## June 22, 2006

### Varieties and Schemes for Dummies, Part II

#### Posted by urs

More details on how varieties are a subcategory of schemes.

Last time I ended after having sketched how to identify the structure sheaf of a variety with that of the affine scheme that is the spectrum of the variety’s coordinate ring.

While I indicated how to associate regular functions on $\mathrm{Spec}A\left(X\right)$ to regular functions on the variety, I did not say anything about the underlying continuous map between the underlying topological spaces.

So here is how to do that.

The problem to be solved is the following: The variety, being a set of zeros, consists of lots of points in the ordinary sense. In addition, certain subsets of these points are distinguished as being algebraic sets. Maybe I did not even define these last time.

So

Definition. ([Hart, p. 2])

$•$ a subset $Y$ of the affine space ${A}_{k}^{n}$ is an algebraic subset if it is the set of common zeros of some collection of polynomials

(1)$Y=Z\left(\left\{{f}_{1},\cdots ,{f}_{n}\right\}\right)$

$•$ The Zariski topology on ${A}_{k}^{n}$ is that in which precisely the algebraic subsets are closed.

$•$ Any nonempty subset $Y$ is called irreducible if it is not the union of two closed proper subsets of $Y$.

$•$ An affine variety is an irreducible Zariski-closed subset of ${A}_{k}^{n}$.

We want to reformulate everything algebraically in terms of ideals. As a corollary of Hilbert’s Nullstellensatz one gets the dictionary

(2)$\begin{array}{ccc}\text{algebraic sets}& ↔& \text{radical ideals}\\ \text{irreducible algebraic sets}& ↔& \text{prime ideals}\end{array}\phantom{\rule{thinmathspace}{0ex}}.$

Moreover, if the ground field $k$ is algebraically closed (such that every polynomial decomposes into linear factors), we have

(3)$\begin{array}{ccc}\text{maximal ideals}& ↔& \text{single points}\end{array}\phantom{\rule{thinmathspace}{0ex}}.$

Notice that

(4)$\text{maximal ideals}\subset \text{prime ideals}\subset \text{radical ideals}$

are, in general, proper inclusions.

But recall that the spectrum of some ring (which is what we are interested in) is defined to be not the collection of the ring’s maximal ideals (corresponding to points if the ring is that of polynomials over an algebraically closed field), but of the ring’s prime ideals.

For rings of polynomials $k\left[{X}_{1},\cdots ,{X}_{n}\right]$, irreducible polynomials generate prime ideals. For $n=2$ these are known as affine curves, for $n=3$ as surfaces and for higher $n$ generally as hypersufaces (unsurprisingly).

The maximal ideals ${𝔪}_{p}$ (the points), however, are those generated by $n$ linear polynomials ${𝔪}_{p}=\left({X}_{1}+{p}_{1},\cdots ,{X}_{n}-{p}_{n}\right)$.

In a word, $\mathrm{Spec}A$, being the collection of prime ideals of $A$ instead of just that of maximal ideals, contains more than just the ordinary points.

The additional points it contains - the non-maximal prime ideals of $A$ - are known as open points or generic points. That’s because their closure is an irreducible subset consisting of more than a single ordinary point.

That might sound more mysterious than it is. The easiest example is this:

Consider the affine $n$-space ${A}^{1}$ with coordinate ring $A=k\left[X\right]$, the polynomials over the algebraically closed field $k$.

The maximal ideals are those generated by monic polynomials; of the form $X-p$, corresponding to points $p\in k$.

But there is also a prime ideal which is not maximal: the zero ideal $\left(0\right)$, which contains only the zero polynomial.

The set of zeros of the zero polynomial is obviously the entire affine space ${A}^{1}$. Equivalently, every maximal ideal contains the zero ideal. Hence every point $\left(X-p\right)$ sits inside the closure of $\left(0\right)$.

But regarded as an element of $\mathrm{Spec}A$, $\left(0\right)$ is regarded as a point, too. But it is not closed, since, by definition of the Zariski topology on $\mathrm{Spec}A$, closed subsets are of the form $V\left(𝔞\right)$, which is the set of all prime ideals containing a given ideal $𝔞$. So $\left\{\left(0\right)\right\}$ is an open set in the Zariski topology and its closure is $V\left(\left(0\right)\right)$, the set of all prime ideals containing the zero ideal. But that’s the entire space $V\left(\left(0\right)\right)=\mathrm{Spec}A$.

In this sense the “point” $\left(0\right)$ touches every other point in $\mathrm{Spec}A$. Hence we call it “generic”.

The same reasoning applies of course to ${A}_{k}^{n}$ and to all irreducible closed subsets of ${A}_{k}^{n}$.

Consider ${A}_{k}^{2}$. Again $\left(0\right)$ is the generic point of the entire affine plane.

Pick any irrducible polynomial $P\left({X}_{1},{X}_{2}\right)\in k\left[{X}_{1},{X}_{2}\right]$. Its set of zeros is some a curve in ${A}_{k}^{2}$. The ideal it generates is a prime ideal, but not a maximal ideal.

