I am reading course notes on algebraic geometry, where a morphism of varieties is defined as follows ($k$ is an algebraically closed field):

Let $X$ be a quasi-affine or quasi-projective $k$-variety, and let $Y$ be a quasi-affine or quasi-projective $k$-variety. A map $f:X\to Y$ is called a morphism of $k$-varieties if $f$ is continuous, and if for every open subvariety $U$ of $Y$ and every regular function $f:U\to k$ the composition $h\circ f$ is a regular function on $f^{-1}(U)$.

I have trouble seeing the motivation for this definition. The above notion of morphism seems to imply that the 'structure' of a variety (what distinguishes it from a mere set) is the following:

  • A topology, the Zariski topology, which is extremely coarse (weak) compared to the Euclidean topology in the case $k=\mathbb{C}$.
  • For every Zariski-open $U\subseteq X$ a specification of which functions $U\to k$ are considered 'nice', i.e. regular.

Given that this is indeed the structure we want to assign to a variety, I agree with the above notion of morphism. In particular, I can see that two varieties are isomorphic exactly when their structure is 'the same' (so that the difference between them is just that their points have different names).

However, I fail to see why the above two bullets accurately capture what we want them to. In fact, I do not know when we want to consider two varieties as isomorphic, and why. I do not know why we want the curves defined by $x^2+y^2=1$ and $x^2+y^2=2$ to be deemed isomorphic, but not the affine plane and the punctured affine plane (except for just pointing back to the definition I am trying to motivate, and showing that no isomorphism exists algebraically, but that's not enlightening). I do know this in the category of smooth manifolds: I expect two spheres to be diffeomorphic because I can stretch one smoothly to exactly match the other. I expect a sphere and a torus not to be diffeomorphic because no matter how hard I try, I cannot strech the sphere and make it coincide with a torus.

Another example: the affine line and the cusp are not isomorphic, and the difference lies exactly in the singularity of the cusp (its... well, cusp). Is this what we want to encode, the behaviour of varieties near singularities (I suspect this is only part of what we want to encode)? Do we want two varieties to be isomorphic if there is a bicontinuous bijection between them that maps singularities to singularities of the same kind? (Here, I do not know what I mean by the 'kind' of a singularity, and in fact I don't even know what exactly I mean by a singularity.) I expect that isomorphic varieties will have 'analogous' singularities at corresponding points, but I suppose there is more to the structure of a variety than this (indeed, not all smooth varieties are isomorphic).

What do we want the structure of a variety to entail intuitively? What intuitive/geometric information is encoded in the 'structure' as outlined in the bullets above?

Edit. I want to know what is encoded in the structure of a variety on a vague and intuitive level. I do not require mathematical justification for the answers at all (no need to prove that this is what we encode).

  • 17
    $\begingroup$ This is the sort of question that will answer itself once you know more algebraic geometry... The definition you gave is the result of a couple of centuries of refining the notion, so it is quite understandable that it is not immediately compherensibe. The same thing happens with the definition of manifolds or connection, which would have seemed pretty arcane even to Gauss himself. $\endgroup$ Nov 25, 2012 at 20:31
  • 5
    $\begingroup$ The best way to deal with your question is for you to read up a bit ahead, browse books likes Reid's book for undergrads, and so on (as opposed to studying those books, say...), to get a feeling of what people actually do, which is the best way to understand what they want to do. $\endgroup$ Nov 25, 2012 at 20:34
  • 2
    $\begingroup$ @GeorgesElencwajg: I agree with your statement, but I don´t really understand the point you´re making. My problem is not with the definition of an isomorphism, or with any technical issue. My problem is motivating that the structure of a variety carries interesting information, and getting some intuition for what information it carries. $\endgroup$ Nov 25, 2012 at 23:51
  • 2
    $\begingroup$ @MarianoSuárez-Alvarez: You make a fair point. However, it seems to me that there should be some intuitive way of explaining what the structure of a variety encodes without justifying that this is what it encodes. When first reading the axioms defining a topological space, it is completely unclear what the structure (the set of opens satisfying some weird rules) encodes, and it takes a lot of experience and getting-used-to before one realizes that these axioms encode what you want them to. Nevertheless, there are vague but very useful formulations of what the structure of a topological (...) $\endgroup$ Nov 26, 2012 at 0:03
  • 2
    $\begingroup$ (...) space encodes: it encodes how the space is pieced together, and for nice examples will tell you about 'holes' in it. It also tells you whether the space is in some sense made of one piece, or of more than one. A topology also tells you about convergence. It is statements like these I am looking for, without requiring mathematical justification. $\endgroup$ Nov 26, 2012 at 0:05

