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In this paper, F. Kato recollects basic facts on henselian schemes and proves some partial results towards GAGA in the context of henselian schemes.

Let $I$ be a finitely generated ideal in a noetherian ring $A$, such that $(A,I)$ is $I$-adically complete: in particular, a henselian pair.

Let $X$ be a proper and flat $A$-scheme, $X^h$ its $I$-adic henselianization, $X^{\wedge}$ its $I$-adic completion.

There are henselianization and completion functors between categories of coherent modules:

$$\text{Coh}(X)\to\text{Coh}(X^h)\to\text{Coh}(X^{\wedge})$$

that induce maps between coherent cohomologies:

$$(1)\ \ \ H^p(X,\mathcal{F})\to H^p(X^h,\mathcal{F}^h)\to H^p(X^{\wedge}, \mathcal{F}^{\wedge})$$

We know from the theorem on formal functions that their composition is an isomorphism of finitely generated $A$-modules.

The best we can say is that $H^p(X^h,\mathcal{F}^h)\to H^p(X^{\wedge},\mathcal{F}^{\wedge})$ is surjective.

Question 1: Does anyone know of an example where this map is not injective?

Note that this would also give examples where $H^p(X^h,\mathcal{F}^h)$ is not a finitely generated $A$-module. Indeed, if we knew that for any coherent $\mathcal{O}_{X^h}$-module $\mathcal{G}$, $\bigoplus_{p}H^p(X^h,\mathcal{G})$ is a finitely generated $A$-module, then we could apply Theorem A in here and deduce that the two functors above are all equivalences of categories, and that the map $H^p(X^h,\mathcal{F}^h)\to H^p(X^{\wedge},\mathcal{F}^{\wedge})$ is an isomorphism.

In other words, if the henselian analog of the theorem on formal functions fails for proper noetherian henselian schemes, it fails badly and the proper mapping theorem fails too.

Note that section 2.4 in Kato's paper is incorrect, because he quotes Theorem B in a paper about quasi-coherent cohomology on affine henselian schemes that contains a fatal mistake: a counterexample to it was found by de Jong here.

However, his partial GAGA results do not rely on that and are correct. In particular, for $p = 0$ the answer to the above question is yes.

Question 2: Does anyone know a proof or a counterexample to the obvious henselian analog of Grothendieck's formal existence theorem for coherent $\mathcal{O}_{X^h}$-modules on proper henselian schemes?

Remarks:

  • I see a way to reduce these statements to the case $X = \mathbb{P}^n_A$. I was trying to think about $H^1((\mathbb{P}^n_A)^h,\mathcal{O}^h)$. If this doesn't vanish then of course these GAGA statements are false and the map $H^p(X^h,\mathcal{F}^h)\to H^p(X^{\wedge},\mathcal{F}^{\wedge})$ cannot possibly be injective.

  • I don't know exactly what to expect as an answer.

    The fact that coherent cohomology doesn't vanish on affines, doesn't mean these GAGA statements can't be true, although it's certainly not helpful.

    For $p = 0$ they are true and the henselianization functor from proper flat $A$-schemes to proper flat henselian $A$-schemes is even fully faithful.

    If a coherent $\mathcal{O}_{X^h}$-module is algebraizable, then so are its coherent submodules and coherent subquotients.

    These should be hints towards the validity of these GAGA statements.

Any counterexample, argument, reference, or even just a calculation of $H^*((\mathbb{P}^n_A)^h,\mathcal{O}^h)$, would be much appreciated.

  • In such calculations one can't use Cech cohomology due to the non vanishing of coherent cohomology on affines, although this may still work on projective spaces if one shows that for any coherent $\mathcal{O}^h$-module $\mathcal{G}$, and any standard open $D_+(f)\subset(\mathbb{P}^n_A)^h$, $H^p(D_+(f),\mathcal{G})=0$ for $p>0$, using that $D_+(f)$ aren't random affine henselian schemes but they are spectra of henselianized localizations of polynomial $A$-algebras.

  • The cohomology $H^*((\mathbb{P}^n_A)^h,\mathcal{O}^h)$ seems to be the whole point. If one proves it agrees with $H^*(\mathbb{P}^n_A,\mathcal{O})$, and if one proves that every coherent $\mathcal{O}^h$-module on $(\mathbb{P}^n_A)^h$ is a quotient of some $\mathcal{O}^h(k)^{\oplus N}$, then everything else follows from Jack Hall's general GAGA principle, at least for $X$ finitely presented in $\mathbb{P}^n_A$.

