An explicit isomorphism $K^\times/K^{\times p} \cong K^\flat/\wp K^\flat$ where $K \supset \mu_p$ is perfectoid field of mixed characteristic $(0, p)$

Question

Let $K$ be a perfectoid field of mixed characteristic $(0, p)$, i.e. $K$ has characteristic $0$ but its residue field has characteristic $p$. Further suppose that $K$ contains all the $p$^th roots of unity.

A subgroup of $K^\times/K^{\times p}$ corresponds to an abelian extension $L/K$ of exponent $p$ by Kummer theory, which corresponds to an abelian extension $L^\flat/K^\flat$ of exponent $p$ by the tilting correspondence, which in turn corresponds to a subgroup of $K^\flat/\wp K^\flat$ by Artin–Schreier theory.

Combining these correspondences should give us an isomorphism $K^\times/K^{\times p} \cong K^\flat/\wp K^\flat$, up to a choice of isomorphism $\mu_p \cong \Bbb Z/p\Bbb Z$.

My question is: Can we make any direction of this isomorphism explicit?

I have tried a few examples and computations to no avail.

Answer 1

Here is an explicit description of the isomorphism: It takes $a\in K^\flat$ to the class of $1+(\zeta_p-1)^p a^{1/p^n}\in K^\times$ for any large enough $n$ (the image modulo $p$-th powers is independent of the choice).

Here's an explanation. Let's first analyze the quotient $K^\times/(K^\times)^p$. As $K$ is perfectoid, any class in here is represented by an element of $1+p\mathcal O_K$. In fact, thinking a bit more about it, by an element of $1+p^{p/(p-1)-\epsilon} \mathcal O_K$ for any $\epsilon>0$ (where this statement only depends on valuations, so we do not need a literal element $p^{p/(p-1)-\epsilon}$ to make sense of this).

On the other hand, any element of $1+p^{p/(p-1)+\epsilon}\mathcal O_K$ is a $p$-th power. In other words, the quotient $K^\times/(K^\times)^p$ is a quotient of $(1+p^{p/(p-1)-\epsilon}\mathcal O_K)/(1+p^{p/(p-1)+\epsilon}\mathcal O_K)$, where the latter group is isomorphic via $x\mapsto 1+(\zeta_p-1)^p x$ (where $(\zeta_p-1)^p$ has the same valuation as $p^{p/(p-1)}$) to a quotient of the additive group $p^{-\epsilon}\mathcal O_K/p^\epsilon \mathcal O_K$. To figure out the quotient, we have to look at $p$-th powers of elements of the form $1+(\zeta_p-1)y$, with $y\in p^{-\epsilon}\mathcal O_K$, which gives modulo $1+p^{p/(p-1)+\epsilon}$: $$1+p(\zeta_p-1)y+(\zeta_p-1)^py^p\equiv 1+(\zeta_p-1)^p (y^p-y).$$

Using $p^{-\epsilon} \mathcal O_K/p^\epsilon \mathcal O_K\cong t^{-\epsilon} \mathcal O_{K^\flat}/t^\epsilon \mathcal O_{K^\flat}$, this quickly gives the desired isomorphism to $\mathrm{coker}(x^p-x|K^\flat)$ (using a similar analysis of the latter, to see that all elements can be represented by elements of $t^{-\epsilon}\mathcal O_{K^\flat}$ -- concretely, replace any $a$ by $a^{1/p^n}$ for large $n$ -- and the elements of $t^\epsilon \mathcal O_{K^\flat}$ are trivial).

Why is this the right isomorphism? Look at a Kummer extension $T^p=1+(\zeta_p-1)^px$ with $x\in p^{-\epsilon}\mathcal O_K$. Changing coordinates as $T=1+(\zeta_p-1)U$, this translates into an equation $$1+(\zeta_p-1)^p(U^p + (\zeta_p-1)[\ldots] - U) = 1 + (\zeta_p-1)^p x,$$ i.e. $$ U^p-U = x + (\zeta_p-1)[...]. $$ Removing the terms divisible by $\zeta_p-1$ does not change the extension for $\epsilon$ small enough, and the tilt is given by the similar Artin--Schreier equation (by the recipee explained in my paper: once the discriminant of the equation is small enough, one can find the tilt by the naive operation on defining equations).