4 Answers 4


Mariano Suarez-Alvarez's point about understanding the intuition as you learn the theory more is correct, but I'd like to give a partial answer to help guide your intuition. After all, it is possible to spend months or years learning algebraic geometry and come away with little intuition of what the whole subject is about.

First, algebraic varieties are geometric spaces which look locally like affine varieties. In this sense, the theory is developed similar to, say, the theory of manifolds where a manifold is defined to be a space that is locally Euclidean. Of course, that limits the local study of manifolds - any two manifolds are locally isomorphic. Not so for algebraic varieties, as there is a wide variety of affine varieties.

So I think you should begun by restricting your question to affine varieties. And the key is that affine varieties are completely determined by their ring of globally regular functions. In other words, two (irreducible) closed subsets of affine space are isomorphic iff we can find a global 'change of variables' that identifies the global regular functions on the two spaces. Rescaling $(x,y) \mapsto (\sqrt{2}x,\sqrt{2}y)$ yields the isomorphism between $x^2+y^2=1$ and $x^2+y^2=2$.

I'll modify your non-example (because $\mathbb{A}^2 \setminus \{0\}$ is not affine) and explain why $\mathbb{A}^1$ and $\mathbb{A}^1 \setminus \{0\}$ are not isomorphic. Their rings of regular functions are $k[T]$ and $k[T,T^{-1}]$ respectively, which are not isomorphic. So there can be no 'changes of variables' that identifies the two spaces.

One important caveat: when I say there is global 'change of variables' from $X \subset \mathbb{A}^n$ and $X' \subset \mathbb{A}^{n'}$, I am talking about using polynomial maps that are restricted from the respective affine spaces, but they only needed to be defined on the spaces $X$ and $X'$. For example $\mathbb{A}^1 \setminus \{0\}$ (viewed as $t \neq 0$) and $xy=1$ are isomorphic via $t \mapsto (t, 1/t)$ and $(x,y) \mapsto x$. Of course, $1/t$ is only a valid change of variables when $t \neq 0$, but fortunately we are only looking at points where $t \neq 0$.

The global story is similar, except that we cannot just compare globally regular functions. (For example, the only globally regular functions on any projective variety are the constant functions, yet intuitvely there ought to be many different projective varieties up to isomorphism.) So now we require a global 'change of variables' so that regular functions on local pieces match up with the regular functions on the corresponding local pieces.

I am not sure if this explanation is what you are looking for. Algebraic geometry is very much a function oriented theory. We compare spaces by looking at the functions on them. One can take such an approach to manifolds as well. But for manifolds we also have an intuition for what the possible change of variables are ('stretching' and 'twisting' and the like). It's much harder to tell such a story in algebraic geometry because algebraic varieties are so much more diverse. There are still some basic intutions such as you can't have an isomorphism between a smooth variety and a singular variety because isomorphisms give rise to (vector space) isomorphisms of tangent spaces. But there are lots of possible singularities, and getting a hold on them is a major on-going project in the field. For example, you could study plane curves in depth and learn to tell apart singularities in this case (using blowups). But then you'll quickly discover the singularities on surfaces are more complicated and those on higher dimensional varieties still more complicated and hard to get a handle on.

