[Math] Two definitions of Zariski Topology

algebraic-geometrycommutative-algebra

$\DeclareMathOperator{\Spec}{Spec}$
Given a ring $R$ the closed sets of the Zariski Topology on the Spectrum of $R$, $\Spec(R)$, is:
$$
V(I)= \{P\in \Spec(R) \mid I \subset P\} \, .
$$

Given an affine space $A^n$ the closed sets of the Zariski Topology of $A^n$ is:
$$
V(I)= \{x \in A^n \mid f(x)=0, \forall f \in I \}
$$

What is a concrete relationship between the two definitions if any? I.e., is there some $R$ so that we have a one to one correspondence between closed sets?

Best Answer

Absolutely! This is a fundamental idea in Grothendieck's modern algebraic geometry.

Let $F$ be an algebraically closed field, and denote by $\mathbb A_F^n$ the $n$-dimensional affine space with points in $F$.

Let $R = F[x_1, \ldots, x_n]$. There is a function $$\Phi: \mathbb A_F^n \to \operatorname{Spec}(R)$$ given by $$(a_1, \ldots, a_n) \mapsto (x_1-a_1, \ldots, x_n-a_n).$$

The Nullstensatz Theorem says that:

  • $\Phi$ maps onto the maximal ideals of $R$, i.e., the closed points in $\operatorname{Spec}(R)$, and
  • $\Phi$ defines a homeomorphism onto its image.

When $F$ is not algebraically closed, $F[x_1, \ldots, x_n]$ has more maximal ideals than those given by the image of $\Phi$, which makes the connection to geometry a little bit less explicit.

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