Let $S$ be a Noetherian affine scheme. Let $S'\to S$ be a flat surjective morphism of affine schemes. Let $X\to S$ be a morphism such that $X\_{S'}\to S'$ is projective. Is $X\to S$ projective? It is not difficult to see that this holds when $S'\to S$ is a field extension.


The analogous statement for proper morphisms is true because being proper is fpqc local on the base. Hironaka's examples show that the same argument can not work for projective morphisms.



Projective after fpqc base change
Hironaka's example is a locally projective birational morphism $f \colon \tilde X \to X = \mathbf P^3$ where $\tilde X$ is not projective; in particular $f$ is not projective. This already shows that the property of being projective is not Zariski-local on the target, so in particular not fppf-local.


But you want the base to be *affine*, so let's do a little bit more.


**Lemma.** *Let $U \subseteq \mathbf P^3$ be any open containing the intersection $\{p,q\} = C \cap D$, and let $\tilde U = f^{-1}(U)$. Then $f|\_{\tilde U} \colon \tilde U \to U$ is not projective.*


In other words, the non-projectivity of $f$ is "purely concentrated above $p$ and $q$", and is not a global phenomenon. The idea is that away from $p$ and $q$ you can concretely write down what the $f$-ample line bundles are.


*Proof.* Suppose it were, and $\mathscr L$ on $\tilde U$ is a relatively ample line bundle. Let $Z = X\setminus U$. By the sequence
\begin{align\*}
\mathbf Z &\to \operatorname{Pic}(\tilde X) \to \operatorname{Pic}(\tilde U) \to 0 \\
1 &\mapsto [f^{-1}(Z)]\\
\end{align\*}
we can choose an extension of $\mathscr L$ to $\tilde X$. We claim that any such extension (which we will also denote by $\mathscr L$) is $f$-ample. (This implies that $\mathscr L \otimes f^\*( \mathcal O(n))$ is ample on $\tilde X$ for $n \gg 0$.)


Indeed, let $V = \mathbf P^3 \setminus \{p,q\}$ and let $\tilde V = f^{-1}(V)$. By assumption, $U \cup V = \mathbf P^3$. Moreover, $f|\_{\tilde V} \colon \tilde V \to V$ is just the blowup in the disjoint union of curves $C \cup D$. Hence, we have
$$\operatorname{Pic}(\tilde V) = f^\*\operatorname{Pic}(V) \oplus \mathbf Z [E\_C] \oplus \mathbf Z [E\_D],$$
where $E\_C \to C$ and $E\_D \to D$ are the exceptional divisors. A line bundle $f^\*\mathscr M \otimes \mathcal O\_{\tilde V}(aE\_C + bE\_D)$ on $\tilde V$ is $f$-ample if and only if $a < 0$ and $b < 0$: clearly these are $f$-ample by construction of the blowup, and conversely the restriction of an $f$-ample to a nontrivial fibre $f^{-1}(x) \cong \mathbf P^1$ for $x \in C \cup D$ should be ample, forcing $a < 0$ and $b < 0$ (since $E\_C|\_{f^{-1}(x)} = \mathcal O\_{\mathbf P^1}(-1)$ for $x \in C$). The same goes for line bundles on $\tilde V \cap \tilde U$. Write
$$\mathscr L|\_{\tilde V} = f^\*\mathscr M \otimes \mathcal O\_{\tilde V}(aE\_C + bE\_D).$$
Restricting this decomposition to $\tilde V \cap \tilde U$ gives
$$\mathscr L|\_{\tilde V \cap \tilde U} = f^\*\left(\mathscr M|\_{V \cap U}\right) \otimes \mathcal O\_{\tilde V \cap \tilde U}(aE\_C + bE\_D).$$
Since $\mathscr L|\_{\tilde U}$ is $f$-ample, the same goes for $\mathscr L\_{\tilde V \cap \tilde U}$, so $a < 0$ and $b < 0$. This in turn implies that $\mathscr L|\_{\tilde V}$ is $f$-ample. Since the same holds for $\mathscr L|\_{\tilde U}$ by assumption, we conclude that $\mathscr L$ is $f$-ample. $\square$




---


This gives the required counterexample: let $S = U \subseteq \mathbf P^3$ be an affine open containing $p$ and $q$, and let $S' = U\_1 \amalg U\_2$ for affine opens $U\_1, U\_2 \subseteq U$ each containing exactly one of $p$ and $q$ such that $U = U\_1 \cup U\_2$. (Something like this can be arranged: take $U\_1$ the complement of a plane $H\_1 \subseteq \mathbf P^3$ through $p$ but not $q$, and $U\_2$ the opposite, and let $U$ be the complement of a plane containing $H\_1 \cap H\_2$ but not $p$ or $q$. Or allow yourself more opens to cover $U$ if you don't want to fidget like this.)


The lemma shows that $f$ is not projective over $U$, and Hironaka showed that $f$ is projective over $U\_1$ and $U\_2$, hence over $S'$. $\square$


