Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Somewhat related to this question, but almost infinitely more basic.

A Confession

I am, should classification prove essential, a differential geometer and a topologist by inclination and by training: as an undergraduate I shunned any ring that wasn't $\mathbb{Z}_n$ or a ring of differential operators and held close the differentiable and the non-singular. It did not seem to matter then that these exotic 'schemes' and their exciting projective morphisms were beyond me, and to a certain extent it does not seem to matter now; but increasingly my old uni friends, fellow MOers (and, hey, even math.stack exchangers) are talking about nothing else but schemes.

In recent months (after a frighteningly eye-opening MO question) I have found myself becoming more amenable to rings, and am less daunted by my paucity of understanding than previously. In spite of this, though, I remain entirely in the dark about schemes.

Where I Sit

I am not a complete novice. I completed an undergraduate course in algebraic geometry: illuminating, interesting, but all classical beyond belief. I have read and re-read the wikipedia page on schemes several times- taking in all of the neccessary components: the spectrum of a ring, a locally ringed space et al, but have no idea how these fit together to make the objects I fiddled with over a semester two years ago.

I have made numerous guesses about generalised nulstellensatze and structure sheaves, but to explain any would probably be to complicate matters further unneccessarily. I am aware there are probably brilliant texts that do exactly what I am asking for, but I am not currently affiliated to a university and my current library would require ordering in, which for the sort of toe-dipping excercise I intend here would be overkill. So I ask:

Can anyone provide me with a canonical example of a scheme, pointing along the way the topology and the spectra associated to each open set. Perhaps deeper, if it pleases: what I am looking for is a sort of 'scheme jargon safari'.

I am aware this is silly, and perhaps asking for a verbatim quotation of page 2 of any decent algebraic geometry text, but I would be ever so grateful. Can anyone help?

share|improve this question
It is not really clear to me what kind of details you want. What part of the construction and its relation to the classical theory, exactly, don't you understand? –  Qiaochu Yuan Aug 5 '10 at 23:45
I literally have no idea how it all fits together. What I am asking is as basic as: "given a classical variety, what are the open sets and how does one associate the spectrum of a ring to these open sets, and how does this characterize the variety?" –  Tom Boardman Aug 5 '10 at 23:50
I'm the poster of the question you linked to. I've found Ravi Vakil's notes (math.stanford.edu/~vakil/0708-216) supplemented by Eisenbud and Harris's Geometry of Schemes to be a very good combination for learning about this stuff. –  Eric O. Korman Aug 6 '10 at 0:53

5 Answers 5

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.

share|improve this answer
this is helpful, i don't know why it never occurred to me that these sheaves in AG are significantly smaller than the ones i would think of naturally. –  Sean Tilson Dec 6 '10 at 22:18

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

Really. :)

share|improve this answer
This is a very good suggestion! You might also try these notes: math.upenn.edu/~chai/624_08/math624_08.html –  Matt E Aug 6 '10 at 0:46
If reading an entire book seems intimidating, let me just say that I think reading just Chapter I of Eisenbud-Harris is probably sufficient for a lot of what Tom is asking for. –  Kevin H. Lin Aug 6 '10 at 5:22
It is a small book, anyways :D –  Mariano Suárez-Alvarez Aug 6 '10 at 5:52

$\text{Spec } \mathbb{C}[x]$ 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 $a \in U$.

share|improve this answer
To really grok this example, compare it to $\mathrm{spec}\mathbf{R}[x],\mathrm{spec}\mathbf{Q}[x]$ and $\mathrm{spec}\mathbf{Z}[x]$! –  Sam Lichtenstein Aug 25 '10 at 4:13

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.")

share|improve this answer
+1 because I love P^1 –  BBischof Aug 6 '10 at 3:03

The picture on Mumford's Red book for the arithmetic surface $Spec\ \mathbb Z[x]$ is very enlightening. The book itself is a fantastic read.

Lieven le Bruyn has a copy of that picture.

The book ought to be read. What is equally enlightening is the functorial point of view emphasized by Grothendieck, that schemes are representing their functor of points and thus can be thought of as the functor instead. This is an easy consequence of Yoneda lemma. A delightful book with this point of view is Mumford, "Lectures on curves on an algebraic surface".

share|improve this answer
Maybe you could explain parts of the picture. I understand the ideals being listed and $V(I)$ but not exactly what the lines or curves are doing or how to interpret their intersection. –  Sean Tilson Dec 6 '10 at 22:14
Well, is the description by Lieven Le Bruyn at neverendingbooks.org/index.php/mumfords-treasure-map.html enough? –  user1119 Dec 6 '10 at 22:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.