  • 2
    $\begingroup$ Yes, this is quite helpful. There are two things I struggle with in your answer. First (detail, I suppose): you write 'polynomial' and then use 1/t. This confuses me. Second: I understand that manifolds are locally euclidean, and that varieties are locally affine varieties, but in the latter the Zariski topology is used (correct?). For me, 'locally in the Zariski topology' is intuitively very different from 'locally' as 'near a point' because the Zariski topology is a very unusual topology (at least for me, as I'm used to working in spaces that are at least Hausdorff). Any thoughts? $\endgroup$ Dec 4, 2012 at 15:09
  • 2
    $\begingroup$ 'Polynomial' is the loose way of describing regular maps - functions which can locally (in the Zariski topology) be represented as ratios of polynomials. I was trying not to be sloppy, but I wasn't careful enough in my writing. For the second point, yes the Zariski topology is unintuitive at first, but that gets alleviated in part through experience. Many things that may seem intuitive now were not always so. Most of us have at least come to terms with the abstract definition of compactness via open covers, even though 'closed and bounded' is a lot more intuitive characterization ... $\endgroup$ Dec 4, 2012 at 15:39
  • 4
    $\begingroup$ for Euclidean spaces. But we learn the abstraction both for the ability to prove things more easily with the abstract definition as well as to provide a theory which generalizes past Euclidean spaces. Similarly, the Zariski topology feels more natural once you learn the various algebraic tricks to recover intuition as much as possible in this more abstract setting. And over time you will appreciate the power of this framework for its generality. And eventually you will come full circle to the sticking point that there are not enough open sets in the Zariski topology, so ... $\endgroup$ Dec 4, 2012 at 15:42
  • 2
    $\begingroup$ you'll learn about things like the etale topology and, more generally, Grothendieck topologies to address that problem while staying in an algebraic setting. Ultimately, this is a long-winded attempt at saying that there are many geometric problems where it is both helpful and natural to stay with an algebraic framework. It takes a good amount of time and study to appreciate the reasons. $\endgroup$ Dec 4, 2012 at 15:47
  • 3
    $\begingroup$ And one last comment on the Zariski topology: it's the only topology that is universal enough to be able to work with algebraic varieties defined over any (algebraically closed, if you like) field. There is no way to generalize the other open sets in the classical Euclidean topology over $\mathbb{R}$ or $\mathbb{C}$ to work over arbitrary fields, though there are generalizations to some fields (such as the $p$-adic numbers, $\mathbb{Q}_p$) which are very important in their own right. $\endgroup$ Dec 4, 2012 at 15:51

One theorem you might have come across is the set of morphisms $X \to Y$, for $X$ and $Y$ affine varieties, is in canonical bijection with the set of homomorphisms $A(Y) \to A(X)$ of their coordinate rings. In particular, up to isomorphism, there is a unique affine variety with a given coordinate ring (so long as that ring is actually of the right form to appear as a coordinate ring, that is, a finitely-generated, reduced algebra over the base field $k$). More generally, if $Y$ is affine but $X$ may not be, the morphisms $X \to Y$ are given by the homomorphisms of rings from $A(Y)$ to the ring of global regular functions on $X$.

The reason that this is is that we've defined affine varieties to have just enough structure to be controlled by their coordinate rings. The Zariski topology and the definition of a morphism are just ways of formalizing this. Now, not all varieties are affine, but they're all locally affine, so again, a morphism of varieties $X \to Y$ can be thought of as a bunch of ring homomorphisms from affine coordinate rings of affine subsets of $Y$ that agree on the necessary intersections. An affine variety can be thought of as telling you (1) a geometric object and (2) what polynomial functions on that geometric object look like.

Under this point of view, the reason that the affine line and a cuspidal curve aren't isomorphic is that their coordinate rings aren't isomorphic. The coordinate ring of the line is just $k[t]$, but, for example, the coordinate ring $k[x,y]/(y^2 - x^3) \cong k[t^2,t^3]$ of the cuspidal curve $y^2 - x^3$ in $\mathbb{A}^2$ differs by not being integrally closed. That is, there is a rational function (here $y/x$) that behaves like a regular function on the curve, but doesn't make sense as a regular function when extended to the plane. The non-isomorphism between the two coordinate rings precisely tells you about the cusp and the various problems it causes.

At this point, there are two motivations. One is that a lot of geometric constructions have nice, algebraic definitions. For instance, the actions of passing to a closed subvariety or omitting a closed subvariety, which in terms of the Zariski topology are just taking open and closed subsets, correspond respectively to taking a quotient or localization of a ring. Even this simple example shows why the algebraic structure is nice to have: while it's obvious what a closed subvariety should look like over $\mathbb{C}$, what its codimension should be, et cetera, you lose the nice picture as soon as you move to a non-topological field. If you care about solving polynomials over fields other than $\mathbb{C}$, you need a theory that mimics the geometric features of $\mathbb{C}$ (and thus allows you to talk about dimension, tangent vectors, smoothness, and so on) without using its analytic topology.

