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Let $R$ be a commutative ring, $spec(R)$ be the set of all prime ideals on $R$. For any ideal $I$ on $R$, we define the $V_I$ to be the set of all prime ideals containing $I$. We define the Zariski topology on $spec(R)$ via the closed sets $\{V_I:I\textrm{ is an ideal of }R\}$.

I am still wrapping my mind around this topology. Can someone tell me if it is a cofinite topology, i.e. the open sets are complements of finite sets, or not?

Paolo
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Simon_Peterson
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2 Answers2

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There can be infinite closed sets besides the whole space.

In $\Bbb Q[x,y]$, for example, $(y)$ is contained in infinitely many prime ideals, so $V_{(y)}$ is a closed, but not finite set.

Another way: the cofinite topology always makes its space $T_1$, but the spectrum is $T_1$ in the Zariski topology iff prime ideals are maximal, and that only occurs for certain rings.

Another way: The cofinite topology on a finite space is the discrete topology, but the Zariski topology on a ring with a finite spectrum need not be discrete. Consider, for example, $Spec(\Bbb Q[[x]])$, where the spectrum is $\{0, (x)\}$ and the closed sets are $\{\emptyset, \{x\}, \{0,(x)\}\}$

I've spent some time in the same boat as you on this topic. It's especially disorienting if most of your topological experience is derived from thinking about metrizable spaces.

Community
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rschwieb
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If $R$ is $F[x]$ for $F$ a field then yes they should be the same. If $R=F[x,y]$ and $F$ is an infinite field then no, it's not the cofinite because it includes the complement of curves which are not finite.

Gregory Grant
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    They are not the same for $F[x]$. Every non-empty open set of the spectrum of $F[x]$ contains the zero ideal. So not all finite subsets are closed. – Rene Schoof Jan 10 '16 at 14:42
  • The maximal spectrum of $F[x]$ has the cofinite topology, but not the spectrum itself, as $(0)$ is a non-closed point. – Watson May 18 '18 at 17:27