I am trying to understand the MathSciNet review written by Mark Kisin of the paper "Almost etale extensions" of Faltings. There Kisin illustrates Faltings' approach to the almost purity theorem with the following proof of the Zariski-Nagata purity theorem in characteristic $p$.

We have a finite map $g\colon A \rightarrow B$ of rings of characteristic $p$ with $A$ regular and $B$ normal such that $g$ is etale at the height $1$ primes. We want to show that, in line with the Zariski-Nagata purity, $g$ is actually finite etale. The key point is to show that $g$ is flat, since then the ramification locus will be generated by the discriminant (so would be a divisor, as desired), and for this, by the Auslander-Buchsbaum formula or by miracle flatness, it suffices to prove that $B$ is Cohen-Macaulay as an $A$-module. In particular, since normality implies $(S_2)$, we may assume that $A$ is of dimension $\ge 3$ and local and, arguing by induction, that $B$ is Cohen-Macaulay away from the closed point $\mathfrak{m}$ of $\mathrm{Spec}(A)$.

The last assumption implies, by SGA 2, Expose VIII, Corollary 2.3, that local cohomology modules $H^i_{\mathfrak{m}}(B)$ for $i < \mathrm{dim}(A)$ are finitely generated $A$-modules, so, since they are supported at $\mathfrak{m}$, they have finite length. But somehow the Frobenius $\varphi$ of $A$ and $B$ induces an isomorphism $$H^i_{\mathfrak{m}}(B) \otimes_{A,\varphi} A \rightarrow H^i_{\mathfrak{m}}(B)$$ described in the review by the mystifying formula $h \otimes a \mapsto \varphi(h)$. If we have this isomorphism, then we are done: by the regularity of $A$, its Frobenius is flat, so the length of the source of the isomorphism is $p^{\mathrm{dim}(A)}$ times the length of the target. Then these lengths are $0$ and $H^i_{\mathfrak{m}}(B) = 0$ for each $i < \mathrm{dim}(A)$, so that $B$ really is Cohen-Macaulay as an $A$-module.

I have two questions:

  1. Why is there an isomorphism as displayed?
  2. This argument doesn't seem to use the etaleness of $B$ at the height $1$ primes. Why wouldn't it prove that any finite map $g: A \rightarrow B$ with $A$ regular of characteristic $p$ and $B$ normal is actually flat? (Which is false: $\mathbb{F}_p[x, y, z]$ has finite normal covers that are not flat.)
  • 4
    $\begingroup$ For #2, it is the step of deducing the discriminant divisor is empty (by checking at height-1 points) where etaleness at height-1 primes is used. For #1, the "mystifying formula" is a typo and should read $h \otimes a \mapsto a \varphi_B(h)$ (for clarity about which Frobenius). For $i\ge 2$, the local cohomology coincides with cohomology of the structure sheaf on the open complement of the closed point, where $\varphi_A$ and $\varphi_B$ form a Cartesian diagram by etaleness there, and the formation of cohomology of quasi-coherent sheaves (on qcqs schemes) commutes with flat base change. $\endgroup$
    – nfdc23
    Sep 28, 2017 at 19:48
  • $\begingroup$ Your explanation about $i \ge 2$ is precisely what I was missing, thank you. This clearly explains the inductive reasoning and the importance of the etaleness through the Cartesianness of the diagram that involves the Frobenii. The latter is also the aspect that would fail if we tried to use the same argument to prove the mere flatness of $g$ as in scenario 2. $\endgroup$
    – Lisa S.
    Sep 28, 2017 at 20:38


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