4
$\begingroup$

Maybe, I am being stupid, but when I consider ramified extension of a perfectoid field with the characteristic $0$, I cannot find the correspondent field with characteristic $p$. Let me put it more precisely.

Let $K$ be the completed field of $\mathbb{Q}_p(p^{-\infty})$. Consider $\zeta_p$ be a $p$th root of unity, and let $L=K(\zeta_p)$. According to Scholze's theorem (or Fontaine-Winterberger maybe enough), there exists a correspondent field $L^\flat$ which is of degree $p-1$ over $K^{\flat}$. However, I cannot find such $L^\flat$.

I tried to follow the definition to construct $L^\flat$ which is

$$L^\flat=\varprojlim L^{\circ}/p$$

where the transition map is Forbenius, and then I meet the problem that I cannot find the $p$th root of $\zeta_p$ in $L$. Similar problem happens if I consider $K$ to be the completion of $\mathbb{Q}_p(\zeta_{p^{\infty}})$ and adjoint $p$th root of $p$. However, the extension of a perfectoid field is a perfectoid field according to Scholze, so I do not think this should happen.

I would like to thanks all of you for your help.

Improve this question
$\endgroup$
5
  • $\begingroup$ You don't need to find a $p$th root of $\zeta_p$; rather, you just need to find a $p$th root of some element that is congruent to $\zeta_p$ mod $p$. $\endgroup$
    – dgulotta
    Commented Mar 27, 2017 at 18:21
  • $\begingroup$ Ahhh! Thanks for reminding me that. I do know what I should find, but just doesn't realize the construction is not that hard. However, do you know what $L^\flat$ is? More precisely, if I let $K^\flat$ be the completion of $\mathbb{F}_p((t^{-\infty}))$ and $L^\flat=K^{\flat}(\alpha)$ for some $\alpha$, then could we write down the primitive polynomial of $\alpha$? The only thing I know is that this is a separable ramified extension of degree $p-1$. Thanks for your kindly help again. $\endgroup$
    – Wayne Peng
    Commented Mar 27, 2017 at 19:42
  • 1
    $\begingroup$ It is possible to find $p$-power roots of $\zeta_p-1$ mod $p$ using the following idea. Let $u=\zeta_p-1$. The quantity $(-p)^{-1/p} u$ is close to being a $p$th root of $u$. Given a polynomial $f_n \in \mathcal{O}_K[x]$, we can find $f_{n+1} \in \mathcal{O}_K[x]$ so that $f_n(u) - f_{n+1}(u) \equiv f_n((-p)^{-1/p}u)^p \pmod p$. For large enough $N$, $f_{N+1}(u)$ will be divisible by $p$, and then $\sum_{n=0}^N f_n((-p)^{-1/p} u)$ will be a $p$th root of $f_0(u)$ mod $p$. $\endgroup$
    – dgulotta
    Commented Mar 27, 2017 at 21:22
  • $\begingroup$ To give a more specific example, let $p=5, u=\zeta_5-1, v=(-5)^{-1/5} u$. Then $v^5 = u+2u^2+2u^3+u^4 \equiv u+2u^2+2u^3 \pmod 5$, $(v^2)^5 \equiv u^2+4u^3 \pmod 5$, and $(v^3)^5 \equiv u^3 \pmod 5$. So $(v-2v^2+6v^3)^5 \equiv u \pmod 5$. $\endgroup$
    – dgulotta
    Commented Mar 27, 2017 at 21:34
  • $\begingroup$ these are rather exotic objects. I can't even begin reading about them. I heard they interpolate between finite fields and cyclotomic extensions $\endgroup$
    – john mangual
    Commented Mar 27, 2017 at 23:01

0

Reset to default

You must log in to answer this question.