# Algebraic Geometry - Definition of a Morphism

What's the best (or your favourite) definition of a morphism of a quasi projective variety? I've seen a huge number of equivalent definitions, and I'd like to know which is the best to memorise!

I'd prefer to have a definition that doesn't mention sheaves/schemes or rational maps (since I normally define a rational map as a morphism on an open set, and don't want my definitions to be circular).

-
A natural transformation between the functors of points... – Qiaochu Yuan Mar 22 '12 at 20:19
Thanks Qiaochu, but I was looking for something perhaps a little more concrete. – Edward Hughes Mar 22 '12 at 20:27
@Edward: I don't see what's not concrete about Qiaochu's answer. Think of a variety as something like a system of equations. The "functor of points" just tells you what solutions that system has, in any given ring. A morphism is just a systematic way of turning solutions of one system into solutions of another. Don't be scared by the word "functor"! – Tom Leinster Mar 22 '12 at 22:21
I voted Yuan's comment up, but I was laughing at the time. – roy smith Mar 23 '12 at 2:15
I suggest you to read Mumford's red book on varieties and schemes. You'll get intuition both for Ruadhai's definition and for the functor of points. – Leo Alonso Mar 23 '12 at 15:13

A regular map $\phi: X \to Y$ of quasi-projective varieties is a continuous map with respect to the Zariski topology such that for $V \subset Y$ an open set and $f$ a regular function on $V$, we have $f\circ \phi$ is regular on $\phi^{-1}V$. This seems to me to be to be exactly what you would want and quite intuitive and understandable.

-
By your regular function f, do you mean a regular map from V to A^1? In which case this seems a little circular... Sorry if I've misunderstood! – Edward Hughes Mar 22 '12 at 21:04
Ah, it would be circular if you didn't know what a regular function was. A function $f: X \to \mathbb{A}^1$ on a quasi-projective variety $X$ is called regular if $\forall x\in X$ there is a neighbourhood $U$ of $x$ such that $f|_U$ is the quotient of two homogeneous polynomials $f|_U = F/G$ of the same degree such that $G$ has no zeroes on $U$. A good reference for this level of algebraic geometry is I.R. Shafarevich - Basic algebraic geometry I. – Ruadhai 0 secs ago – Ruadhaí Dervan Mar 22 '12 at 21:12
Marvellous that's a fantastic definition now, and easy to remember! Thanks a lot. – Edward Hughes Mar 22 '12 at 22:18
@Ruadhai - would it be equivalent to say that a map $\phi:V \rightarrow W$ is a morphism iff for all $p \in V$ there exist affine open sets $U$ of $p$ and $U'$ of $\phi(p)$ s.t. $\phi(U) \subset U'$ and $\phi|_U$ is a morphism of affine varieties? – Edward Hughes Mar 22 '12 at 22:41
This thread should be helpful, it's essentially the same question. mathoverflow.net/questions/1397/… – Ruadhaí Dervan Mar 22 '12 at 23:50

I'll leave an answer mainly because I don't want to discourage people from asking basic questions. Actually, I don't have a favourite definition, but I'll give you my least favourite. What I mean by that is that it is clumsy, and not obviously well defined. However, there should be no doubt that it captures the correct meaning.

1. A map of affine varieties is a morphism if it is expressible by polynomials.
2. A map $f:X\to Y$ of general (quasiprojective or not) varieties is a morphism if is locally a morphism of affine varieties. To make this more precise, we should require that $f$ is continuous, $Y$ has an open cover $\lbrace U_i\rbrace$ by affine varieties, $X$ has an affine open cover $\lbrace V_{j_i}\rbrace$ refining $\lbrace f^{-1}U_i\rbrace$, and $f:V_{j_i}\to U_i$ is a morphism in the sense of 1.

You can treat this as a yardstick for measuring the correctness of other more elegant definitions (mentioned above and elsewhere). This process takes time, and, unfortunately, there really isn't any shortcut.

