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:  

Prescheme in EGA=Scheme nowadays  

Scheme in EGA=Separated scheme nowadays.  

Thanks a lot to R. Van Dobben de Bruyn for making me aware of my initial confusion.



Can you give an example of two projective morphisms of schemes whose composition is not projective?
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. $\square$




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**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](https://stacks.math.columbia.edu/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.


