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:

Proposition 1: Let $S$ be a scheme and $f:X\to Y$ a surjective map of $S$ -schemes. Assume that $X\to S$ is universally closed, then $Y\to S$ is universally closed. In particular, if $X\to S$ is proper and $Y\to S$ is separated and finite type then $Y\to S$ 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 $T\to S$ be a morphism. We need to show that the map $Y_T\to T$ is closed. But, if $C\subseteq Y_T$ is closed then so is $f_T^{-1}(C)\subseteq X_T$ where $f_T:X_T\to Y_T$ is the base change map. But, since $X_T\to T$ is closed (since $X\to S$ is universally closed) we know that the image of $f^{-1}(C)\subseteq T$ is closed. But, this coincides with the image of $C$ under $Y_T\to T$ and the conclusion follows. $\blacksquare$

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

Question 2(a): Let $S$ be a scheme and $f:X\to Y$ a surjective map of $S$ -schemes. If $Y\to S$ is separated and of finite type and $X\to S$ is projective, then is $Y\to S$ projective?

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

Question 2(b): Let $k$ be a field and let $f:X\to Y$ be a surjective map of separated $k$ -varieties. If $X$ is projective, then is $Y$ projective?

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

Of course, this only gives a negative answer to Question 2(b) if one is able to produce a proper $k$ -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 $k$ be a field and let $X$ be a proper $k$ -variety. Then, $X$ is projective in each of the following situations:

The dimension of $X$ is $1$ .

(Zariski–Goodman) The dimension of $X$ is $2$ and $X$ is smooth.

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

Lemma 4: Let $k$ be a field and $K/k$ an extension. Suppose that $X$ is a $k$ -variety such that $X_K$ is projective. Then, $X$ 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 $K/k$ is finite we actually discuss a generalization of this later in the form of Corollary 8. $\blacksquare$

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 ‘ $H$ -projective’ and ‘projective’ do not differ). For 2. see [Bad, Theorem 1.28]. $\blacksquare$

An example of a smooth proper non-projective $3$ -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 $f$ is finite in Question 2(b) that one obtains an affirmative answer. Namely, one has the following:

Question 2(c): Let $k$ be a field and let $f:X\to Y$ be a surjective finite map of separated $k$ -varieties. If $X$ is projective, is $Y$ 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 $k$ be a field and let $f:X\to Y$ be a finite flat surjection of separated $k$ -varieties. Then, if $X$ is projective then $Y$ is projective.

To explain the proof it is useful first to define the ‘norm map’. Namely, let’s suppose that $X$ and $Y$ are arbitrary schemes and $f:X\to Y$ is a finite flat map locally of finite presentation (this last condition is superflous if $Y$ is Noetherian). Then, by standard theory, the quasi-coherent $\mathcal{O}_Y$ -module $f_\ast\mathcal{O}_X$ is locally free. Then, there is naturally a norm map

$N:(f_\ast\mathcal{O}_X)^\times \to \mathcal{O}_Y^\times$

which is the globalization of the notion of the norm $B^\times\to A^\times$ for $B$ a finite free $A$ -algebra—-this is just the map $b\mapsto \mathrm{det}(L_b)$ where $L_b$ is the left multiplication map $L_b:B\to B$ (where we are thinking of $B$ as a free $A$ -module and so the determinant makes sense). Note that $(f_\ast\mathcal{O}_X)^\times=f_\ast\mathcal{O}_X^\times$ and we prefer to use the latter notation.

We thus get an induced map

$H^1_\mathrm{et}(Y,f_\ast\mathcal{O}_X^\times)\to H^1_\mathrm{et}(Y,\mathcal{O}_Y^\times)\cong \mathrm{Pic}(Y)$

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

$H^1_\mathrm{et}(X,\mathcal{O}_X^\times)\xrightarrow{\approx} H^1(Y,f_\ast\mathcal{O}_Y^\times)$

But, since this first group is just $\mathrm{Pic}(X)$ (as we’ve already mentioned) we see that putting everything together we get a map

$N:\mathrm{Pic}(X)\to \mathrm{Pic}(Y)$

called the norm map.

