Spectrum of the coordinate ring of an affine variety

Question

My question is rather simple for an algebraic-geometer maybe and because I'm not, it confuses me a lot and has to do with the following. Sometimes, I see authors indentify an affine variety $V \subset \mathbb{A}^{n}$ over an algebraically closed field say $\mathbb{K}$, with the spectrum $Spec(\mathbb{K}[V])$, and they are referring to the latter as the the affine variety with coordinate ring $\mathbb{K}[V]$. (where as far as I know the last spectrum it is called affine scheme in the literature)

So, it's obvious by definition that those two sets are not equal, hence it's about an identification (natural or if you prefer categorical) between them. Even if $V$ is irreducible variety (which means that we don't have other primes except the maximals and the trivial $\{ 0 \}$ and due to Nullstellenstaz we have a "good" correspondence between the points of $V$ and elements in $Spec(V)$) there is something missing, namely the generic point we add because of the non-maximal prime ideal $\{ 0 \}$. I would understand the identification of $V$ with the maximal spectrum $mSpec(\mathbb{K}[V])$ in that case but the previous one doesn't make any sense. Can you please explain me how this identification comes in and what's the idea behind the "equality" $V=Spec(\mathbb{K}[V])$?

Answer 1

If you take the affine variety with its Zariski topology, it is (among other things) a topological space $V$.

Now given a topological space $V$, we can construct a new topological space $X$ whose points are (by definition) the irreducible closed subsets of $V$, and whose open sets are in bijection with the open sets of $V$ by mapping an open set $U$ in the latter to the set of irreducible subsets of $V$ which have non-empty intersection with $U$.

There is a map from $V$ to $X$ which sends a point in $V$ to its closure, and by construction the topology on $V$ is obtained by pull-back from the topology on $X$ (i.e. the open sets in $V$ are precisely the preimages of the open sets in $X$).

So: two points of $V$ map to the same point of $X$ if and only they have the same closure, and hence $V \to X$ is injective iff $V$ is $T_0$ (i.e. two points with the same closure coincide); in this case $V$ is a topological subspace of $X$.

The map $V\to X$ is a homeomorphism if and only if every irreducible subset of $V$ has a unique generic point, i.e. if and only if $V$ is sober.

Affine schemes are sober, so this construction does nothing in the case of an affine scheme.

But affine varieties are not sober (unless they are zero-dimensional), and the construction $V\mapsto X$ in this case gives rise to the corresponding affine scheme. Given $X$, we can recover $V$ as the subset of closed points in $X$.

(If we want to be more sophisticated and think about structure sheaves, we can do that too: the structure sheaf on the scheme $X$ is the pushforward of the structure sheaf on $V$, and the structure sheaf on $V$ is the restriction of the structure sheaf on $X$.)

So there is a completely functorial, purely topological mechanism for moving from the affine variety $V$ to the affine scheme $X$, and back again, and so the two objects carry identical information. But sometimes it is convenient to work explicitly on $X$, so that all the generic points are available; it often simplifies sheaf-theoretic arguments (but any argument using the generic points can be rephrased in a way that works entirely on $V$, via the above discussion). And of course the affine scheme $X$ sits in a wider world of all schemes, not all of which correspond to affine varieties, or to varieties at all, and this is often useful too.

Answer 2

You are right that the identification is not a literal one. The points of the variety correspond to the maximal ideals. In addition to this Grothendieck is adding some new points. Not just a generic point for the variety itself but a generic point for every irreducible subvariety. The idea of a generic point is a classical one and has been nicely incorporated into scheme theory. This is part of the genius of the idea.

Answer 3

$\def\bbA{\mathbb{A}} \def\spec{\operatorname{Spec}} \def\sA{\mathcal{A}} \def\sO{\mathcal{O}}$Suppose $k$ is algebraically closed, and let $X\subset\bbA_k^n$ be a classical affine variety, equipped with its sheaf of regular functions. Denote $\sA(X)=k[x_1,\dots,x_n]/\mathcal{I}(X)$ to the $k$-algebra of polynomial functions on $X$ (here, $\mathcal{I}(X)$ is the ideal of polynomials vanishing on $X$), the so-called coordinate ring of $X$. (There is a result that says that $\Gamma(X,\sO_X)=\sA(X)$, i.e., that a regular function defined on an classical affine variety is a polynomial.)

Then there is a canonical morphism $f:X\to\spec \sA(X)=X'$ of locally ringed spaces over $\spec k$. On spaces, this map is a quasi-homeomorphism (meaning $U\mapsto f^{-1}(U)$ is a bijection on open sets), and it is a homeomorphism onto $\operatorname{Spm}\sA(X)$. On structure sheaves, the morphism $\mathcal{O}_{X'}\to f_*\mathcal{O}_{X}$ is an isomorphism. Moreover, $f$ is universal among morphisms of locally ringed spaces $X\to Y$ (over $\spec k$ and not necessarily over $\spec k$), with $Y$ a scheme. In this very precise sense, $X'=\spec \sA(X)$ is the “schematization” of $X$.

I wrote the proof of all of this and more in The Classical-Schematic Equivalence (M. Haiman discusses this ideas in this text as well). It's basically a thorough account of the construction hinted by the user 'tracing' in their answer (the assignment they denote as $V\mapsto X$ is called the soberification functor), while proving in detail that everything works.