In this post we discuss a weird example of a finite map of varieties which doesn’t preserve projectivity.

# Image of proper is proper and motivation

We start by recalling a very well-known, and easy fact:

Proposition1:Let be a scheme and a surjective map of -schemes. Assume that is universally closed, then is universally closed. In particular, if is proper and is separated and finite type then is proper.

This is one-hundred percent intuitive if we think of proper as being a sort of ‘relative compactness’–it just says that the image of a compact set is compact. The proof is also quite simple:

*Proof (Proposition 1):* Let be a morphism. We need to show that the map is closed. But, if is closed then so is where is the base change map. But, since is closed (since is universally closed) we know that the image of is closed. But, this coincides with the image of under and the conclusion follows.

In practice though properness is not always enough for applications and so one might ask the following question:

Question 2(a):Let be a scheme and a surjective map of -schemes. If is separated and of finite type and is projective, then is projective?

Or, perhaps, a better question to first ask in the ‘non-relative’ situation:

Question 2(b):Let be a field and let be a surjective map of separated -varieties. If is projective, then is projective?

In fact, the answer to this question is *obviously no *even if you require to be projective and birational by Chow’s theorem which says that if is a proper -variety then there exists a projective birational map with a projective -variety.

Of course, this only gives a negative answer to Question 2(b) if one is able to produce a proper -variety which is not projective. Such objects are always somewhat tricky to discuss, so let’s begin by recalling when properness and projectivity do agree:

Theorem 3:Let be a field and let be a proper -variety. Then, is projective in each of the following situations:

- The dimension of is .
- (Zariski–Goodman) The dimension of is and is smooth.

Before giving references for these proofs we note the following well-known lemma:

Lemma 4:Let be a field and an extension. Suppose that is a -variety such that is projective. Then, is projective.

*Proof: *For example, one can see [EGA II, Corollaire 6.6.5]. Since one can, by the standard tricks, reduce the case when is finite we actually discuss a generalization of this later in the form of Corollary 8.

With this lemma in hand we can proceed to give references for Theorem 3:

*Proof(Theorem 3):* For 1. one can see this (note that by this the notion of ‘-projective’ and ‘projective’ do not differ). For 2. see [Bad, Theorem 1.28].

An example of a smooth proper non-projective -fold is the famous example of Hironaka (e.g. see this exposition). We will see later an example of a non-smooth surface which is proper but not projective.

Now, while Question 2(b) (and thus Question 2(a)) have negative answers, one might hope that if one imposes that is finite in Question 2(b) that one obtains an affirmative answer. Namely, one has the following:

Question 2(c):Let be a field and let be a surjective finite map of separated -varieties. If is projective, is projective?

# The norm map and a positive answer

To explain why Question 2(c) *might* have a affirmative answer we explain a proof of the following slight modification:

Theorem 5:Let be a field and let be a finite flat surjection of separated -varieties. Then, if is projective then is projective.

To explain the proof it is useful first to define the ‘norm map’. Namely, let’s suppose that and are arbitrary schemes and is a finite flat map locally of finite presentation (this last condition is superflous if is Noetherian). Then, by standard theory, the quasi-coherent -module is locally free. Then, there is naturally a norm map

which is the globalization of the notion of the norm for a finite free -algebra—-this is just the map where is the left multiplication map (where we are thinking of as a free -module and so the determinant makes sense). Note that and we prefer to use the latter notation.

We thus get an induced map

where the subscript ‘et’ denotes etale cohomology, and the latter isomorphism is the usual one. One then uses the fact that finite maps are acyclic for the etale topology (e.g. see [Fu, Proposition 5.7.4]) to see that we have a canonical isomorphism

But, since this first group is just (as we’ve already mentioned) we see that putting everything together we get a map

called the *norm map*.

In words, what this really means is that if we start with a line bundle on this can be thought of as gluing together local pieces of via transition maps in . We can then imagine as being line bundle on obtained by gluing together locally using the norm of the transition maps.

*Remark 6:* Introducing etale cohomology is not at all necessary. It just is the approach that makes the most sense to me (and shows why etale cohomology is useful even for ‘Zariski constructions’). Namely, I used in the above that is zero where is taken in etale category, but it vanishes even in the Zariski category (e.g. see this).

The reason that the norm map is useful, besides being a natural way to produce line bundles, is the following

Lemma 7:Let be a finite flat map locally of finite presentation. Let be an ample line bundle on , then is an ample line bundle on .

The proof of this lemma is not too hard, and we leave it to the reader as an excercise (cf. [EGA II, Proposition 6.6.1] or this).

In particular, we deduce the following:

Corollary 8:Let be a field and let be a finite flat surjection of separated -varieties. Then, if is projective then is projective.

*Proof:* By Proposition 1 we know that is proper. Since is projective it has an ample line bundle and then is an ample line bundle on . The conclusion then follows from standard theory (i.e. use the result from this link to say that there is a dense open embedding in to a projective -scheme but use the fact that is proper to deduce that the image is closed, and thus its an isomorphism).