In words, what this really means is that if we start with a line bundle $\mathcal{L}$ on $X$ this can be thought of as gluing together local pieces of $\mathcal{O}_X$ via transition maps in $\mathcal{O}_X^\times$ . We can then imagine $N(\mathcal{L})$ as being line bundle on $Y$ obtained by gluing together $\mathcal{O}_Y$ 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 $R^1 f_\ast\mathcal{O}_X^\times$ is zero where $R^1f_\ast$ 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 $f:X\to Y$ be a finite flat map locally of finite presentation. Let $\mathcal{L}$ be an ample line bundle on $X$ , then $N(\mathcal{L})$ is an ample line bundle on $Y$ .

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 $k$ be a field and let $f:X\to Y$ be a finite flat surjection of separated $k$ -varieties. Then, if $X$ is projective then $Y$ is projective.

Proof: By Proposition 1 we know that $Y$ is proper. Since $X$ is projective it has an ample line bundle $\mathcal{L}$ and then $N(\mathcal{L})$ is an ample line bundle on $Y$ . 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 $k$ -scheme but use the fact that $Y$ is proper to deduce that the image is closed, and thus its an isomorphism). $\blacksquare$

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

Proposition 9: Let $k$ be a fieldand let $f:X\to Y$ be a finite etale surjection of connected separated $k$ -varieties such that there is a dominating Galois cover of degree invertible in $k$ . Then, if $X$ is projective then $Y$ is projective.

Remark 10: This assumption that there is a dominating Galois cover whose degree is invertible in $k$ is automatically true if characteristic of $k$ 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 $f:X\to Y$ is Galois with degree invertible in $k$ . Let $G$ be the Galois group of $f:X\to Y$ . We then claim that $(f_\ast\mathcal{L})^G$ is an ample line bundle. Let us begin by noting that $(f_\ast \mathcal{L})^G$ is in fact a line bundle. We can check this etale locally on $X$ (e.g. we’ve already implicity used thsi fact, but it follows, for instance from [Ols, Proposition 4.3.8]). But, since $X\times_Y X\to X$ is isomorphic, as an $X$ -scheme, to $X\times \underline{G}\to X$ this is obvious. Thus, it suffices to show that $(f_\ast\mathcal{L})^G$ is ample.

But, by [Har, Proposition III.5.3] it suffices to show that for every coherent $\mathcal{O}_Y$ -module $\mathcal{F}$ there exists some $n_0>0$ such that for each $i>0$ and $n\geqslant n_0$ we have that $H^i(Y,\mathcal{F}\otimes (f_\ast\mathcal{L})^G)=0$ . But, by our assumption that $\mathcal{L}$ is ample there is such an $n_0$ for $f^\ast\mathcal{F}$ . We claim that this $n_0$ also works for $\mathcal{F}$ . Indeed, note that we have the Hochschild-Serre spectral sequence (e.g. see [Mil, Theorem 2.20])

$H^p\left(G,H^q\left(X,f^\ast\left(\mathcal{F}\otimes \left((f_\ast\mathcal{L})^G\right)^{\otimes n}\right)\right)\implies H^{p+q}(Y,\mathcal{F}\otimes \left(f_\ast\mathcal{L})^G\right)^{\otimes n}\right)$

(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 $H^q\left(X,f^\ast\left(\mathcal{F}\otimes \left((f_\ast\mathcal{L})^G\right)^{\otimes n}\right)\right)$ is a $k$ -space and $|G|$ is invertible on $k$ that

$H^p\left(G,H^q\left(X,f^\ast\left(\mathcal{F}\otimes \left((f_\ast\mathcal{L})^G\right)^{\otimes n}\right)\right)\right)=0$

for $p>0$ (e.g. see Corollary 12 and the suceeding discussion in this post). Thus, we deduce that

$H^{i}\left(Y,\mathcal{F}\otimes \left(f_\ast\mathcal{L})^G\right)^{\otimes n}\right)\cong H^q\left(X,f^\ast\left(\mathcal{F}\otimes \left((f_\ast\mathcal{L})^G\right)^{\otimes n}\right)\right)^G \qquad (\ast)$

