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 ?
Best Answer
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 $(0,1)$: $$[0,1]=\overline {(0,1)}=\bigcap_{ \{f\in C(\mathbb R)\vert f (x)=0 \operatorname {for all} x\in (0,1)\} }f^{-1}(0)$$