A slightly more complicated example is flatness. A ring homomorphism $R \to S$ is flat if, whenever $M \to N$ is an injection of $R$-modules, $S \otimes_R M \to S \otimes_R N$ is an injection of $S$-modules. Serre found that this slightly arcane algebraic condition is actually the best way to talk about deformations of varieties. Namely, given a morphism $f:X \to Y$ of varieties, you can think of the various fibers $f^{-1}(y)$ for $y \in Y$ to be deformations of each other; if the morphism locally corresponds to a flat ring homomorphism, then these deformations are actually nicely behaved (they'll all be the same dimension, and so on).

If you're interested in solving polynomial equations over fields, this should hopefully be enough motivation. This is a fascinating subject on its own, of course. For example, Serre's GAGA established deep relations between the algebraic geometry of varieties over $\mathbb{C}$ and their analytic geometry as manifolds; statements like Fermat's Last Theorem are naturally statements about solutions to polynomials over fields; and I hear that solutions to polynomials over finite fields are important to modern-day cryptography, among other things.

If this isn't enough, here's the second piece of motivation. Even if you don't care about varieties, there are a lot of reasons to care about rings. Varieties can be thought of as nicer versions of rings, ones that we can patch together and use geometric intuition on, and some statements purely about rings can be cleanly proved in the language of varieties. The one caveat is that only certain rings appear as affine coordinate rings of affine varieties. For example, if you're interested in number theory or diophantine equations, you might want to study rings that aren't algebras over any field; other areas of math naturally give rise to non-noetherian rings, which then don't arise as finitely generated algebras over fields. A major program of Grothendieck in the 1960's, which has greatly influenced modern-day algebraic geometry, was to generalize varieties to more general objects, called schemes, that do for arbitrary rings what varieties do for finitely generated, reduced algebras over a field. Now, a scheme is much harder to picture than a variety, but it can be tremendously helpful in terms of applying some sort of geometric intuition to arbitrary rings.

So, in summary, varieties have rings attached to them given by the polynomials 'defined on' that variety, and a morphism of varieties is supposed to pull polynomials on its codomain back to polynomials on its domain (which Michael Joyce's answer explained quite well). The structure of a variety is meant to encode precisely this information, and thus to allow us to use our geometric intuition over fields where analytic reasoning is no longer possible. More generally, defining schemes allows us to use this geometric intuition to understand arbitrary rings.

Algebraic geometry is an old, wide, and fascinating field, and I guarantee you that if you continue to study it, you will find some key idea that puts all this into context and makes everything make sense for you personally. A few good sources to begin that journey are Harris's Algebraic Geometry: A First Course (for the variety-centric point of view), Hartshorne's Algebraic Geometry and Vakil's excellent online notes (if you're interested in schemes), and Eisenbud and Harris's The Geometry of Schemes (only after you've gained a bit of experience -- it attempts to provide motivation and geometric intuition for many ideas about schemes). Best of luck with it!

  • 1
    $\begingroup$ Thank you Paul, I like your answer. I particularly appreciate the description of flatness. I have decided to accept Michael's answer because it was helpful, but award you the bounty because your answer comes closer to the intuition I am looking for. $\endgroup$ Dec 6, 2012 at 15:37

Just to add to the discussion, I would say that there is geometric content in the "extra data" of being a variety, but this data can be incredibly subtle.

For instance, lets consider a smooth, projective surface $X$, a complex variety whose underlying space $|X|$ is an oriented manifold of real dimension 4. For any algebraic curves $Y_i$ on $X$, we have the associated oriented $2$ dimensional real manifolds $|Y_i|$ on $|X|$. Topologically, we can isotope (perturb) any two $|Y_i|$, $|Y_j|$ to having tranverse intersection, a finite number of points in $|X|$, which we count with signs accordinging to orientation, to obtain the "intersection number", denoted $\langle Y_i,Y_j\rangle$. Note that since we are counting with orientation taken into account, this can be negative, and it turns out this number is independent of how we perturb our surfaces $|Y_i|$ and $|Y_j|$.

