# Zariski topology on prime $\mathrm{Spec}$ of a ring $R$

Let $R$ be a commutative unital ring. Let $\mathrm{Spec}(R) = \{ \mathfrak p \subset R \mid \mathfrak p \text{ a prime ideal of } R \}$. We define a set $C$ to be closed in this space if and only if there is an ideal $I$ such that $C(I) = \{\mathfrak p \mid I \subset \mathfrak p, \mathfrak p \text{ a prime ideal of } R \}$.

Now I'd like to show that these sets form a topology on $\mathrm{Spec}(R)$:

(i) For the zero ideal we get $C(0) = \mathrm{Spec}(R)$ and for $R$ we get $C(R) = \varnothing$.

(ii) Arbitrary intersections are closed again: $\bigcap_\alpha C(I_\alpha) = \{\mathfrak p \mid \mathfrak p \text{ a prime ideal of } R, I_\alpha \subset \mathfrak p \text{ for all } \alpha \} = \{\mathfrak p \mid \mathfrak p \text{ a prime ideal of } R, \sum_\alpha I_\alpha \subset \mathfrak p \} = C(\sum_\alpha I_\alpha)$.

(iii) We want to show that finite unions are closed again. It's enough to show it for two ideals $I,J$: $C(I) \cup C(J) = \{\mathfrak p \mid \mathfrak p \text{ a prime ideal of } R, \text{ such that either } I \subset \mathfrak p \text{ or } J \subset \mathfrak p \}$

Now I'm stuck. How do I express the "either or" in terms of operations on ideals? Thanks for your help.

• Hint: So far, you've not once used that the $\mathfrak p$ are prime ideals. – Thomas Andrews Jul 31 '12 at 20:38
• Dear Matt, you should replace $\bigoplus_\alpha I_\alpha$ by $\sum_\alpha I_\alpha$: a sum of ideals is practically never direct. – Georges Elencwajg Jul 31 '12 at 21:50
• Dear Matt, As a complement to Thomas's hint: look at the definition of prime ideal and see if "either or" appears in it! Regards, – Matt E Jul 31 '12 at 23:28
• Dear @GeorgesElencwajg, thank you. You are right. That's what I meant, actually. : ) – Rudy the Reindeer Aug 1 '12 at 5:41
• Dear @MattE Thank you for the hint. – Rudy the Reindeer Aug 1 '12 at 5:55

Suppose $\mathfrak p \in C(IJ)$ then $IJ \subset \mathfrak p$. Since $\mathfrak p$ is prime, $I \subset \mathfrak p$ or $J \subset \mathfrak p$. Thus $\mathfrak p \in C(I) \cup C(J)$.

Suppose $\mathfrak{p} \in C(I) \cup C(J)$ then $I \subset \mathfrak p$ or $J \subset \mathfrak p$. Since ideals are closed under products $IJ \subset \mathfrak p$. Hence $\mathfrak p \in C(IJ)$.

• You kind of have to prove that $IJ$ contained in the ideal means that one of them is. An easy step, but you can't really assume it. – Thomas Andrews Aug 1 '12 at 1:19

Extension of my hint above.

If $I\not\subset \mathfrak p$ and $J\not\subset\mathfrak p$, is it possible that $IJ\subset\mathfrak p$?

There's actually an interesting parallel here. While you can take arbitrary sums of ideals, you can only take the product of a finite number of ideals.

Since I'm stuck on which one of the two helpful answers to accept, I'm posting an answer myself:

Claim: $C(I) \cup C(J)$ is closed again

Proof: $C(I) \cup C(J) = \{\mathfrak p \mid \mathfrak p \text{ a prime ideal of } R, \text{ such that either } I \subset \mathfrak p \text{ or } J \subset \mathfrak p \}$.

We claim that: $\text{ either } I \subset \mathfrak p \text{ or } J \subset \mathfrak p$ if and only if $IJ \subset \mathfrak p$.

$\implies$ Is clear since $IJ \subset I$ and $IJ \subset J$.

$\Longleftarrow$ Let $IJ \subset \mathfrak p$, that is, $\{ \sum^n i_k j_k \mid i_k \in I, j_k \in J \} \subset \mathfrak p$, in particular, $ij \in \mathfrak p$ for all $i \in I$ and all $j \in J$. Now assume that $\text{ neither } I \subset \mathfrak p \text{ nor } J \subset \mathfrak p$. Then there are $i$ and $j$ such that $i \notin \mathfrak p$ and $j \notin \mathfrak p$. But we have $ij \in \mathfrak p$ and since $\mathfrak p$ is prime, either $i \in \mathfrak p$ or $j \in \mathfrak p$, which is a contradiction. Hence $\text{ either } I \subset \mathfrak p \text{ or } J \subset \mathfrak p$.

Hence $C(I) \cup C(J)$ is closed since $C(I) \cup C(J) = C(IJ)$.

• Dear Matt, Nicely done. Just as a side, cultural, remark: the result that is implicit in your second implication is actually one of the standard characterizations of prime ideals: an ideal $\mathfrak p$ is prime if and only if for any two ideals $I$ and $J$, the product $IJ$ is contained in $\mathfrak p$ if and only if one of the factors $I$ or $J$ is contained in $\mathfrak p$. (This way of phrasing things reflects a general tendency as you develop further in commutative algebra: to argue less with individual elements in the ring, and more with higher level objects such as ideals, ... – Matt E Aug 1 '12 at 13:29
• modules, etc.) Regards, – Matt E Aug 1 '12 at 13:30
• Dear @MattE, thank you very much for pointing this out and for checking my work. – Rudy the Reindeer Aug 1 '12 at 14:15