-
Hm, in some sense this is my favorite definition... :-) – Karl Schwede Mar 23 '12 at 13:21
this reveals the fact that different definitions are useful for different purposes. E.g. if one wants to display an actual morphism, this one is hard to avoid. for abstract proofs, others may appeal more. – roy smith Mar 24 '12 at 19:22
e.g. See which definition helps most to find a morphism from the intersection curve of the quadric surfaces Q : {x^2 – xz – yw =0} and Q’ : {yz –xw-zw=0} in P^3, to the plane cubic C: {y^2.z = x^3–xz^2} in P^2. (hint: try the “obvious” one.) – roy smith Mar 24 '12 at 20:08
Donu, is this definition complete? I.e. the notion of a covering by "affine varieties" seems to require the prior definition of an isomorphism, hence of a morphism. What am I missing? – roy smith Apr 2 '12 at 16:14
Dear roy, for what it's worth in my mind I was imagining the definition of a variety to be some affine varieties and gluing data (which can be expressed as isomorphisms of affine varieties). – Karl Schwede Apr 2 '12 at 17:56

I confess I really like this question, because it troubled me a lot when trying to teach (and understand) beginning algebraic geometry. There seem to be two questions here: 1) which is the more fundamental notion, morphism or rational map; and 2) what special definition is preferred in the case of quasi projective varieties? I finessed these difficulties for years by working largely in the category of complex analytic varieties and appealing to Chow's theorems, so I am not expert, but I have thought about it. I will share my necessarily naive conclusions, so I can learn from the responses.

As to 1), the notion of rational map seems more fundamental except in the one case of polynomial maps of affine varieties. I.e. if one wants the concept of regularity to be local, one must define it using rational functions. This arises as soon as one defines a regular function on even a quasi affine variety. So it appears as if regular functions are naturally a subclass of rational ones. I.e. first one must understand rational functions, and the set on which such a function is regular.

If the varieties are embedded in projective space as closed sets, one could use the standard affine open cover and stick to polynomials, but if they are abstract or even quasi projective, it seems to require a notion of isomorphism, or of the sheaf of regular functions to define an affine open set.

Note Ruadhai's definition above requires the use of rational functions to define that of regular ones, after which the definition of morphism appears to involve only the concept "regular".

In case 2) we are dealing with projective space, which has two representations, as a space with a natural affine open cover, but also as a space which is a group quotient of an open subset of affine space. Hence there are two natural definitions of regular map, locally regular in terms of the affine cover, but also as a regular map of the big affine space which is constant on orbits. the latter definition yields a map by homogeneous polynomials of the same degree. This definition is the one most convenient for writing down examples.

Again, Ruadhai's definition of regular locally rational function uses the homogeneous representation, so that definition of morphism is an amalgam of abstract sheaf theoretic concepts and concrete projective ones.

The fully concrete version of Ruadhai's definition of morphism is thus: a map locally defined by homogeneous polynomials of the same degree with non vanishing denominators. This is the definition actually used in practice to write down morphisms of projective and quasi projective varieties.

Several basic books give both definitions (locally rational and regular for the affine open cover, and local homogeneous representation) in the quasi projective case, Shafarevich, Harris, and Reid, I believe. This seems to me a helpful practice.

Bill Fulton's lovely little book on curves, which is intended to prepare one for schemes, uses only the abstract open cover approach as definition, but then leaves as exercise to the reader to check that such innocent maps as [x:y]-->[x:y:z] from P^1 to P^2 are regular.

To summarize, it seems one cannot avoid local definitions, and if one wants to avoid fractions altogether, the only definition I know of a morphism of quasi projective varieties is as a map given locally by homogeneous polynomials of the same degree without common zeroes.

