# Can you give an example of two projective morphisms of schemes whose composition is not projective?

Grothendieck and Dieudonné prove in $$EGA_{II}$$ (Proposition 5.5.5.(ii), page 105) that if $$f:X\to Y, g:Y\to Z$$ are projective morphisms of schemes and if $$Z$$ is separated and quasi-compact, or if the underlying topological space $$\operatorname {sp}(Z)$$ is noetherian, then the composition $$g\circ f:X\to Z$$ is also projective.
Question: Is there an example where $$Z$$ satisfies neither of the two sufficient conditions above and where $$g\circ f:X\to Z$$ is not projective?

Edit I have corrected my initially wrongly stated sufficient conditions on $$Z$$, caused by the change in terminology:
Thanks a lot to R. Van Dobben de Bruyn for making me aware of my initial confusion.

• Slight correction in the conditions: either $Z$ is quasi-compact and separated (stacks project: quasi-compact and quasi-separated), or $\operatorname{sp}(Z)$ is Noetherian. – R. van Dobben de Bruyn Nov 13 '19 at 4:50
• The most obvious (and probably most interesting) way for this to fail would be a non-quasi-compact $Z$ such that the power $a$ needed for $\mathscr L \otimes f^* \mathscr M^{\otimes a}$ to be $(g \circ f)$-ample (see Tag OC4K) is unbounded if you run over all affine opens in $Z$. But then you need a reason why you couldn't have chosen some other line bundle (so $Z$ a disjoint union of points is not going to work). – R. van Dobben de Bruyn Nov 13 '19 at 6:16
• It seems to me that quasi-separatedness is only used in Tag 07RM to show that $(g \circ f)_* \mathscr L$ is a filtered colimit of finite type quasi-coherent $\mathcal O_Z$-modules. This also holds for example when $Z$ is locally Noetherian because the projective morphisms $f$ and $g$ preserve coherent sheaves under pushforward. In general however it could happen that $g \circ f$ is quasi-projective and proper, but not projective (but again you would need some reason why there isn't some other projective embedding). – R. van Dobben de Bruyn Nov 13 '19 at 6:20
• Thanks a lot for your comments, dear @ R. van Dobben de Bruyn. And I have modified the two possible conditions on $Z$ according to your first comment. – Georges Elencwajg Nov 13 '19 at 10:23

Here is a locally Noetherian separated counterexample. I also give some motivation for this construction afterwards.

Definition. Let $$Z$$ be an infinite chain of affine lines: $$Z = Z_1 \amalg_{p_1} Z_2 \amalg_{p_2} \ldots$$, where $$Z_i \cong \mathbf A^1_{\mathbf C}$$ and $$p_i$$ is the point $$1$$ in $$Z_i$$ and the point $$0$$ in $$Z_{i+1}$$. Let $$Y = \mathbf P^2_Z$$, which is clearly projective over $$Z$$. Write $$Y_i = Z_i \times_Z Y$$ for the irreducible component of $$Y$$ above $$Z_i$$, with structure map $$g_i \colon \mathbf A^1 \times \mathbf P^2 \cong Y_i \to Z_i \cong \mathbf A^1$$.

Finally, let $$X$$ be obtained by blowing up $$Y_i$$ at $$i$$ collinear points $$W_i \subseteq \mathbf A^1 \times \mathbf P^2$$ in the fibre $$g_i^{-1}(2)$$ for all $$i$$ (the only thing that matters is that we choose a point $$2$$ that is not one of the glueing points $$0$$ and $$1$$). Then $$f \colon X \to Y$$ is projective because $$X \subseteq \mathbf P(\mathcal I_W)$$ is a closed immersion, where $$W = \bigcup_i W_i \subseteq Y$$ is the closed subvariety we're blowing up and $$\mathcal I_W$$ its ideal sheaf (which is of finite type because that's a local condition).

Notation. Write $$h = g \circ f \colon X \to Z$$, and once again write $$X_i = Z_i \times_Z X$$ for the irreducible component of $$X$$ above $$Z_i$$. Denote the maps $$X_i \to Y_i$$ and $$X_i \to Z_i$$ by $$g_i$$ and $$h_i$$ respectively.

Proposition. The map $$X \to Z$$ is not projective.

We will show that there is no $$h$$-ample line bundle on $$X$$ by computing all line bundles on $$X$$. Note that for any open $$U \subseteq Z_i \cong \mathbf A^1$$, we have $$\operatorname{Pic}(U \times \mathbf P^2) \cong \operatorname{Pic}(\mathbf P^2)$$ by pullback.

