## Proper Maps and (Quasi) Projective Varieties

I’m back!  So let’s get down to business.

What I really want to talk about is the Riemann Roch Theorem.  You may wonder why, since it has already been discussed here , but there have been vast generalizations of this theorem throughout the last century.  I want to focus on Grothendieck’s version, but I plan to start slow.  The material I’m following will mostly be taken from the classic Serre-Borel paper (recently pointed out to me).  Today, we’ll content ourselves with talking about proper maps.   I know that Charles already talked about a complete variety here and even proved Chow’s Lemma which is great.  It tells us, that all complete varieties are “close” to being projective, so today we’ll focus on the projective ones.

First with the formalities: all varieties will be assumed to be defined over an algebraically closed field.  For today, a variety will mean a quasi projective one.  Recall that this means it is (isomorphic to one which is) a locally closed subvariety of projective space.  A variety is called projective  if it (isomorphic to one which is) a closed subvariety of projective space.  The letters X and Y will always denote varieties, P and P’ will denote projective spaces.  The notation of the ground field will be omitted!

Now, Charles already outlined in his post why the projection $\mathbb{P}^n \times U \rightarrow U$ takes closed sets to closed sets.  Recall this means that projective space is complete.  Recall also that given a morphism $f: X \rightarrow Y$, we denote $\Gamma_f \subset X \times Y$ to be the graph of f.  It’s easy to see that the graph is closed inside the product.  Let’s prove two quick facts:

Lemma: Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be two morphisms, where X and Y are subvarieties of the projective spaces P and P’.  Suppose that $\Gamma_f$ is closed in $P \times Y$ and $\Gamma_g$ is closed inside $P' \times Z$.  Then $\Gamma_{gf}$ is closed inside $P \times Z$.

Proof: We have that $\Gamma_f \subset P \times Y = P \times \Gamma_g \subset P \times P' \times Z$, and each inclusion is closed in the next.  In particular, $\Gamma_f$ is closed inside $P \times P' \times Z$.  Note that when we project $\Gamma_f$ down to $P \times Z$, we get exactly the graph of $\Gamma_{gf}$, which is then closed because projective space is complete.

Lemma: Suppose that $f: X \rightarrow Y$ is a morphism, and that X can be embedded into two projective spaces P and P’.  If $\Gamma_f \subset P \times Y$ is closed, then it is also closed inside $P' \times Y$.

Proof:  Here we apply the previous Lemma to $X \rightarrow X \rightarrow Y$, where the first map is the identity and the second map is f.  We must only check that the graph of the identity is closed in $P' \times X$, but we’ll leave that as an easy exercise!

This Lemma justifies, Definition: A morphism $f: X \rightarrow Y$ is called proper if the graph $\Gamma_f$ is closed inside $P \times Y$ for some projective space P containing X.

We immediately have the following equivalence which shows how proper morphisms behave much like complete varieties.  (Indeed, it implies that the structure map of a variety is proper if and only if the variety is complete).

Proposition:  A morphism $f: X \rightarrow Y$ is proper if and only if for each variety Z, and each closed subset T of $X \times Z$, the image of T in $Y \times Z$ is closed.

Proof:  Let P be a projective space containing the embedded X.  Suppose f is proper, that is, that $\Gamma_f \subset P \times Y$ is closed.  Then also $\Gamma_f \times Z$ is closed inside $P \times Y \times Z$ and T can be considered closed in this triple product.  By the completeness of projective space, the projection of T down to $Y \times Z$  is closed.  Conversely, suppose the property is always verified, and now take Z = P and T to be the diagonal of $X \times X \subset X \times P$.  The image of T inside $Y \times Z = Y \times P$ is none other than $\Gamma_f$, which must be closed, so that f is proper.

We now list some properties of proper morphisms, which are generally quite easy to verify using the definition and/or the propsition:

Properties

1. The idenity map is proper.

2. The composition of two proper maps is proper.

3. The product of two proper maps is proper.

4. The image of a closed subvariety under a proper map is proper.

5. An injection $Y \rightarrow X$ is proper if and only if Y is closed in X.

6. Morphisms from a projective variety are proper.

7. A projection $X \times Y \rightarrow X$ is proper if and only if Y is projective.

For example, to verify property 4, suppose that $f: X \rightarrow Y$ is a proper morphism and T is closed in X.  We can apply the proposition where Z is a point so that T is a closed subvariety of $pt \times X = X$.  The image, according to the proposition, is closed.  The others are as easy.  As a corollary, we can prove:

Corollary: A morphism $f: X \rightarrow Y$ is proper if and only if it factors through$X \rightarrow P \times Y \rightarrow Y$ where the first map is a closed immersion and the second is the projection map.