In my example above of a map from the intersection of two quadrics to a plane cubic the map [x:y:z:w] to [x:y:z] is undefined at [0:0:0:1]. By using the locally rational regular map definition in affine coordinates, it seems that point goes to [-1:0:1], as given by the hopefully equivalent map [w(y-w) : (x-z)(y-w) : w^2 ].

Edit:This is my attempt to understand Donu’s definition above. Now I think I understand it, as well as why it is a benchmark, and yet not his favorite. Of course I cannot speak for him. One of its beauties as he said is that it does mimic closely the familiar chart/coordinate system definition from analytic and differential geometry.

Its shortcoming is perhaps its cumbersome nature. E.g. it requires an infinite family of coordinate charts even to discuss regular functions on (open subsets) of affine space. As to whether it enables one to dispense with rational as opposed to polynomial maps, well yes and no. It is true that all regular maps are locally polynomial in terms of the charts, but the trick is that the charts themselves are defined by rational functions.

E.g. Suppose I want to glue two copies of A^1, along the complement of their origins, to get P^1. Can this gluing be done by polynomial functions? Well yes, as Karl said, but it depends somewhat on your point of view. I may not think the regular function 1/z is a polynomial function on the set {z≠0} in A^1. But to understand Karl’s point, I must realize that z is not an affine coordinate system for that subset, so I should not expect all regular functions there to be polynomials in z. But the pair of functions (z,1/z) = (z,w) is an affine coordinate system there, and then 1/y is a polynomial in w, i.e. 1/y is a polynomial in the variable 1/y! In this way one can make any regular locally rational function look locally polynomial. Here are some details as I understand them.

Define an abstract variety as a topological space with a basis of open sets {Uj} such that each Uj is equipped with a homeomorphism fj:Uj-->Vj where Vj is a Zariski closed subset of some affine space. Then require that every inclusion map Ui into Uj becomes a polynomial map of the corresponding affine varieties (fj)o(fi)^(-1):Vi-->Vj.

In this category Donu’s definition of morphism makes perfect sense. I.e. a continuous map of abstract varieties is a morphism if it is locally polynomial in some collection of coordinate systems. In particular a continuous k valued function is regular if and only if it is locally defined by polynomials in some coordinate cover. Nothing could be conceptually simpler or more natural.

What is the catch? With this definition it is not immediately obvious that any familiar (non finite) example at all is an abstract variety, not even k ≈ A^1. I.e. it takes an infinite number of affine coordinate charts even to give affine space itself the structure of an abstract variety. Moreover these charts are defined by rational functions.

If we define a coordinate system in U as a finite set of regular functions such that every regular function in U is a polynomial in terms of those functions, then it is sufficient but not necessary to be affine in order to possess a coordinate system. Fortunately every affine variety has a topological basis of affine open sets, thus we can put a structure of abstract variety on every quasi projective variety.

The difference between the algebraic case and the analytic and differential cases is that the restriction of an affine coordinate system may not be an affine coordinate system. Thus we cannot use the same coordinate system on every open subset of affine space, as we would in differential geometry.

Lemma: The principal open subsets Uf = {f≠0} define a structure of abstract affine variety on affine space A^n. Proof: It suffices to use polynomials f which have no repeated prime factors. Define the coordinate map Uf-->A^(n+1) by sending x-->(x,1/f(x)) = (x,T). Then the image Vf = {1-f.T = 0} is a closed affine set. The coordinate map itself is defined by regular rational functons on Uf, and is a homeomorphism. Moreover, Ug is contained in Uf if and only if g = fh, for some polynomial h. If we map Ug-->Vg by x-->(x,1/g(x)) = (x,W), then Vg = {1-g.W=0}. Hence the inclusion map from Ug to Uf becomes in affine coordinates, the map (x,W)-->(x,W.h(x)) from Vg to Vf, a polynomial map in the coordinates (x,W). QED.

With this lemma it seems one can use restrictions to define a structure of abstract variety on every quasi affine and quasi projective variety.

