Let $S$ be a connected scheme, let $\pi : \mathbb{P}_{S}^{r} \to S$ be projective $r$-space over $S$, and let $\mathcal{E}$ be a flat and locally finitely presented $\mathcal{O}_{\mathbb{P}_{S}^{r}}$-module. Is $\pi_{\ast}(\mathcal{E}(n))$ (nonzero and) flat and locally finitely presented for $n \gg 0$?

Remarks: If $S$ is a Noetherian scheme, the answer is "yes" using usual cohomology and base change theorems. I included "connected" because I am assuming we could otherwise construct an example with $S$ being an infinite disjoint union of fields. The case when $S$ is quasi-compact reduces to the case when $S$ is affine. Say $S = \operatorname{Spec} A$; we may write $A$ as a filtered colimit $A = \varinjlim_{\lambda \in \Lambda} A_{\lambda}$ where each $A_{\lambda}$ is a finite type $\mathbb{Z}$-subalgebra of $A$. Set $S_{\lambda} := \operatorname{Spec} A_{\lambda}$. Then we can descend $\mathcal{E}$ to vector bundles $\mathcal{E}_{\lambda}$ on $\mathbb{P}_{S_{\lambda}}^{r}$ for large enough $\lambda$. Let's fix some $\lambda$. By Serre vanishing [1, III, Theorem 5.2], we may replace $\mathcal{E}_{\lambda}$ by $\mathcal{E}_{\lambda}(n)$ to assume that $\mathrm{H}^{i}(\mathbb{P}_{A_{\lambda}}^{r},\mathcal{E}_{\lambda}) = 0$ for $i > 0$. By [2, Section 4, second Theorem] there is a finite complex (the "Grothendieck complex") $$ K^{\bullet} = \{0 \to K^{0} \to \dotsb \to K^{n} \to 0\} $$ of finitely generated projective $A_{\lambda}$-modules and, for each $i \ge 0$, a functorial isomorphism $$ \mathrm{H}^{i}(\mathbb{P}_{B}^{r} , \mathcal{E}_{\lambda}|_{\mathbb{P}_{B}^{r}}) \simeq \mathsf{h}^{i}(K^{\bullet} \otimes_{A_{\lambda}} B) $$ for every $A_{\lambda}$-algebra $B$. Now $\mathsf{h}^{i}(K^{\bullet}) = 0$ for $i > 0$ which means $K^{\bullet}$ is a direct sum of complexes of the form $\{\mathrm{id}_{P} : P \to P\}$ and a complex consisting of a single finitely generated projective $A_{\lambda}$-module at degree $0$. Thus we may assume that $K^{i} = 0$ if $i \ne 0$. Then $K^{0} \simeq \Gamma(\mathbb{P}_{A_{\lambda}}^{r} , \mathcal{E}_{\lambda})$, and the above gives $\Gamma(\mathbb{P}_{A}^{r} , \mathcal{E}) \simeq \Gamma(\mathbb{P}_{A_{\lambda}}^{r} , \mathcal{E}_{\lambda}) \otimes_{A_{\lambda}} A$. (I guess this argument should work if we replace $\mathbb{P}_{S}^{r}$ by a flat projective finitely presented morphism $X \to S$ that descends to some $S_{\lambda}$.)

I am guessing that the difficulty is "flatness" but I also don't know whether the "locally finitely presented" part is true, thus I will include the following subquestion:

Let $S$ be an affine scheme, let $\pi : \mathbb{P}_{S}^{r} \to S$ be projective $r$-space over $S$, and let $\mathcal{E}$ be a locally finitely presented $\mathcal{O}_{\mathbb{P}_{S}^{r}}$-module. Is $\pi_{\ast}\mathcal{E}$ necessarily finitely presented?


[1] Hartshorne, Algebraic Geometry

[2] Mumford, Abelian Varieties, Tata Institute of Fundamental Research Studies in Mathematics (1970)

Keywords: Noetherian approximation, cohomology and base change, higher direct images, not Noetherian

  • $\begingroup$ In the quasi-compact case, can you not modify the descending induction argument for Serre vanishing followed by the flatness argument? I think that the general case might be false (even for $S$ connected), because $n$ might vary depending on the affine open you choose. However, I don't know an example of this phenomenon. $\endgroup$ Commented Sep 11, 2018 at 4:33
  • $\begingroup$ @R.vanDobbendeBruyn Thanks, your comment reminded me of the Grothendieck complex and I think I now have an argument for the affine case. $\endgroup$ Commented Sep 11, 2018 at 6:06

1 Answer 1


The answer to the question about flatness is no and here is a counterexample. Let $S_n = \mathbf{A}^1$ with coordinate $t$ for $n\geq 1$ (over some base field). Let $$ \mathcal{A}_n = \mathcal{O}_{\mathbf{P}^1}(-n)\oplus \mathcal{O}_{\mathbf{P}^1}(n). $$ We have a non-split extension $$ 0\to \mathcal{O}_{\mathbf{P}^1}(-n) \to \mathcal{O}_{\mathbf{P}^1}(-n+1)\oplus \mathcal{O}_{\mathbf{P}^1}(n-1) \to \mathcal{O}_{\mathbf{P}^1}(n) \to 0 $$ where the middle term is $\mathcal{A}_{n-1}$. (For $n=1$, this is the Euler sequence twisted by $1$.) Degenerating this extension to the split one as $t\to 0$ (i.e. pulling it back to $\mathbf{P}^1\times S_n$ and pushing out along multiplication by $t$ on the left term) yields a vector bundle of rank two $\mathcal{E}_n$ on $\mathbf{P}^1_{S_n}$ whose fiber as $t=0$ (resp. $t\neq 0$) is $\mathcal{A}_n$ (resp. $\mathcal{A}_{n-1}$).

(Taking the associated projective bundles, we obtain the well-known degeneration of the Hirzebruch surface $F_{2n-2}$ to $F_{2n}$.)

Now let $S$ be an infinite chain of lines obtained by taking the disjoint union of $S_n=\mathbf{A}^1$ for all $n\geq 1$ and identifying $0\in S_n$ with $1\in S_{n-1}$. The above construction yields a vector bundle of rank two on $\mathbf{P}^1_S$ such that $\pi_*\mathcal{E}(m)$ does not have constant rank for every $m$.

The bottom line is that the statement you want is false for the stack of vector bundles of rank two and degree zero on $\mathbf{P}^1$; above, we exhibited a suitable scheme over that stack.

Note that $S$ is locally noetherian, so all the push-forwards are of course locally finitely presented.

  • $\begingroup$ Great, thanks. Just to make sure, if I wanted to avoid gluing along closed subsets, I would do your construction on $\mathbb{P}^{1}$ over the coordinate axes $\operatorname{Spec} k[s,t]/(st)$? $\endgroup$ Commented Sep 11, 2018 at 19:29
  • $\begingroup$ Yes, or rather $k[s, t]/(s(t-1)) [1/(s-1), 1/t]$. $\endgroup$ Commented Sep 11, 2018 at 20:38

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