One can give a more concrete proof in the case when is a Galois cover whose degree is invertible in .

Proposition9:Let be a fieldand let be a finite etale surjection of connected separated -varieties such that there is a dominating Galois cover of degree invertible in . Then, if is projective then is projective.

*Remark 10:* This assumption that there is a dominating Galois cover whose degree is invertible in is automatically true if characteristic of is zero. This is also probably non-essential, but certainly saves us some headache–there’s probably just ean easier way to check this (e.g. just prove that ampleness is etale local using the fact that affineness can be checked etale locally), but I like this proof. Note also that there is always a dominating Galois cover (e.g. see [Fu,Proposition 3.2.10]).

*Proof:* Clearly we may assume without loess of generality that is Galois with degree invertible in . Let be the Galois group of . We then claim that is an ample line bundle. Let us begin by noting that is in fact a line bundle. We can check this etale locally on (e.g. we’ve already implicity used thsi fact, but it follows, for instance from [Ols, Proposition 4.3.8]). But, since is isomorphic, as an -scheme, to this is obvious. Thus, it suffices to show that is ample.

But, by [Har, Proposition III.5.3] it suffices to show that for every coherent -module there exists some such that for each and we have that . But, by our assumption that is ample there is such an for . We claim that this also works for . Indeed, note that we have the Hochschild-Serre spectral sequence (e.g. see [Mil, Theorem 2.20])

(note that we have used Zariski and not etale cohomology, but there is no difference since our objects are quasi-coherent). But, note that since is a -space and is invertible on that

for (e.g. see Corollary 12 and the suceeding discussion in this post). Thus, we deduce that

But what is this right-hand side? Namely, note that

where we have used the equality which can be checked after finite etale cover which is easy to do, by hand, on which is isomorphic, as an -scheme, to . But, then evidently the right-hand side of vanishes by the way we chose and thus the left-hand side also vanishes. The conclusion follows.

# The weird counterexample

We now explain how to produce a weird surface with the property that its proper but not projective but has normalization . Of course, since the normalization map is finite (as we will see by hand in our case) this gives us an explicit counter example to Question 2(c).

*Remark 11: *This example is taken from this mathoverflow post.

To construct this example let’s take to be an algebraically closed field (of characteristic just to be safe). We now fix two curves and in with the former having degree and the latter having degree and which intersect tangentially at a point . We know that and are abstractly isomorphic (e.g. see [Vak, §19.3]) and we can assume that maps to so we can take the quotient variety that identifies and .

More rigorously, let us fix an isomorphism and consider the pushout

where is the closed embedding and is the finite surjection which is the identity on and the isomorphism on . One can show that the scheme exists by applying [Sch, Theorem 3.4] affine locally (or one can see the above cited mathoverflow post for more details). We then, by definition, get a surjection which is finite. One can see this by the explicit construction or by pure power of thought. Indeed, it’s not hard to see that is separated (by the explicit gluing criterion) and since is proper we know that is proper. It’s also clear that is quasi-finite from which the conclusion follows (e.g. see [Vak, Theorem 29.6.2]). Finally, it’s clear that is an open embedding and so is birational. Thus, we know that is the normalization map (e.g. see [GW, Proposition 12.44]).

So, our finite normalization map is finite and, since is separated, we know that is proper (e.g. by Proposition 1). We claim though that is *not* projective. Indeed, suppose that it were. Then, evidently there would exist an ample line bundle on . Note then by standard theory (e.g. see this post) that would be an ample bundle on . But, note that is trivial for all on . Indeed, say that . Note then that by standard theory (e.g. see this and apply Bezout’s lemma) if denotes the tautological closed embeddings then and . But, note that if is our tautological closed embedding then by our definition of we have that

which implies that and so . Thus, can never be ample, and thus neither can be ample. Thus, is not projective.

# References

[Bad] Badescu, L., 2013. *Algebraic surfaces*. Springer Science & Business Media.

[EGA II] Grothendieck, A., 1961. Éléments de géométrie algébrique (rédigés avec la collaboration de Jean Dieudonné): II. Étude globale élémentaire de quelques classes de morphismes. *Publications Mathématiques de l’IHÉS*, *8*, pp.5-222.

[Fu] Fu, L., 2011. *Etale cohomology theory* (Vol. 13). World Scientific.

[GW] Görtz, U. and Wedhorn, T., 2010. *Algebraic geometry*. Wiesbaden: Vieweg+ Teubner.

[Mil] Milne, J.S. and Milne, J.S., 1980. *Etale cohomology (PMS-33)* (No. 33). Princeton university press.

[Ols] Olsson, M., 2016. *Algebraic spaces and stacks* (Vol. 62). American Mathematical Soc.

[Sch] Schwede, K., 2005. Gluing schemes and a scheme without closed points. *Contemporary Mathematics*, *386*, p.157.

[Vak] Vakil, R., 2017. *The Rising Sea: Foundations of Algebraic Geometry* (Ver. Nov. 18 2017). http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf