Let $R$ be a Noetherian commutative unital ring. Let $f:X\rightarrow Y=\mathrm{Spec}\,R$ be a proper flat morphism of schemes with geometrically reduced fibers. We want to prove that the induced map $Z=\mathrm{Spec}\, H^0(X, O_X)\rightarrow Y$ is flat.

Some thoughts. Since $f$ is qcqs, $f_*O_X$ is a quasi-coherent $O_Y$-module, i.e. associated to an $R$-module. Since flatness of a module can be checked on localizations, and properness and flatness are preserved by base change, we reduce to the case $R$ is local. I am not sure what to do next.

  • $\begingroup$ Related: mathoverflow.net/a/107603/3847 . I guess the examples there have nonreduced fibers. Why do you think this should be true? $\endgroup$ – Piotr Achinger Mar 27 '19 at 5:30
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    $\begingroup$ Welcome new contributor. If the geometric fibers are reduced, then the smooth locus of the morphism $f$ is dense in every fiber. Thus, 'etale locally on $Y$, there exists sections of $f$ meeting every component of every geometric fiber. Since formation of pushforward $f_*$ is compatible with flat base change, in particular it is compatible with 'etale base change. Thus, 'etale locally, $Z$ factors a union of sections of $f$. Thus, also $Z$ is 'etale over $Y$. $\endgroup$ – Jason Starr Mar 27 '19 at 7:33

The claim is correct, and, actually, even more is true. The map $R \to \mathrm{H}^0(X, \mathcal O_X)$ is (finite) etale, but the difficult part is really to prove flatness. Before going to the actual proof of this statement, let me discuss the main idea of the proof.

Step 1: Reduce to the case of a local noetherian ring $R$ (the main input is local nature of flatness).

Step 2: Reduce to the case of a local noetherian ring with algebraically closed residue field (the main input is EGA 0_{III} 10.3.1 and faithfully flat descent)

Step 3: Reduce to the case of a local noetherian complete ring with algebraically closed residue field (the main input is faithfully flat descent)

Step 4: Reduce to the case of Artinian local ring with algebraically closed residue field (the main input is Formal Function Theorem and Cohomology and Base Change)

Step 5: Leverage the fact that for any scheme $X$ over an artinian local ring we have a homeomorphism of underlying topological spaces $|X|$ and $|X_s|$ (fiber over a closed point) to prove the claim (with a help of Cohomology and Base Change result from Step 4).

Being said this, let us start proving our claim.

Theorem: Let $R$ be a noetherian ring and let $f: X \to \operatorname{Spec} R$ be a proper flat morphism with geometrically reduced fibres, then a natural map $$ g: \operatorname{Spec}(f_*\mathcal O_X) \to \operatorname{Spec} R $$ is a finite etale map.

First of all, note that Proper Mapping Theorem (applied to the morphims $f$) guarantees that $A:=\mathrm{H}^0(X, \mathcal O_X)$ is a finite $R$-algebra. Also we know that fibers of the morphism $g$ are geometrically reduced (since fibers of $f$ are), therefore once we can show that $A$ is a flat $R$-algebra, we automatically conclude that $A$ is a finite etale $R$-algebra. Now we go to the proof of this flatness result.

We want to make a bunch of reductions to simplify the ring $R$ in the problem as much as possible. To make these reduction steps we observe that Stein factorization (for qcqs morphisms) commutes with flat base change. This means that it suffices to prove the statement for all localizations of $R$ (since flatness can be checked on the level of local rings). So we can assume that $R$ is local. But still, a ring $R$ can be pretty bad, we want to simplify even more by making its residue field algebraically closed. To actually do this, we use EGA 0$_{III}$ 10.3.1 to find a faithfully flat local extension of local rings $R \to R'$ with a residue field of $R'$ being algebraically closed. Since flatness descends along faithfully flat morphisms (and Stein factorization commutes with flat base change!) we can reduce the question to the case of a local ring with algebraically closed residue field. Finally, we look at the completion map $R \to \widehat{R}$, which is also faithfully flat (by noetherianness assumption), to reduce the question to the case of a local complete noetherian ring $R$ with algebraically closed residue field.

Now we are in the situation as above and we want to show that $A=\mathrm{H}^0(X, \mathcal O_X)$ is $R$-flat. Recall that there is the Cohomology and Base Change Theorem that says the following:

Theorem (Cohomology and Base Change): Let $f:X \to \operatorname{Spec} R$ be a proper flat morphism, and pick an arbitrary point $y\in \operatorname{Spec} R$ such that a natural morphism $$ \phi^0(y):\mathrm{H^0}(X, \mathcal O_X)\otimes_R k(y) \to \mathrm{H^0}(X_y, \mathcal{O}_{X_y}) $$ is a surjective morphism. Then there is a neighborhood $y\in U$ such that for all $z\in U$ the natural map $\phi^0(z)$ is an isomorphism and $(f_*\mathcal O_X)|_U$ is locally free and commutes with arbitrary base change.