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

$\begin{aligned} f^\ast\left(\mathcal{F}\otimes \left((f_\ast\mathcal{L})^G\right)^{\otimes n}\right) &\cong f^\ast\mathcal{F} \otimes f^\ast\left( \left((f_\ast\mathcal{L})^G\right)^{\otimes n}\right)\\ &\cong f^\ast\mathcal{F} \otimes \left( f^\ast\left((f_\ast\mathcal{L})^G\right)\right)^{\otimes n}\\ &\cong f^\ast\mathcal{F}\otimes \mathcal{L}^{\otimes n}\end{aligned}$

where we have used the equality $f^\ast\left((f_\ast\mathcal{L})^G\right)\cong \mathcal{L}$ which can be checked after finite etale cover which is easy to do, by hand, on $X\times_Y X\to X$ which is isomorphic, as an $X$ -scheme, to $X\times \underline{G}$ . But, then evidently the right-hand side of $(\ast)$ vanishes by the way we chose $n_0$ and thus the left-hand side also vanishes. The conclusion follows. $\blacksquare$

The weird counterexample

We now explain how to produce a weird surface $X$ with the property that its proper but not projective but has normalization $\mathbb{P}^2_k$ . Of course, since the normalization map $\nu:\mathbb{P}^2_k\to X$ 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 $k$ to be an algebraically closed field (of characteristic $0$ just to be safe). We now fix two curves $C_1$ and $C_2$ in $\mathbb{P}^2_k$ with the former having degree $1$ and the latter having degree $2$ and which intersect tangentially at a point $p$ . We know that $C_1$ and $C_2$ are abstractly isomorphic (e.g. see [Vak, §19.3]) and we can assume that $p$ maps to $p$ so we can take the quotient variety $\varphi:\mathbb{P}^2_k\to X$ that identifies $C_1$ and $C_2$ .

More rigorously, let us fix an isomorphism $C_2\to C_1$ and consider the pushout

$X:= \mathbb{P}^2_k\sqcup_{C_1\cup C_2} C_1$

where $C_1\cup C_2\to \mathbb{P}^2_k$ is the closed embedding and $C_1\cup C_2\to C_1$ is the finite surjection which is the identity on $C_1$ and the isomorphism $C_2\to C_1$ on $C_2$ . One can show that the scheme $X$ 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 $\nu:\mathbb{P}^2_k\to X$ 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 $X$ is separated (by the explicit gluing criterion) and since $\mathbb{P}^2_k$ is proper we know that $\nu$ is proper. It’s also clear that $\nu$ is quasi-finite from which the conclusion follows (e.g. see [Vak, Theorem 29.6.2]). Finally, it’s clear that $\nu\mid_{\mathbb{P}^2_k-(C_1\cup C_2)}$ is an open embedding and so $\nu$ is birational. Thus, we know that $\nu$ is the normalization map (e.g. see [GW, Proposition 12.44]).

So, our finite normalization map $\nu:\mathbb{P}^2_k\to X$ is finite and, since $X$ is separated, we know that $X$ is proper (e.g. by Proposition 1). We claim though that $X$ is not projective. Indeed, suppose that it were. Then, evidently there would exist an ample line bundle $\mathcal{L}$ on $X$ . Note then by standard theory (e.g. see this post) that $\nu^\ast\mathcal{L}$ would be an ample bundle on $\mathbb{P}^2_k$ . But, note that $\nu^\ast\mathcal{L}$ is trivial for all $\mathcal{L}$ on $X$ . Indeed, say that $\nu^\ast\mathcal{L}=\mathcal{O}(k)$ . Note then that by standard theory (e.g. see this and apply Bezout’s lemma) if $\iota_i:C_i\hookrightarrow\mathbb{P}^2_k$ denotes the tautological closed embeddings then $\deg \iota_1^\ast\nu^\ast\mathcal{L}=k$ and $\deg \iota_2^\ast\nu^\ast\mathcal{L}=2k$ . But, note that if $\iota:C_1\hookrightarrow X$ is our tautological closed embedding then by our definition of $\nu$ we have that

$\deg \iota^\ast\mathcal{L}=\deg \iota_1^\ast \nu^\ast\mathcal{L}=\deg \iota_2^\ast\nu^\ast\mathcal{L}$

which implies that $k=2k$ and so $k=0$ . Thus, $\nu^\ast\mathcal{L}$ can never be ample, and thus neither can $\mathcal{L}$ be ample. Thus, $X$ 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

Hard Arithmetic

Weird example: finite maps don’t preserve projectivity

Image of proper is proper and motivation

The norm map and a positive answer

The weird counterexample

References

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Image of proper is proper and motivation

The norm map and a positive answer

The weird counterexample

References

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