# $\operatorname{Spec} (A)$ as a topological space satisfying the $T_0$ axiom

I have been spending a few days now proving the last bit of the following problem of Atiyah Macdonald:

Prove that $X = \operatorname{Spec}(A)$ as a topological space with the Zariski Topology is $T_0$.

Now since the Zariski Topology is specified in terms of closed sets, I thought it might be easier to prove that given $x,y \in X$ such that $x \neq y$, there exists a closed set $U$ such that $x \in U$, $y \notin U$, or else there exists a closed set $V$ such that $y \in V$, $x \notin V$.

So I tried to follow my nose by explicitly producing such a closed set, namely

$$\overline{\{x\}} = V(\mathfrak{p}_x)$$

following the notation of Atiyah Macdonald. Now one of the things I tried from here was to use the fact that closed subsets of compact sets are compact. Hence $V(\mathfrak{p}_x)$ is compact but this approach did not work out. I realised while typing this problem that if I were actually able to prove it like this, then switching the roles of $x$ and $y$ I would have proved $T_1 -$ ness. However this cannot be possible for the finite point set $\{x\}$ is not closed when $\mathfrak{p}_x$ is not a maximal ideal.

What should I look at now to try to solve the problem? Please do not post complete solutions. Thanks.

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Dear Benjamin: Hint: Express the $T_0$ axiom in term of closed subsets (instead of open subsets). – Pierre-Yves Gaillard Feb 12 '12 at 13:53
@Pierre-YvesGaillard C'est ce que j'ai dit. – user38268 Feb 12 '12 at 13:54
@Pierre-YvesGaillard I mentioned above how I expressed the $T_0$ axiom in terms of closed sets. – user38268 Feb 12 '12 at 13:58
Cher Benjamin: En effet! Désolé... On a intérêt à ce que l'ensemble fermé $U$ contenant $x$ soit le plus petit possible... (For the others: The closed subset $U$ containing $x$ has better be as small as possible.) – Pierre-Yves Gaillard Feb 12 '12 at 14:02
$X$ is $T_0$ iff $x \in \overline{\{y\}}$ and $y \in \overline{\{x\}}$ implies $x=y$. This might help. – Henno Brandsma Feb 12 '12 at 14:02

(expanding my comment to a (partial) answer)

Use that $X$ is $T_0$ iff for all $x,y \in X$, if $x \in \overline{\{y\}}$ and $y \in \overline{\{x\}}$ then $x = y$.

Proof: from left to right: If $x \neq y$, the $T_0$ gives us an open set such that $x \in U, y \notin U$, say (the other case is similar). But then $x \in \overline{\{y\}}$ implies $y \in U$, as $U$ is a neighbourhood of $x$, so intersects $\{y\}$. Contradiction.

From right to left: if $x \neq y$, then $x \notin \overline{\{y\}}$ or $y \notin \overline{\{x\}}$, say the former. Then $U = X \setminus \overline{\{y\}}$ is open, contains $x$, but not $y$ (in the other case, of course, the other way around).

You already know $\overline{\{x\}} = V(\mathfrak{p}_x)$, so you need to prove $$V(\mathfrak{p}_x) = V(\mathfrak{p}_y) \rightarrow x=y$$

which is something that you might know how to do (I don't as I don't have this book).

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Like I said the implication follows immediately from the previous exercise in AM :D – user38268 Feb 16 '12 at 23:50

Here is an alternate proof to Henno's observation:

A topological space $X$ is $T_0$ iff $x \in \overline{\{y\}}$ and $y \in \overline{\{x\}}$ $\implies$ $x=y$.

$\textbf{Right to left:}$ Suppose $X$ is not $T_0$. Then there are points $x,y$ with $x \neq y$ such that for all closed sets $U$ about $x$ and $V$ about $y$, $U$ must contain $y$ and $V$ must contain $x$. In particular taking $U = \overline{\{x\}}$ and $V = \overline{\{x\}}$ this means that $y \in \overline{\{x\}}$ and $x \in \overline{ \{y\}}$. But then by assumption this implies that $x = y$, contradicting $x \neq y$.

$\textbf{Left to right:}$ Suppose $\exists x,y$ such that $x \in \overline{\{y\}}$ , $y \in \overline{\{x\}}$ and $x \neq y$. Then it is easy to see that $$\overline{\{x\}} = \overline{\{y\}}.$$

Now since $x\neq y$ and $X$ is $T_0$, we may assume there is either a closed set $U$ about $x$ that does not contain $y$, or a closed set $V$ about $y$ that does not contain $x$.

Without loss of generality assume the former. Then clearly $x \in \overline{\{x\}} \subseteq U$ by definition of the closure. As $\overline{\{x\}} \subseteq U$ by the contrapositive this means that $y \notin \overline{\{x\}}$. But then by our observation before this means that $y \notin \overline{\{y\}}$ which is ridiculous.

$\hspace{6in} \square$

I know Henno's proof above is much slicker but this is one that I came up with myself.

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Let $f$ in $A$. It defines a principal open set $U_f$. Namely : $$U_f = \{ x\in X \mid f\not\in \mathfrak p_x \}$$

Mouse over the grey box only if you want more than the above hint.

If you have two different points of $X$, you can certainly find a $f$ in $A$ which is in one and only one of the two ideals $\mathfrak p_x$ and $\mathfrak p_y$. Then $U_f$ contains exactly one of both points $x$ and $y$.

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