An example of a scheme in the language of schemes

Solution 1:

I am going to consider everything here over the field $\mathbb C$. You can replace $\mathbb C$ by any algebraically closed field (or even any field) with essentially no changes, but working over $\mathbb C$ is the natural starting point, and has the advantage that one can connect with the kind of geometry/topololgy with which you are more familiar.

Given a classical variety $V$, you can consider all the closed subvarieties. These satisfy the axioms of a topology, called the Zariski topology. For definiteness, let's say that our variety is an affine variety, so it is being cut out by polynomials in $\mathbb C^n$, for some $n$. The usual topology on $\mathbb C^n$ induces a topology on $V$, which has many more open sets than the Zariski topology (unless $V$ is $0$-dimensional). The point is that to be closed in the Zariski topology, you have to really be the zero locus of some polynomial, i.e. another variety, so it is hard to be Zariski closed, and hence similarly hard to be Zariski open. (Just to be absolutely clear, let's look at an example: the real line is closed in $\mathbb C$, but is not Zariski closed; there is no polynomial in one variable over $\mathbb C$ that vanishes precisely on the points of the real line; indeed, such a polynomial either vanishes at only finitely many points, or else is identically zero, and so vanishes everywhere.)

You also have the notion of rational function on the variety (just think of the restriction of a ratio of polynomials in $n$-variables to $V$, such that the denominator does not vanish identically on $V$); a rational function is called regular at a point $P$ of the variety if it has no singularity at that point. Being a singularity is a Zariski closed condition (singularities occur where the denominator of the rational function, which is a polynomial, vanish), so being regular at a point is a Zariski open condition. If we fix a Zariski open set in advance, we can look at the ring of all rational functions that are regular on that open set.

These form a sheaf on $V$ (with its Zariski topology). It is much "smaller" than the sheaves you are used to, like smooth or continuous functions. Not only are there many fewer open sets to think about (just the Zariski open ones), but on a given open set, there will be incredibly more continuous or smooth functions than regular functions, just because being the ratio of polynomials is a very restrictive condition on a function.

If we look at the global sections of this sheaf, i.e. the functions that are regular on the whole of $V$, we in particular get a ring which is called the affine ring of $V$. If I just hand you this ring (as a $\mathbb C$-algebra), it turns out that you can recover $V$, namely $V$ is the maximal spectrum of this ring (i.e. point of $V$ are in natural bijection with maximal ideals of $V$). The map one way is easy: given a point, we can look at all regular functions on $V$ that vanish at the point; this gives a maximal ideal in the ring of all regular functions. That this is a bijection is harder, and is essentially equivalent to the Nullstellensatz.

To see the role of the entire spectrum of the ring (i.e. the prime ideals as well as the maximal ideals) one has to say and think about more, but this is probably enough for now.

To learn more, you should google "affine ring of a variety" or similar expressions, and you should find troves of information, at a great range of levels. Once you understand this basic connection, it makes sense to look at schemes in more detail.

Solution 2:

You should read David Eisenbud and Joe Harris's The Geometry of Schemes.

Really. :)

Solution 3:

$\text{Spec } \mathbb{C}[x]$ is probably as basic as it gets. As a set, this is the collection of maximal ideals $(x - a), a \in \mathbb{C}$ together with the prime ideal $(0)$. As a topological space, this is $\mathbb{C}$ in the cofinite topology (the closed sets are the finite ones) together with a point $(0)$ whose closure is the entire space (and which is in every open set except the empty set). The local ring at $(x - a)$ is the subring of $\mathbb{C}(x)$ of rational functions which are defined at $a$, and the local ring at $(0)$ is all of $\mathbb{C}(x)$; more generally, the section of the structure sheaf over a Zariski-open set $U$ is the subring of $\mathbb{C}(x)$ of rational functions which are defined at every $a \in U$.

Solution 4:

Well, Qiaochu lists the affine line, so I'm going to do the projective line. (Everything here is over the complex numbers.)

The projective line $P^1$, or equivalently the Riemann sphere. This is the space of all lines through the origin in $\mathbb{C}^2$. Recall that the Riemann sphere can be obtained by gluing two copies of the complex plane. Namely, if $S^2 = \mathbb{C} \cup { \infty }$, then one copy is just the subset $\mathbb{C}$. The other copy is $\mathbb{C}^* \cup {\infty}$, which is identified with the complex numbers by taking reciprocals.

This is obviously a Riemann surface. In fact, the maps defining this "gluing" can be checked to be algebraic (they're just reciprocals), so it is in fact a scheme, and a toy example which isn't affine.

What's a regular function over an open subset of $S^2$? Well, one defines a holomorphic function over an open subset of $S^2$ by saying that the pull-back to each chart is holomorphic. Now here we say that the pull-back to each chart is regular in the sense of its being a rational function.

Interestingly, there are only constant functions which are regular on all of $P^1$. Indeed, such a function (considered as a map $P^1 \to \mathbb{C}$) would be a holomorphic map from a compact Riemann surface, hence constant. Here's an algebraic argument. Consider the open set $\mathbb{C}^*$. On this (affine) open set, it can be checked that the only regular functions are polynomials in $z $ and $1/z$ (or otherwise the denominator would blow up). However, if the function is regular everywhere, then $1/z$ can't occur (that would blow up at the origin) and $z$ can't occur (that isn't defined at ${\infty}$).

The projective line is important because it is compact in the complex topology. The algebraic version of this is that it is proper over $\mathbb{C}$. In particular, any map out of it into a complex variety is a emph closed map. However, with affine varieties, you don't have this closed property anymore. For instance, the hyperbola $xy=1$ in the affine plane (clearly a closed set in the Zariski topology) projects to the non-open complement of the origin of $A^1$ via $(x,y) \to x$. This doesn't happen for the projective line. ("Proper" is an algebraic analog of "compact," just as "separated" is an analog of "Hausdorff.")