coincide, i.e. Spec $A$ is Hausdorff.

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Hello @Matt: that is very interesting. I had not heard about spectral spaces before, and, to be honest, I didn't think anything like this would be possible! Thanks a lot for the proof and the interesting link(s). –  InvisiblePanda Feb 14 '12 at 12:33

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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I doubt you can show it without going through (i). The good news is that $T_1$ immediately implies that no prime ideal is properly contained in another prime ideal, i.e. (ii) every prime ideal is maximal.

To find the "canonical" disjoint open subsets that separate two points, take two distinct primes $\mathfrak p, \mathfrak q$, and choose $x \in p$, but $x \notin q$. Let $U_x=\{\mathfrak p' \in X|x \notin \mathfrak p'\}$. Then $\mathfrak q \in U_x$ and $\mathfrak p \notin U_x$. Since $\mathfrak q$ is maximal, $\mathfrak q + (x)=A=(1)$. This means that $y+ax=1$ for some $y \in \mathfrak q$ and $a \in A$. Then $y=1-ax \notin \mathfrak p$, so set $U_y=\{\mathfrak p' \in X|y \notin \mathfrak p'\}$, for which $\mathfrak p \in U_y$ and $\mathfrak q \notin U_y$.

The problem is why should $U_x \cap U_y=U_{xy}=\varnothing?$ The only way I know how to do this is to go through (i), which shows that $xy \in \mathfrak N(A)$.

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Hello @Parsa: when I proved 1 $\Rightarrow$ 4, I used similar open subsets (in fact, I had $U_x$ and $U_{ax-1}$ for some fitting $a$. Maybe there is something wrong with my solution, I'll check every step again!). I could have thought myself about using the same construction to find open subsets... sometimes (often) I'm blind ;) But I understand why one is stuck there. Thank you very much! –  InvisiblePanda Feb 14 '12 at 12:39
Your solution is probably right, I'm just pointing out that I couldn't do it directly, i.e. without using (i). However, Matt gave a good direct solution above using a purely topological argument, which actually answers your initial question. –  Parsa Feb 14 '12 at 22:52