Remark: Cohomology and Base Change Theorem is actually a more general statement. Look at Corollary 8.3.11 and Remark in FGA Explained for the most general form.

Let us now go back to our situation. We know that $R$ is local in our case, so it suffices (for our purposes) to show that the natural map $\mathrm{H}^0(X, \mathcal O_X) \otimes_R k(\mathfrak m) \to \mathrm{H}^0(X_s, \mathcal O_{X_s})$ is surjective for the maximal ideal $\mathfrak m \subset R$ (and a corresponding closed point $s\in \operatorname{Spec} R$). Note that geometric fibers of the map $X \to \operatorname{Spec} R$ are still geometrically reduced, this implies that $\mathrm{H}^0(X_s, \mathcal{O}_{X_s})$ is isomorphic to a finite product of fields $k=k(\mathfrak m)=R/\mathfrak m$ (here we are using that $k$ is algebraically closed!). In other words, $$ \mathrm{H}^0(X_s, \mathcal{O}_{X_s}) \cong \prod_{i \in I} ke_i, $$ where $e_i\in \mathrm{H}^0(X_s, \mathcal{O}_{X_s})$ and each $i$ is labeled by a connected component of $X_s$. In particular, $\mathrm{H}^0(X_s, \mathcal{O}_{X_s})$ is generated by idempotents as a $k$-vector space. This implies that in order to show surjectivity of the map $\mathrm{H}^0(X, \mathcal O_X) \to \mathrm{H}^0(X_s, \mathcal O_{X_s})$ (which is equivalent to the surjectivity of $\mathrm{H}^0(X, \mathcal O_X) \otimes_R k(\mathfrak m) \to \mathrm{H}^0(X_s, \mathcal O_{X_s})$), it is sufficient to show that each idempotent of $\mathrm{H}^0(X_s, \mathcal O_{X_s})$ can be lifted to $\mathrm{H}^0(X, \mathcal O_X)$. This is a pretty standard question on the Formal Function Theorem.

The Formal Function Theorem guarantees that $$ \mathrm{H}^0(X,\mathcal O_X)\hat{} \cong \lim_n \mathrm{H}^0(X_n, \mathcal O_{X_n}), \text{where } X_n = X\times_{R} \operatorname{Spec} R/\mathfrak{m}^{n+1}. $$ Since $R$ is complete(!) we conclude that $$ \mathrm{H}^0(X,\mathcal O_X)\hat{} =\mathrm{H}^0(X,\mathcal O_X) \otimes_R R\hat{} =\mathrm{H}^0(X,\mathcal O_X). $$ Therefore, we see that $\mathrm{H}^0(X,\mathcal O_X) \cong \lim_n \mathrm{H}^0(X_n, \mathcal O_{X_n})$. So, in order to lift idempotents from $\mathrm{H}^0(X_s, \mathcal O_{X_s})$ to $\mathrm{H}^0(X, \mathcal O_{X})$ it is actually sufficient to lift idempotents from each $\mathrm{H}^0(X_n, \mathcal O_{X_n})$ to $\mathrm{H}^0(X_{n+1}, \mathcal O_{X_{n+1}})$.

But this is pretty easy to do, the key is that the underlying topological spaces of schemes $|X_{n}|$ and $|X_{n+1}|$ are the same. Choose an idempotent $e\in \mathrm{H}^0(X_n, \mathcal O_{X_n})$, then a vanishing locus $V(e)$ is a union of connected components of $X_n$, and $e$ is a function, which is equal to $0$ on them and $1$ on all other connected components. Since the underlying topological space of $X_{n+1}$ is equal to $X_n$, we can choose the function $e'\in \mathrm{H}^0(X_{n+1}, \mathcal O_{X_{n+1}})$ such that it is equal to $0$ on the same connected components and it is equal to $1$ on all other connected components. Then $e'$ lifts $e$, so we are done!

This finishes the proof of the claim that $\mathrm{H^0}(X, \mathcal O_X)\otimes_R k(\mathfrak m) \to \mathrm{H^0}(X_s, \mathcal{O}_{X_s})$ is surjective. And this implies flatness of $A$ by Cohomology and Base Change!

UPD: By the way, the result also holds without any noetherianess assumption, if you assume that $f$ is proper and finitely presented. The proof is (as always) a reduction to the noetherian case using the spreading out techniques from EGA IV$_3$. But I guess it is not that interesting.

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  • $\begingroup$ I think we clicked "submit" at the same time. $\endgroup$ – Jason Starr Mar 27 '19 at 7:39
  • $\begingroup$ in the final conclusion are you using the fact that in an affine scheme (in your case Noetherian, but I don't think it matters), an open subset containing all closed points is the whole space? $\endgroup$ – user74900 Mar 27 '19 at 10:08
  • $\begingroup$ @Aknazar Kazhymurat You may phrase it in this way. But really I am using the fact that the only open subset of $X=\operatorname{Spec} R$ ($R$ is a local ring), containing the closed point, is the whole $X$. $\endgroup$ – gdb Mar 28 '19 at 4:17

You can find this result in EGA, III.7.8.6.

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