Update:

  • the posted answer shows that for $p = 1$, $A$ a $t$-adically complete dvr of generic characteristic zero, $X = \mathbf{P}^1_A$ and $F = \mathcal{O}$, the maps in equation (1) read $0\to H^1((\mathbb{P}^1_A)^h,\mathcal{O}^h)\to 0$ and the middle module is nonzero. In particular, the answer to Question 2 is "no".

  • when $A$ is adically complete, noetherian and of pure characteristic $p>0$, I was able to show that the maps in equation (1) are isomorphisms. I was also able to construct, for any coherent $\mathcal{O}^h$-module $F$ on $(\mathbb{P}^n_A)^h$, a surjection to $F$ from an $\mathcal{O}^h$-module of the form $\mathcal{O}^h(k)^{\oplus N}$ for appropriate $k,N$. In my argument I crucially use the fact that $A$ is an $\mathbf{F}_p$-algebra. This takes care of the existence part of GAGA too, so there seems to be a GAGA principle for "projective" henselian schemes in characteristic $p>0$.

  • The example given in the answer still does not mean that a GAGA principle for henselian schemes of generic characteristic zero cannot hold. Full faithfulness of the functor $\text{Coh}(X)\to\text{Coh}(X^h)$ is true and was proved by F. Kato for $X$ proper of arbitrary characteristic, because the maps in equation (1) are isomorphisms for $p=0$. This shows that coherent submodules and subquotients of an algebraizable coherent $\mathcal{O}^h$-modules are algebraizable. If one shows that for $X = \mathbf{P}^n_A$, and for any coherent $\mathcal{O}^h$-module $F$ on $(\mathbb{P}^n_A)^h$, a surjection to $F$ from an $\mathcal{O}^h$-module of the form $\mathcal{O}^h(k)^{\oplus N}$ for appropriate $k,N$ exists, then GAGA holds, even though the "henselian analog" of the theorem on formal functions fails.

    This last part of the question remains open.

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  • $\begingroup$ Welcome new contributor. The map of cohomology that you write is surjective only after you tensor the domain $A^h$-module with $\widehat{A}$. $\endgroup$ Feb 12, 2019 at 11:24
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    $\begingroup$ @JasonStarr Hi! I'm assuming $A^h = A^{\wedge}$ precisely for this reason. $\endgroup$
    – user135665
    Feb 12, 2019 at 11:25
  • $\begingroup$ Now I see that $A$ is $I$-adically complete. $\endgroup$ Feb 12, 2019 at 11:27
  • $\begingroup$ If the finiteness aspect is true in the complete situation, one can reduce the case where $A$ is only henselian along $I$ by base changing to $A^{\wedge}$ and using faithful flatness of $A^h\to A^{\wedge}$ anyway. So the case when $A$ is complete is really the case where one should test all these statements first. $\endgroup$
    – user135665
    Feb 12, 2019 at 11:27

1 Answer 1

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OK, it turns out that $H^1((P^1_A)^h, O^h)$ is nonzero in general. A counter example can be found in a blog post on the very blog you mention in your post. Here is a link.

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    $\begingroup$ I was able to prove the whole thing in characteristic $p >0$. In this situation one can construct, for any projective henselian scheme $X$ over a Noetherian, complete adic $\mathbf{F}_p$-algebra, and any coherent $O^h$-module $F$, a surjection from $O^h(k)^{\oplus N}$ for suitable $k,N$. This uses $p>0$ in a crucial way (at least the way I do it). Then I was able to compute $H^*((\mathbb{P}^n)^h, O^h)$ to be correct and reduce the full faithfulness statement in GAGA to this calculation. The bottom line is that the finiteness aspect holds in char $p>0$, and the rest of the assumptions $\endgroup$
    – user135665
    Feb 14, 2019 at 5:14
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    $\begingroup$ in Jack Hall's GAGA paper are easier to check as a consequence of basic properties of henselianizations. The finiteness aspect was the whole point of course. So it looks like there is a good theory of henselian schemes in char $p>0$, at least when they're "henselian finitely presented" and equipped with a projective embedding. One can imagine to bootsrap the comparison theorem to the proper case. I'll write this up when I find some time. This originated mostly as a curiosity, byproduct of some loosely related work. Thanks for this link! $\endgroup$
    – user135665
    Feb 14, 2019 at 5:17

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