There is the Zariski topology on the set $k^n$. There is also the Zariski topology on the set of prime ideals of $k[x_1,\dots,x_n]$. I was wondering if anyone could explain, on a basic level, without appealing to coordinate rings or tensor products, how these two are related? I do understand that if $k$ is algebraically closed, the points of $k^n$ correspond bijectively to maximal ideals of the polynomial ring. Wouldn't it be natural to define the topology on the set of maximal ideals? Why does one enlarge the set (from maximal ideals to prime ideals) on which the topology is being defined? What is the motivation for the definition of the Zariski topology on the set of prime ideals (how to come up with such definition?) And by the way, are $k^n$ and the set of maximal ideals of $k[x_1,\dots,x_n]$ homeomorphic if each of them is given the Zariski topology?

  • $\begingroup$ They are in fact homeomorphic $\endgroup$ – leibnewtz Jan 22 '19 at 2:07

One of the reasons we use the prime spectrum is because of the following: if $\varphi:R\to S$ is a ring homomorphism and $\mathfrak{p}$ is a prime ideal of $S$, then $\varphi^{-1}(\mathfrak{p})$ is a prime ideal of $R$. This allows us to view $\mathrm{Spec}$ as a functor $\mathrm{CommRing^{op}}\to\mathrm{LocRing}$, where $\mathrm{LocRing}$ is the category of locally ringed spaces. One thing we get is an equivalence of categories $\mathrm{CommRing^{op}}\simeq\mathrm{AffSch}$ (the other direction is given by the global sections functor).

It is natural to consider $\mathrm{MaxSpec}$, but not all the rings we want to consider are algebraically closed fields. In addition, we lose extra information like (non)closed points.

The motivation for the Zariski topology is sort of extending the Zariski topology from algebraically closed fields to all commutative rings. Given $\mathrm{Spec}\,R$, we can form an open basis of subsets $D(f)$ for $f\in R$, $D(f)=\{\mathfrak{p}\in\mathrm{Spec}\,R : f\not\in\mathfrak{p}\}$. These $D(f)$ can be thought of where $f$ does not vanish.

As for why $\mathrm{MaxSpec}\,k[x_1,\dots,x_n]\simeq k^n$, this is one way to state Hilbert's Nullstellensatz.


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