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Let $K$ be a complete nonarchimedean field. The classical Tate algebra $K\langle T \rangle$ has lots of automorphisms, e.g., any substitution $T\mapsto a_1T+a_2T^2+\cdots$, where $a_1\in \mathcal{O}_K^\times$, $a_n\in\mathfrak{m}_K$ for $n\geq 2$, and $a_n\to 0$ as $n\to \infty$.

This question is about perfectoid spaces. Let $K$ be a perfectoid field (in particular $K$ is a non-discrete valued field). Let $A$ be the ``perfectoid Tate algebra'' $K\langle T^{1/p^\infty}\rangle$. This is obtained by inverting $p$ on the $p$-adic completion of $\mathcal{O}_K[T^{1/p^\infty}]=\bigcup_{n\geq 1} \mathcal{O}_K[T^{1/p^n}]$.

If the characteristic of $K$ is $p$, then $K$ is perfect. In that case, the automorphisms of $K\langle T\rangle$ from the first paragraph extend uniquely to $A$. Other automorphisms arise by composing these with $p$th powers and roots.

My question is, are there any other automorphisms? For instance, suppose $K$ has characteristic $p$ and $\varpi\in K$ has positive valuation. Then the substitution $$ T\mapsto T+\varpi T^{1/p}+\varpi^2 T^{1/p^2}+\cdots $$ is certainly an endomorphism of $A$, but I cannot decide if it is an automorphism.

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  • $\begingroup$ The one you mention seems tricky, but what about something like $T\rightarrow T+\sum_{n=1}^{\infty} \omega^n T^{n+\frac{1}{p^n}}$? Seems like theres not the same issues here, so its an automorphism and not one of the ones you mentioned. $\endgroup$
    – jacob
    Jan 29, 2014 at 23:29

1 Answer 1

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Recall the following elementary fact in commutative algebra:

Fact: Say $f:R \to S$ is a map of commutative rings. Assume there exists a $p \in R$ that is a non-zero divisor in both $R$ and $S$, and that $R$ and $S$ are $p$-adically complete. Then $f$ is an isomorphism if and only if $f_1$ is so; here $f_n$ is reduction modulo $p^n$ of $f$.

[ Proof: Indeed, it is elementary, using the torsion freeness, to go from $f_1$ being an isomorphism, to $f_n$ being an isomorphism for all $n$, and then one takes a limit. ]

This immediately implies that the map mentioned in the question is an isomorphism (as it is defined on the integral perfectoid Tate algebra, and coincides with the identity modulo $\varpi$). In particular, there indeed are more automorphisms than the ones coming from Frobenius or the classical Tate algebra.

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  • $\begingroup$ Thank you. I had foolishly assumed that non-noetherianness would foil this kind of argument, but all that was necessarily was a finitely-generated (indeed principal) ideal of definition. $\endgroup$ Jan 30, 2014 at 16:33

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