This is so far, all very geometric, we are wiggling surfaces in a four dimensional manifold, and counting points of intersection. So lets imagine now collapsing all of these surfaces to a single point, building a (potentially singular) space $\overline{X}$. So now we can ask, when can we lift this operation to the level of varieties? When can we collapse the curves $Y_i$ to a single point, but this time within the category of algebraic varieties? Essentially, when does this collapsed space and collapsing map arise from a map of varieties?

It turns out this is possible if and only if the symmetric matrix $M_{i,j}=\langle Y_i,Y_j\rangle$ is negative definite, and this is a theorem due to Grauert. So the "variety structure" forces some very subtle constraints on what kind of maps of topological spaces are allowed. This is a special case of a much more general theorem, the decomposition theorem, which is a result due to Deligne and others, which to quote Macpherson “contains as special cases the deepest homological properties of algebraic maps that we know.”


Question: "What do we want the structure of a variety to entail intuitively? What intuitive/geometric information is encoded in the 'structure' as outlined in the bullets above?"

Answer: If $X \subseteq \mathbb{P}^n_{\mathbb{C}}$ is a complex manifold and $U \subseteq X$ is an open subset (in the strong topology), we may define the ring $\mathcal{O}_X(U)$ of holomorphic functions $f: U \rightarrow \mathbb{C}$. If $X,Y$ are complex manifolds, a morphism

$$\phi: X \rightarrow Y$$

is "holomorphic" iff $\phi$ is continuous and for any open set $V \subseteq Y$ it follows for any function $f\in \mathcal{O}_Y(V)$, the composed function $f \circ \phi:U:=\phi^{-1}(V) \rightarrow \mathbb{C}$ is a holomorphic function on $U$: $f \circ \phi \in \mathcal{O}_X(U)$. This gives a map of sheaves

$$\phi^{\#}: \mathcal{O}_Y \rightarrow \phi_*\mathcal{O}_X$$

of $\mathbb{C}$-algebras, and the pair $(\phi, \phi^{\#})$ is a "map of complex manifolds". A "similar" construction exists for algebraic varieties and schemes.

Algebraic varieties: In Chapter I.3 in Hartshorne, the structure sheaf $\mathcal{O}_X$ is defined for any quasi projective algebraic variety $X \subseteq \mathbb{P}^n$ over any algebraically closed field $k$. A morphism of algebraic varieties

$$(\phi, \phi^{\#}): (X ,\mathcal{O}_X) \rightarrow (Y, \mathcal{O}_Y)$$

is defined similarly: The map $\phi: X \rightarrow Y$ is continuous in the Zariski topology and the map of sheaves

$$\phi^{\#}: \mathcal{O}_Y \rightarrow \phi_*\mathcal{O}_X$$

has the same properties as for complex manifolds: It is a map of sheaves of $k$-algebras.

Schemes: If $(f,f^{\#}):(X, \mathcal{O}_X) \rightarrow (Y, \mathcal{O}_Y)$ is a map of schemes (over a scheme $S$) it follows for any open set $U \subseteq Y$ there is a map of rings $f^{\#}_U: \mathcal{O}_Y(U) \rightarrow \mathcal{O}_X(f^{-1}(U))$. If $S:=Spec(A)$, the map $f^{\#}_U$ will be a map of unital $A$-algebras. This is a "more abstract" construction, since the ring $\mathcal{O}_X(U)$ may not be interpreted as the "ring of functions" on $U$ taking values in a field as in the case of complex manifolds and algebraic varieties over an algebraically closed field.

Edit. "I want to know what is encoded in the structure of a variety on a vague and intuitive level. I do not require mathematical justification for the answers at all (no need to prove that this is what we encode)."

Answer: An algebraic variety is a topological space ($X, \tau)$ and a sheaf of rings $\mathcal{O}_X$ on $X$ in $\tau$. The functions you study in the case of algebraic varieties are rational functions - quotients of polynomials. If you want to "classify" algebraic varieties up to "isomorphism" you must specify what are the "isomorphisms of algebraic varieties", and this involves constructing the structure sheaf $\mathcal{O}_X$.