Hence, in $\mathrm{Spec}k\left[{X}_{1},{x}_{2}\right]$, the ideal $\left(P\left({X}_{1},{X}_{2}\right)\right)$ becomes a point which is not closed. Its closure $V\left(P\left({X}_{1},{X}_{2}\right)\right)$ is the set of all ordinary points which lie on the curve. So now the “point” $P\left({X}_{1},{X}_{2}\right)$ is the generic point of that curve.

To summarize this, the closed points of $\mathrm{Spec}k\left[{X}_{1},\cdots ,{X}_{n}\right]$ are in bijection with the ordinary points of the affine variety ${A}_{k}^{n}$. In addition, $\mathrm{Spec}k\left[{X}_{1},\cdots ,{X}_{n}\right]$ contains an open generic point for every irreducible algebraic subset (curve, surface, etc.) of ${A}_{k}^{n}$, such that the closure of this generic point coincides with this subset.

For these reasons we cannot expect that a variety itself can be isomorphic to the spectrum of its coordinate ring. It does not contain the generic points. But we can easily remedy this by

$•$ constructing for every variety $X$ the topological space $t\left(X\right)$ which consists of all irreducible closed subsets of $X$ (points, curves, surfaces, etc),

$•$ by inducing a suitable topology on $t\left(X\right)$

$•$ such that there is a natural continuous map $\alpha :X\to t\left(X\right)$

$•$ which allows us to push the structure sheaf of $X$ along it over to $t\left(X\right)$

$•$ thus obtaining the locally ringed space $\left(t\left(X\right),{\alpha }_{*}{𝒪}_{X}\right)$

$•$ which then has a good chance of being an affine scheme.

Indeed, this is how it works. David Mumford gives a formulation on p. 124 of his book. I’ll stick to Robin Hartshorne’s book, p. 78.

So that’s how we associate a scheme to a variety $X$: we first push forward the structure sheaf ${𝒪}_{X}$ to the set of all generlized points $t\left(X\right)$ along $\alpha$. Then, as described in the previous entry, we identify sections of $\mathrm{Spec}A\left(X\right)$ with those of ${\alpha }_{*}{𝒪}_{X}$ by evaluating them at stalks over ordinary points and dividing out the maximal ideal.

Hence on objects, the functor we are looking for works like

(5)$\begin{array}{ccccc}t& :& \mathrm{Var}\left(k\right)& \to & \mathrm{Sch}\left(k\right)\\ & & \left(X,{𝒪}_{X}\right)& ↦& \left(t\left(X\right),{\alpha }_{*}{𝒪}_{X}\right)& \simeq \mathrm{Spec}A\left(X\right)\end{array}\phantom{\rule{thinmathspace}{0ex}}.$

In Hartshorne’s book it is an exercise (p. 81) to show that the natural induced map on morphisms

(6)$t:{\mathrm{Hom}}_{\mathrm{Var}\left(k\right)}\left(X,Y\right)\to {\mathrm{Hom}}_{\mathrm{Sch}\left(k\right)}\left(t\left(X\right),t\left(Y\right)\right)$

is bijective, so that $t$ is indeed fully faithful.

Being sloppy by nature, I have all along missed to say how the categories $\mathrm{Var}\left(k\right)$ and $\mathrm{Sch}\left(k\right)$ are defined in detail. Maybe now it’s time to do so.

Definition.([Hart, pp. 15]) The category of varieties over $k$, $\mathrm{Var}\left(k\right)$ has objects being varieties with respect to polynomials in $k$ and morphisms being smooth map between them, such that regular functions pull back to regular functions.

Definition.([Hart, pp. 78]) The category of schemes, $\mathrm{Sch}$, has schemes as objects and morphisms between them being morphisms of locally ringed spaces. $\mathrm{Sch}\left(k\right)$ is the slice category over $\mathrm{Spec}\left(k\right)$, called the category of schemes over $k$

In the end, we want to identify a subcategory of the category of schemes over $k$, which is equivalent to the category of varieties.

In David Mumford’s book this is done by identifying a candidate subcategory of $\mathrm{Sch}\left(k\right)$ and explicitly constructing an equivalence with $\mathrm{Var}\left(k\right)$.

Following Hartshorne, we can make use of the fact that two categories are equivalent precisely if there is a fully faithful and essentially surjective functor between them ($\to$). Since we already have a fully faithful functor $t$ from varieties into schemes, all that remains to be done is to identify the image of $t$ in $\mathrm{Sch}\left(k\right)$.

The result is (for $k$ algebraically closed, as always)

Fact.([Hart, prop. 4.10]) The sub-category in question is precisely that of quasi-projective integral schemes over $k$.

In particular, for any variety $X$, the scheme $t\left(X\right)$ is

$•$ integral,

$•$ seperated,

$•$ and of finite type

over $k$.

So I guess I should next say what all these attributes mean.

Posted at June 22, 2006 12:58 PM UTC

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