Proof: This is necessary by definition (take X embedded in some projective space P) and sufficient by Properties 2,5, and 7.

As a last fact, perhaps you have heard before that proper maps in algebraic geometry are akin to proper maps in topology… that is indeed the case as the following Proposition explains:

Proposition: Suppose that the ground field is the complex numbers.  A morphism $f: X \rightarrow Y$ to be proper (in the algebraic sense) if and only if it is proper (in the topological sense) when X and Y are given their “usual” topologies.

Proof : Suppose that f is proper in the algebraic sense and that K is compact in Y in the topological sense.  Let X be embedded in a projective space P, which is well known to be compact.  Then we have that $f^{-1}(K) = \Gamma_f \cap (P \times K)$, which is again compact, so that f is proper in the topological sense.  Suppose conversely that the topological property is satisfied.  Suppose T is Zariski closed inside $X \times Z$ as in the previous proposition (we must show its image is Zariski closed in $Y\times Z$).  Its image is closed in the usual topology because the map is topologically closed.  Then there are some standard comparison facts which allow us to conclude that the image is also Zariski closed.  Let W be the Zariski closure of the image.  Then there is a Zariski open set U which is Zariski dense in W (this is standard I think?).  This open U is also topologically dense in W (this is basically an exercise).  This in turn implies that the Zariski and topologicaly closures of the image coincide, but since the image is topologically closed it is also Zariski closed, and so f is (algebraically) proper.

Note that we have only cheated a little bit by using our definition of proper, which depends on having quasi-projective varieties.  In general (for example, in Hartshorne’s book), the definition is taken to be the criteria in the first proposition (that is being universally closed, along with separated, and finite type – which are automatic here).  All the properties and propositions go through as long as we’re careful about replacing “projective” by “complete” (and applying Chow’s Lemma as explained in the earlier post by Charles).

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### 12 Responses to Proper Maps and (Quasi) Projective Varieties

1. Soarer says:

May I know if the corollary means that on quasi-projective varities, projective morphism and proper maps are the same thing?

Moreover, in the scheme case, is there any motivation for the definition of proper maps?

Thanks!

2. Matt DeLand says:

Hi,

Projective morphisms are always proper. In general, suppose that $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ are such that the composition is (quasi) projective. If $g$ is separated, then the morphism $f$ is (quasi) projective also. Then the answer to your first question is the case when $Z = Spec(k)$.

As a reminder, in the “scheme” case a morphism of schemes is called proper if it is separated, of finite type, and universally closed. The motivation for this definition, is supposed to be exactly the post above! The “important” part of the definition is the universally closed part. For the definition given above, this is exactly the first proposition. Proper morphisms should take closed sets to closed sets. The second proposition gives more motivation, since our definition agrees with the topological notion in case the variety is also a complex manifold.

The separated and finite type conditions will rule out certain pathologies and insure things are well behaved. For example, we don’t want to consider the projective line with the origin doubled a projective variety since it can’t be embedded in projective space. The finite type condition allows us to rule out things like infinitely many copies of $Spec(k)$ being projective over $Spec(k)$. Hope that helped.

3. Soarer says:

So am I right to say that completeness of variety is a nice property that we want, so we try to mimic it for the case of maps, thus “universally closed” comes up.

Then separatedness is something like Hausdorff to rule out ugly cases, and finite type is something to keep us working in “finite dimension”.

Is this correct? Thanks!

• Matt DeLand says:

You are a right in a high-level way. Completeness is a “nice” property (whether one wants it or not depends on one’s tastes I guess). Separated is a condition that’s always satisfied locally and passes to open and closed subvarieties (e.g., all quasi projective varieties are separated). Being non-separated is a consequence of how the open affine coverings are “glued” together. The finiteness condition is fine.

4. Matteo says:

if f:Y\rightarrow Z is proper and let X\subset Y such that the restriction of f to X is again proper, is the embedding of X into Y proper?

5. Matt DeLand says:

See Hartshorne’s “Algebraic Geometry”, II.4.8

• Matteo says:

Ok.
Thank you.

• Matteo says:

Isn’t there a proof of this fact which does not use scheme theory?

I mean: let’s suppose that X, Y and Z are quasi projective varieties; is it possible to proof this fact using only the definition (and the properties) written in this page?

Thank you

6. Matt DeLand says:

Yes. If i is the inclusion of X into Y, look at the graph of i inside P \times Y (where P is some projective space containing X) and compare it to the graph of “f” inside P \times Z.

• Matteo says:

Thanks a lot!!!

7. Bertie says:

I have exactly what info I want. Check, please. Wait, it’s free? Awoesme!