Finally, as Donu remarked, it is not obvious that one can check regularity using any coordinate cover. I.e. there might be one cover by affine coordinate systems in which a given map is locally polynomial, and yet another cover by different affine coordinate systems in which it is not. One must prove the usual theorem, via the nullstellensatz, that a locally polynomial function on an affine variety is globally polynomial.

This is a beautiful, conceptually natural point of view on what it means to be a morphism of varieties. I would advocate, after giving this definition, proving that a map of quasi projective varieties is a morphism in this sense if and only if it has the property in the accepted answer above, if and only if it can be defined locally by sequences of homogeneous polynomials with no common zeroes. I.e. it is hard to be prepared for all situations with just one characterization.

The last (homogeneous polynomial) point of view can then lead to the important idea that a morphism of a variety to projective space is also defined by a line bundle and a sequence of regular sections without common zeroes, an approach not yet mentioned. I.e. maps of a variety to projective space are much more special than maps to arbitrary varieties, and this special case is well worth understanding.

If we want to discuss the meaning of the common zeroes of sections, as in the example of the intersection curve of two quadrics, note the restriction of a linear polynomial to this curve must vanish 4 times, so the image of the map defined by linear polynomials should have degree 4. Since the image curve has degree three there must be a point where the rational map is not defined. I.e. the line bundle on the intersection curve defining the morphism is O(1) restricted to the curve tensored with O(-p) where p is the point [0:0:0:1]. Since we extended the map using quadratic polynomials, each vanishing 8 times on the curve, presumably those polynomials have 5 common zeroes on our curve.

-
Roy, thanks for sharing your thoughts on this. – Donu Arapura Apr 4 '12 at 17:06

If I understand correctly you are looking for the most elementary definition possible. I assume that you are working over algebraically closed filed and you think of quasi-projective varieties as subsets of the projective space. In this case I'll suggest the following definition:

a morphism $f:X \to Y$ is a Zariski closed subset of $\Gamma(f) \subset X \times Y$ s.t. the projection $\Gamma(f) \to X$ is a bijection.

You may ask what is a Zariski closed subset of $X \times Y$. Here are 2 definitions:

1. A subset $Z$ s.t. for any open (quasi) affine sets $U \subset X$, $V \subset Y$ the intersection $Z \cup U \times V$ is Zariski closed in $U \times V$. (I assume that you know what is a Zariski closed subset of (quasi) affine variety)

2. There is a well known embedding of a product two projective spaces into a larger projective space. This shows that $X \times Y$ is quasi-projective (or gives a quasi-projective structure on $X \times Y$, depending on you framework). I assume that you know what is a Zariski closed subset of quasi-projective affine variety

-
I like this definition too, but somehow it seems a bit heavy to require us to consider product spaces to define a morphism. That's just personal preference, btw, so you may disagree! I guess I'm just looking for the most intuitive definition, and thus far I reckon Ruadhai has it. Any other suggestions? – Edward Hughes Mar 22 '12 at 22:24
Does this definition work when X is not normal? E.g. the graph of the normalization of a curve with a cusp seems to project bijectively in both directions, but presumably is not a morphism in one of them. In Mumford's yellow book, there is an attempt to define a morphism using the graph, but he explicitly requires not just that there be only one image of a given point of the domain, but that the graph be locally there the graph of a regular map, defined in terms of locally rational, regular functions. – roy smith Apr 2 '12 at 16:25
Roy, I afraid that you are write (and I'm wrong) Let me think how to fix it. Sorry – Rami Apr 2 '12 at 21:29
Rami, well it works for normal varieties, hence for open sets in affine space. So it should suffice to fill the detail I was worried about in applying Donu's definition to quasi affine sets, namely how to find a basis of open affines, by showing a principal open set in k^n is isomorphic to a hypersurface. I.e. this principal open is smooth, hence normal. For principal opens in arbitrary closed affine sets we can restrict this isomorphism. ??? – roy smith Apr 3 '12 at 0:18