# Algebraic Varieties and Zariski Topology

Given a set $$S \subset F[x_1, \dotsc ,x_n]$$ of polynomials, an affine variety is defined by $$S$$ is the set

$$V(S) := \{a \in A^n \mid f(a) = 0\ \ \ \forall f \in S\}$$

And the zariski topology defines closed sets to be exactly $$V(S)$$. But, the definition of closed sets in the Zariski topology that I have seen and worked with is $$V(I) = \{P: P \in \text{Spec}(R), I \subseteq P \},$$ where $$R$$ is a ring, $$I$$ is any ideal of $$R$$ and $$P$$ is a prime ideal.

Why are these two definitions equivalent? How do we get from sets of functions (presumably functions over $$R$$?) to prime ideals of $$R$$?

## 1 Answer

The definitions aren't equivalent. They are related though.

If $$F$$ is algebraically closed, the maximal ideals of $$R=F[x_1,\ldots,x_n]$$ are of the form $$\left$$ where $$(a_1,\ldots,a_n)$$ runs through $$F^n=\Bbb A^n(F)$$. We can define $$\text{mSpec}(R)$$, the maximal ideal spectrum of a commutative ring $$R$$ as the set of its prime ideals. Then we can identify $$\Bbb A^n(F)$$ with $$\text{mSpec}(F[x_1,\ldots,x_n])$$.

Of course, $$\text{mSpec}(R)$$ is a subset of $$\text{Spec}(R)$$, the collection of prime ideals of $$R$$. You have defined the Zariski topology on $$\text{Spec}(R)$$ and the subspace topology defines a Zariski topology on $$\text{mSpec}(R)$$. In the case $$R=F[x_1,\ldots,x_n]$$ this is the same as the Zariski topology you have defined on $$\Bbb A^n(F)$$. You defined $$V(S)$$ for a set of polynomials, but $$V(S)=V(\left)$$ where $$\left$$ is the ideal generated by $$S$$, so there's no loss in just considering $$V(I)$$ for ideals $$I$$. So the $$V(I)$$ for $$\Bbb A^n(F)$$ is in effect $$V(I)\cap\text{mSpec}(F[x_1,\ldots,x_n])$$ where now the $$V(I)$$ is given by your formula for arbitrary rings.

• I think you have a typo: the maximal ideal spectrum is the set of maximal ideals. – Jess Dec 12 '19 at 17:07