# Why is Faltings' "almost purity theorem" a purity theorem?

My understanding of purity theorems is that they come in several flavors:

1) Those of the form "this Galois representation is pure, i.e. the eigenvalues of $Frob_p$ are algebraic numbers all of whose absolute values have size $p^{w/2}$". I don't think that this is the kind of purity I'm interested in.

2a) Purity in algebraic geometry 1: on a smooth algebraic variety the ramification locus of a morphism is a pure codimension 1 subvariety (Zariski, Nagata etc).

2b) Purity in algebraic geometry 2: absolute cohomological purity. Basically -- if $Y$ is a pure codimension $d$ subscheme of $X$ then the local cohomology groups $H^i_Y(X,\mathbf{Z}/n\mathbf{Z})$ should vanish away from $i=2d$ (under various hypotheses, e.g. $X$ locally Noetherian, $n$ invertible everywhere etc) and should be $\mathbf{Z}/n\mathbf{Z}$ etale locally if $i=2d$ (SGA5, Gabber etc).

Faltings proved an "almost purity theorem" and I think that I'm supposed to be regarding it as some sort of analog of a purity theorem above. Faltings' work occurs in the context of "almost mathematics", where one is working over the integers $R$ in a certain type of (non-discrete) valuation ring $K$, so $R$ has a maximal ideal $m$ and the idea is that instead of working in the category of $R$-modules, one works in the category of $R$-modules up to $m$-torsion (some localised category); this is the category of "almost $R$-modules".

Faltings' almost purity theorem (or perhaps some beefed-up version due to Scholze) says something like this:

Theorem: If $K$ is a perfectoid field, $A$ is a perfectoid $K$-algebra, and $B/A$ is finite etale, then $B$ is also perfectoid and $B^o/A^o$ is almost finite etale.

Whatever does this have to do with the purity theorems mentioned at the beginning of this post?

• Dear armpit6, Have you looked at Kisin's review of one of Faltings's papers on MathSciNet, where he explains the analogy between Faltings's method of proof and (one of) the method(s) of proof of purity of the branch locus? Back in grad school, Mark told me that this was the reason for the name. Regards, Jun 4, 2013 at 20:41

This is really just an elaboration of Emerton's comment: You should read Mark Kisins' review of Faltings's paper "Almost etale extensions".

But I wanted to elaborate: Faltings regards the almost purity theorem as an analogue of Zariski-Nagata purity. In Faltings's original setup, it was formulated as follows. Consider the rings

$$R_m = \mathbb{Z}_p[p^{1/p^m},T_1^{\pm 1/p^m},...,T_n^{\pm 1/p^m}]$$

Each of them is smooth over $\mathbb{Z}_p[p^{1/p^m}]$ (in particular regular), and the transition maps are finite and etale after inverting $p$. Let $S_0$ be a finite normal $R_0$-algebra, which is etale after inverting $p$, and let $S_m$ be the normalization of $S_0\otimes_{R_0} R_m$. The idea is that there is ramification of $S_0$ in the special fibre, and you want to get rid of it, by adjoining the chosen tower of highly ramified rings $R_m/R_0$. It is not hard to see that this actually works almost at the generic point of the special fibre: At the generic point, the local ring is a discrete valuation ring, and the statement is that the discriminant of the extension of discrete valuation rings becomes arbitrarily small as $m\to\infty$ (this boils down to some more or less classical ramification theory).

Now, assume that you were lucky, and for some $m$, the ramification at the generic point of the special fibre is not just very close to zero, but actually zero on the nose. Zariski-Nagata purity tells you that the ramification locus of $S_m$ over $R_m$ has to be pure of codimension $1$, but you know that there is no ramification at the codimension $1$ points (which are either in characteristic $0$, or equal to the generic point of the special fibre) -- thus, there is no ramification at all, and $S_m$ over $R_m$ is etale.

The almost purity theorem says that this result extends to the almost world: In the limit as $m\to \infty$, $S_\infty$ over $R_\infty$ is almost etale (in the technical sense defined in almost ring theory).

• Thanks Peter. Your papers are slowly making more sense to me. Jun 9, 2013 at 8:10