Two algebraic varieties $(X ,\mathcal{O}_X), (Y, \mathcal{O}_Y)$ are isomorphic iff there are maps $\phi, \psi$ with

$$f:=(\phi, \phi^{\#}): (X ,\mathcal{O}_X) \rightarrow (Y, \mathcal{O}_Y)$$


$$g:=(\psi, \psi^{\#}): (Y ,\mathcal{O}_Y) \rightarrow (X, \mathcal{O}_X)$$

with $f\circ g=g \circ f"="Id$. (Here I should specify $Id_X, Id_Y$).

Bountied question: "Put another way, what exactly is 'geometric' about the modern, intrinsic algebraic geometry of abstract varieties?"

Example: If $k$ is the field of complex numbers, $X$ is "smooth" and $U \subseteq X$ is a Zariski open set, any function $s\in \mathcal{O}_X(U)$ is holomorphic. If $\tau_Z$ is the Zariski open sets in $X$ and $\tau_s$ is the open subsets in the "strong" topology, there is an inclusion $\tau_Z \subseteq \tau_s$ and a continuous map $id: (X, \tau_s) \rightarrow (X, \tau_Z)$: For any open set $U \in \tau_Z$ it follows $id^{-1}(U):=U \in \tau_s$. Hence the identity map is a continuous map. Let $\mathcal{O}_X^s$ denote the structure sheaf of holomorphic functions on $X$ in $\tau_s$. We get a map of ringed spaces

$$(id, id^{\#}):(X,\tau_s, \mathcal{O}_X^s) \rightarrow (X, \tau_Z, \mathcal{O}_X)$$

defined by sending a local sections $s \in \mathcal{O}_X(U)$ to "itself": We may view $s \in \mathcal{O}_X^s(U)$ since $\mathcal{O}_X(U) \subseteq \mathcal{O}_X^s(U)$ is a sub ring. Hence for any smooth quasi projective algebraic variety $X$ we may construct the corresponding "complex quasi projective manifold" $(X, \tau_s, \mathcal{O}_X^s)$. We may classify $X$ as algebraic variety and as complex manifold, and when doing this we use the structure sheaves $\mathcal{O}_X$ and $\mathcal{O}_X^s$. Changing the structure sheaf from the first to the second means changing from "rational functions" to "holomorphic functions". There is a canonical map

$$id^{\#}: id^{-1}(\mathcal{O}_X) \rightarrow \mathcal{O}_X^s$$

and for any coherent $\mathcal{O}_X$-module $\mathcal{E}$ we may construct


If $\mathcal{E}$ is locally free it follows $\mathcal{E}^s$ is locally free. We get a functor $F: Coh(\mathcal{O}_X) \rightarrow Coh(\mathcal{O}_X^s)$ defined by $F(\mathcal{E}):=\mathcal{E}^s$. When the variety $X$ is smooth and projective we do not "get anything new" when passing to holomorphic functions: This is expressed by saying there an "equivalence of categories" between the category of coherent sheaves on $X$ viewed as algebraic variety and the category of coherent sheaves on $X$ viewed as a complex manifold - the above defined functor is an equivalence of categories. Hence if you study vector bundles on complex projective manifolds you are studying algebraic varieties and algebraic geometry.

Bountied question: "As opposed to the more classical study of sets of polynomial equations - in this case I have a geometric picture, but clearly the intrinsic morphisms don't capture most of that geometry, as for example a line, a circle, and a twisted cubic are somehow abstractly isomorphic, while the cubic curve $y^2=x^3−1$ isn't isomorphic to them - in fact not even birational to them; But geometrically I don't see what differentiates it from the others, nor do I see geometrically what puts the others in the same class."

Example: An isomorphism of algebraic varieties $f:X \rightarrow Y$ would induce isomorphisms of tangent spaces (for any $x\in X$ with $f(x)\in Y$):

$$T_x(X) \cong T_{f(x)}(Y).$$

If $x$ is non-singular and $f(x)$ is singular, the above tangent spaces have different dimensions and cannot be isomorphic. Hence if $X$ is non-singular and $Y$ is singular there can be no such isomorphism: Singularity is preserved under isomorphisms.

