Closure of a point : Zariski Topology.

Question

Let $R$ a ring and $Spec(R)$ the set of prime ideal of $R$. Let $x\in Spec(R)$ and $\mathfrak p_x$ the corresponding ideal. The closure of $x$ is $$\bar x=\{\mathfrak p\in Spec(R)\mid \mathfrak p\supset \mathfrak p_x\}.$$ More generally, if $Y\subset Spec(R)$, then, the closure of $Y$ is $$\bar Y=\left\{\mathfrak p\in Spec(R)\mid \mathfrak p\supset \bigcap_{y\in Y}\mathfrak p_y\right\}.$$

What is the intuition behind this definition ? To me, if $(X,\mathcal T)$ is a topology space and $Y\subset X$, then, $$\bar Y=\{x\in X\mid \forall U\in \mathcal T: U\ni x, Y\cap U\neq \emptyset \}.$$

I can't find an equivalence of the closure defined over and my topological definition of the closure. Any idea ?

Answer 1

An element of $R$ should be seen as a function on $X=\operatorname {Spec} R$.
This is the fantastic vision of Grothendieck, which turns an arbitrary ring into a ring of regular functions on $X$, namely : $R=\Gamma (X,\mathcal O_X)$.
The value of the "function" $r\in R$ at $x\in X$ is $r[x]:=[r]_{\mathfrak p_x}\in \operatorname {Frac}(R/\mathfrak p_x)=\kappa(x)$.
Hence the condition that $r$ vanish at $x$ is: $r[x]=0\iff r\in \mathfrak p_x$, and the condition that $r$ vanish at all points $y\in Y$ is thus $r\in \bigcap_{y\in Y}\mathfrak p_y$.
The Zariski topology is such that its closed sets are given by the vanishing of a family of functions and thus it is quite intuitive that the closure of $Y$ is given by the the vanishing of the functions $r\in \bigcap_{y\in Y}\mathfrak p_y$.
In other words $$\overline Y=\{x\in X\vert r[y]=0 \operatorname {for all } y\in Y\}=V( \bigcap_{y\in Y}\mathfrak p_y)$$ which is your formula.

An Analogy
Actually the formula is quite natural and its analogue already exists in elementary calculus!
The closure of the interval $(0,1)\subset \mathbb R$ is exactly the common zero set of all functions vanishing on $(0,1)$: $$[0,1]=\overline {(0,1)}=\bigcap_{ \{f\in C(\mathbb R)\vert f (x)=0 \operatorname {for all} x\in (0,1)\} }f^{-1}(0)$$

Answer 2

They are equivalent, but the equivalence is maybe better seen more abstractly.

In particular if $S$ is a subset of the points in a topological space and we've chosen a basis for the topology, then:

• $\bar{S}$ is the smallest closed set containing $S$
• $\bar{S}$ is the intersection of all closed sets containing $S$
• $\bar{S}$ is the intersection of all basic closed sets containing $S$

(where by "basic closed set" I mean the complement of a basic open set).

You might have more success showing the first or second is equivalent to the definition you prefer, that each of these characterizations easily imply the next, and that the third better resembles the formula you're asking about.

Alternatively, using the fact the closure is the complement of the exterior, you might opt to work with dual versions of these statements in terms of open sets not containing $S$.

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Alternatively, it is of interest to look at other formulations of the notion of topology — e.g. one can axiomatize topology in terms of the notion of "closure" rather than the notion of "open"; a sample set of axioms is

• $\overline{\varnothing} = \varnothing$
• $A \subseteq \overline{A}$
• $\overline{A \cup B} = \overline{A} \cup \overline{B}$
• $\overline{\overline{A}} = \overline{A}$

The corresponding notion of "closed" is a set equal to its closure, and "open" is the complement of a closed set.

Thus, to show the equivalence of the two formulas you've cited, what you need to do is to show that the given closure operation satisfies these axioms, and that closed sets of the topology defined by that closure operation are precisely the Zariski closed sets.