Lemma. The projection $$\pi \colon Y = X \times \mathbf P^2 \to \mathbf P^2$$ induces an isomorphism $$\pi^* \colon \operatorname{Pic}(\mathbf P^2) \stackrel\sim\to \operatorname{Pic}(Y).$$

Proof. Indeed, $$\pi^*$$ is injective since $$\pi$$ has a section. Every $$\mathscr L$$ on $$Y$$ restricts to some $$\mathcal O_{Y_i}(n_i)$$ on each $$Y_i$$, which has to be the same $$n_i = n$$ for all $$i$$ by restricting to $$Y_i \cap Y_{i+1} = p_i \times \mathbf P^2$$. We can choose identifications $$\mathscr L|_{Y_i} \cong \mathcal O_{Y_i}(n)$$ compatibly for all $$i$$ starting with $$i = 1$$ and moving through the chain. At any stage we might have to modify the chosen identification on $$Y_{i+1}$$ by an element of $$\operatorname{Aut}(\mathscr L|_{p_i \times \mathbf P^2}) = \mathbf C^\times$$, which is harmless because there are no loops. $$\square$$

By the same reasoning, any line bundle on $$X$$ is pulled back from $$\mathbf P^2$$ away from $$\bigcup_i h_i^{-1}(2)$$, so it suffices to glue line bundles on the open covering consisting of the locus $$U = X \setminus \bigcup_i h_i^{-1}(2)$$ where $$f \colon X \to Y$$ is an isomorphism, and the opens $$U_i = X_i \setminus h_i^{-1}\Big(\{p_{i-1},p_i\}\Big)\cong \operatorname{Bl}_{W_i} \bigg(\left(\mathbf A^1 \setminus \{0,1\} \right) \times \mathbf P^2\bigg)$$ consisting of $$X_i$$ minus its intersections with $$X_{i-1}$$ and $$X_{i+1}$$.

Corollary. We have $$\operatorname{Pic}(X) \cong \operatorname{Pic}(\mathbf P^2) \times \prod_{i=1}^{\infty} \prod_{r = 1}^i \mathbf Z E_{i,r}.$$

Proof. We have $$\operatorname{Pic}(U_i) \cong \operatorname{Pic}(\mathbf P^2) \times \prod_{r = 1}^i \mathbf Z E_{i,r}$$ where the $$E_{i,r}$$ are the exceptional divisors above the $$i$$ points of $$W_i$$. The result now from the description above by glueing on $$U$$ and the $$U_i$$, since there is no compatibility condition between the coefficients in $$E_{i,r}$$ and $$E_{i',r'}$$ if $$i \neq i'$$. $$\square$$

Proof of Proposition. Now suppose $$\mathscr L = (n,n_{i,r}) \in \operatorname{Pic}(X)$$ is $$h$$-ample. In particular this implies that $$\mathscr L|_{h^{-1}(z)}$$ is ample for each $$z \in Z$$. Apply this to $$z = 2 \in Z_i$$ to get $$nH + \sum_{r = 1}^i n_{i,r}E_{i,r}$$ ample on $$h_i^{-1}(2)$$. But the fibre $$h_i^{-1}(2)$$ contains the strict transform $$\operatorname{Bl}_{W_i}(\mathbf P^2)$$ of the fibre $$g_i^{-1}(2) \cong \mathbf P^2$$ as a closed subvariety, and the restriction of $$E_{i,r}$$ to $$\operatorname{Bl}_{W_i}(\mathbf P^2)$$ is the $$r^{\text{th}}$$ exceptional divisor $$E_r$$ of $$\operatorname{Bl}_{W_i}(\mathbf P^2) \to \mathbf P^2$$.

So $$nH + \sum_{r = 1}^i n_{i,r}E_r$$ has to be ample on $$\operatorname{Bl}_{W_i}(\mathbf P^2)$$. Since $$E_r \cdot E_{r'} = -\delta_{r,r'}$$ and $$E_r \cdot H = 0$$, we must have $$n_{i,r} < 0$$. If $$\ell$$ is the strict transform of the line through $$W_i$$ (which are collinear by assumption), then $$\ell \cdot E_r = 1$$ and $$\ell \cdot H = 1$$. Hence, $$(nH + \sum n_{i,r}E_r)\cdot \ell > 0$$ forces $$n > -\sum n_{i,r} \geq i$$. Since $$i$$ is arbitrary, this is impossible. $$\square$$

Motivation. The idea behind the construction is as follows. As I commented above, if $$Z$$ is affine, $$\mathscr L$$ is $$f$$-ample on $$X$$ and $$\mathscr M$$ is $$g$$-ample on $$Y$$, there is some $$a$$ such that $$\mathscr L \otimes f^* \mathscr M^{\otimes a}$$ is $$(g \circ f)$$-ample by Tag 0C4K. For $$Z$$ quasi-compact, you can choose this $$a$$ uniformly over all affines, but in general you might need a bigger and bigger $$a$$.

The first thing you try is $$Z$$ an infinite disjoint union of points, where you need a bigger and bigger $$a$$ on each component (for example when you blow up more and more collinear points in $$\mathbf P^2$$). But this doesn't work because in a disjoint union you have too much freedom to choose a completely different line bundle on $$X$$ that does the job ("you can choose a different $$a_i$$ per component").

However, if you put yourself in a situation where

1. you can compute $$\operatorname{Pic}(X)$$, and
2. there is a reason why all the $$a_i$$ have to be the same for any line bundle on $$X$$,

then you can make this into an argument leading to a contradiction.

• Thanks a lot, dear R. I find your construction brilliant and highly non trivial, even though I confess I haven't checked (yet?) the details... – Georges Elencwajg Nov 13 '19 at 13:02