Example: If $p: \mathbb{C}[y] \rightarrow \mathbb{C}[x,y]/(y^2-x^3)$ is the canonical map, we get a map of algebraic varieties

$$(p,p^{\#}): C \rightarrow \mathbb{A}^1_{\mathbb{C}}$$

where $C:=V(y^2-x^3) \subseteq \mathbb{A}^2_{\mathbb{C}}$. The map at the level of topological spaces $p: C \rightarrow \mathbb{A}^1$ is a homeomorphism, but the tangent spaces at $z:=(0,0)$ have different dimensions: $T_z(\mathbb{A}^1)$ is one dimensional and $T_z(C)$ is 2-dimensional. Hence the two algebraic varieties $\mathbb{A}^1$ and $C$ are not isomorphic.

Note: If you ignore the structure sheaves $\mathcal{O}_{\mathbb{A}^1}$ and $\mathcal{O}_C$ in the above construction you end up making mistakes: The underlying topological spaces $\mathbb{A}^1$ and $C$ are homeomorphic, but the ringed spaces $(\mathbb{A}^1, \mathcal{O}_{\mathbb{A}^1})$ and $(C, \mathcal{O}_C)$ are not isomorphic. Hence if you want to work with the problem of classifying algebraic varieties you cannot ignore the structure defined by the structure sheaf.

Example: With the definition in HH.I.3 a morphism $(\phi,\phi^{\#})$ of quasi projective algebraic varieties, induce a map of $k$-algebras

$$ \phi^{\#}_X:\Gamma(Y, \mathcal{O}_Y) \rightarrow \Gamma(X, \mathcal{O}_X),$$

and if $X,Y$ are affine algebraic varieties, it follows $(\phi, \phi^{\#})$ is uniquely determined by $\phi^{\#}_X$. The rings of global sections $\Gamma(Y, \mathcal{O}_Y), \Gamma(X, \mathcal{O}_X)$ are finitely generated $k$-algebras and we can describe all maps between them. Hence $X$ and $Y$ are isomorphic (as varieties over $k$) iff there is an isomorphism of $k$-algebras $\Gamma(Y, \mathcal{O}_Y)\cong \Gamma(X, \mathcal{O}_X)$. If $X,Y$ are projective, it follows $\Gamma(Y, \mathcal{O}_Y)$ and $\Gamma(X, \mathcal{O}_X)$ are isomorphic to $k$ and there is only one $k$-algebra map $\phi: K \rightarrow k$ - the identity map.

Comment: "To summarize, the structure of a variety simply does not encode any geometric information (other than some rather poor geometric notions like the dimension and singularities). Rather, it mostly encodes algebraic information, and this algebraic information can then be used to imbue the variety with geometric information by (for example) choosing an embedding into projective space."

Answer: If you have been active on this forum you will find that a number of students are asking about the relation between the Zariski tangent space $T_x(X):=(\mathfrak{m}_x/\mathfrak{m}_x^2)^*$ and the "embedded tangent space". If you read Mumford's book "the red book.." you wil find a discussion of the relation between the "embedded tangent space" and the Zariski tangent space. (see the attached example)

About the definition of tangent space and tangent cone.

If your variety $X$ is affine, the embedded tangent space $E_x(X)$ is defined in terms of an embedding into affine space, and one has to prove that $E_x(X)$ is independent of choice of embedding. This is why we introduce the Zariski tangent space - It is isomorphic to $E_x(X)$ and does not depend on the choice of embedding. You seem to believe that by choosing an embedding of $X$ into affine or projective space makes things "more geometric". This is not the case - whenever you make a definition using an embedding you must prove that the definition is "independent of choice of embedding". It saves you much effort to instead defined things in terms of intrinsic objects such as the local ring, or the structure sheaf. Anything that can be defined in terms of the structure sheaf is independent of such an embedding.

Example: If $Y\subseteq \mathbb{P}^n_k$ is a projective variety (in the sense of HH, chapter I), the arithmetic genus $p_a(Y)$ is in HH. Ex.I.7.2 defined using the Hilbert polynomial, which depends on an embedding into projective space. In HH. Ex.III.5.3 they prove that this definition is independent of choice of embedding using the structure sheaf $\mathcal{O}_Y$. Are you claiming that for this reason the arithmetic genus is not a "geometric" invariant?

"...other than some rather poor geometric notions like the dimension and singularities"

Your comment: "Only this gives a geometric interpretation to the variety. Looking from this perspective, it is now obvious why essentially all theorems in algebraic geometry which have a clear geometric interpretation (rather than being algebraic theorems) depend on choosing an embedding into projective space - like Bezout's theorem or the 27 lines."

