Two definitions of Zariski Topology

Solution 1:

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.

Solution 2:

If $k$ is an algebraically closed field and $I$ an ideal of $k[X_1,..,X_n]$, the second definition is the maximal spectrum: the set of maximal ideals containing $I$, for the first definition, the points in $V(I)$ are elements of the maximal spectrum and irreducible varieties whose ideal of definition contains $I$.

Solution 3:

To say a bit more about Dustan and Tsemo's answers, there are some things you can see.$\DeclareMathOperator{\Spec}{Spec}$

1. You can take the family of the quotients of the form $R/I$ where $I$ is a radical ideal of $R$ and $R$ is a polynomial algebra over an algebraically closed field $k$. It'll be a category which morphisms will be the $k-$algebra homomorphisms.

2. For each of these algebras $R/I$ and by Nullstellensatz you can associate to them an algebraic set $V=V(I)$, embedded in some affine space $\mathbb{A}^n(k)$. You'll get $I=I(V)$ and then denote $R/I$ as $k[V]$ and call it its coordinate ring. All the algebraic sets will also form a category, and morphisms will be the restrictions of polynomial functions. And this way you can define a "functor" between these two categories where each coordinate ring is mapped to its corresponding algebraic set and each $k-$algebra homomorphism is mapped to some morphism of algebraic sets. In less words: $k[V]\mapsto V$ determines a functor.

3. We know that there is an homeomorphism between $V$ and $\text{maxSpec}(k[V])$, but there is a bit more structure here. In general, we have that the prime spectrum association $R\mapsto\text{Spec}R$ determines a functor from $\textbf{Ring}$ to $\textbf{Top}$ (To each ring homomorphism $f:R\to S$ you can assign a continuous functions $f^{-1}:\Spec(S)\to\Spec(R)$ given that the inverse image of a prime ideal is also prime). In general we can't do that with the maximal spectrum (The inverse image of a maximal ideal is not always maximal). However, if you restrict the rings to coordinate rings and the homomorphisms to $k-$algebra homomorphisms, we get that the inverse image of a maximal ideal is a maximal ideal and then the maximal spectrum $k[V]\mapsto \text{maxSpec} (k[V])$ determines a functor (We won't need the fact that $k$ is algebraically closed here, but it makes the proof easier and we will need it in 4 anyway).

4. And last, the homeomorphism from Nullstellensatz will determine a natural isomorphism between the two functors, $k[V]\mapsto V$ and $k[V]\mapsto\text{maxSpec}(k[V])$.

(Note, all the functors here are contravariant)