Example: If $X$ is any scheme and $L\in Pic(X)$ is any invertible sheaf, a hypersurfaces $H(s)$ in $X$ is by definition the zero scheme $H(s):=Z(s)$ of a global section of $L$. If $i:S\subseteq \mathbb{P}^3_k$ is a cubic surface and if $L:=\mathcal{O}(1)=i^*\mathcal{O}_{\mathbb{P}^3}(1)$, it follows a line in $S$ wrto the embedding $i$ is such a zero scheme: $l:=Z(s)$ for $s$ a global section of $L$. The Picard group $Pic(X)$ is "intrinsic", does only depend on $X$, does not refer to any embedding $i$ and you may construct the line $l$ without reference to $i$. Hence the line $l \subseteq S$ is "intrinsic" in this sense.

Example: In the case of the Bezout theorem, let $S:=\mathbb{P}^2$ and let $L_d:=\mathcal{O}_S(d)$ with $s_d\in H^0(S, L_d)$ a global section, with zero scheme $H_d:=Z(s_d) \subseteq S$. The theorem says that the scheme $H_d \cap H_l$ has $dl$ points "counted with multiplicity". Equivalently: The length of the scheme $H_d \cap H_l$ is $dl$. Hence you must include the notion "multiplicity" when studying the Bezout theorem.

Example: In the construction of the tangent bundle $T(M)$ of a differentiable manifold $M$ you define the tangent space $T_x(M)$ at each point $x∈M$ and take the "disjoint union". Then you prove that $T(M)$ has the structure of a differentiable vector bundle using an atlas. In algebra/geometry for any k-algebra $A$, you defined the module of differentials $Ω:=Ω^1_{A/k}$ and its dual $Ω^∗$ and when $k$ is a field of characteristic zero and $A$ a finitely generated regular $k$-algebra it follows $Ω$ and $Ω^∗$ are finite rank projective $A$-modules. The "fiber" of $Ω^∗$ at a $k$-rational point $m$ is the Zariski tangent space $(m/m^2)^∗$, hence $Ω^∗$ is the "module of sections" of the tangent bundle $T_{X/k}$ where $X:=Spec(A)$. With this definition we do not use an embedding of $X$ into an affine space to define $T_{X/k}$ - the definition is "intrinsic" and independent of choice of coordinates. Some people - when constructing the tangent bundle of $X$ - define the Zariski tangent space at any point $x$ and state that "these tangent spaces patch together to form the tangent bundle". When they do this, they are implicitly using the above construction and use the "hand-waving method" to convince their students.


Note: For affine space (or any affine variety of finite type over a field $k$) the answer is simple: A map $\phi: U:=\mathbb{A}^n_k \rightarrow V:=\mathbb{A}^m_k$ is given by $m$ polynomials $p_i$ in $n$ variables. Equivalently it is given by a map $\phi^*: A(V) \rightarrow A(U)$ of $k$-algebras. For a map of quasi projective varieties $f:X \rightarrow Y$ and any point $x\in X$ with $y:=f(y)$ there are open affine subsets $x\in U:=Spec(B), y\in V:=Spec(A)$ with $f(U) \subseteq V$ and where the induced map

$$f_U: U \rightarrow V$$

is induced by a map of $k$-algebras $f_U^*: A \rightarrow B$.

  • 2
    $\begingroup$ For me, this answer does not bring any geometric intuition. (Example: in the first answer section, defining things in terms of morphisms of $\mathbb{C}$-algebras is exactly the kind of thing the question is not after.) $\endgroup$ Jul 23, 2021 at 18:45
  • $\begingroup$ @DaanMichiels - when studying algebraic varieties you use the Zariski topology, polynomial/rational functions and the structure sheaf of regular functions $\mathcal{O}_X$. When studying complex manifolds you use the "strong topology", holomorphic functions and the sheaf $\mathcal{O}^{hol}$ of holomorphic functions. Hence when defining a "morphism of algebraic varieties" you must include the structure sheaf in the definition, similar for a "morphism of complex manifolds". $\endgroup$
    – hm2020
    Jul 24, 2021 at 10:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .