用户: Fyx1123581347/偏微分方程/变分方法/Hille–Yosida 定理

1Perfectoid Fields

A characteristic $p$ ring $R$ (i.e., a ring $R$ such that $p\cdot 1_R=0$ or synonymously an ${\mathbb{F}}_p$-algebra) is perfect if the Frobenius $\varphi :R\to R,x\mapsto x^p$ is an isomorphism; it is semiperfect if $\varphi$ is merely surjective.

定义 1.1. Let $R$ be a ring of characteristic $p$, $I$ be the poset $\mathbb{N}$, $F:I\to \mathsf{Ring}$ be the functor $F(i)=R$, $F((i,j))={\varphi }^{j-i}$ for $j\ge i$, $G:I^{op}\to \mathsf{Ring}$ be the functor $F(i)=R$, $F((j,i))={\varphi }^{j-i}$ for $j\ge i$. Define $$R_{perf}=\mathrm{colim}F,R^{perf}=\lim G.$$

Notice the colimit is filtered so one can write it down explicitly, and$$R^{perf}=\{(a_i{)}_{i\in \mathbb{N}}\mid a_i=a_{i+1}^p\}.$$

命题 1.2. Both $R_{perf}$ and $R^{perf}$ are perfect. The canonical map $R=F(0)\to R_{perf}$ (resp. $R^{perf}\to G(0)=R$) is universal for maps into (resp. from) perfect rings.

证明. One can argue directly, or notice that $\varphi$ induces the same universal property of the (co)limit.

$\square$

例 1.3. The image of projection $R^{perf}\to R$ consisting of $f\in R$ that admit a compatible system $\{f^{\frac{1}{p^k}}\}$. We shall sometimes call such elements perfect. It follows that $R^{perf}\to R$ is surjective exactly when $R$ is semiperfect.

命题 1.4. Let $I$ be an ideal of characteristic $p$ ring $R$.

1.	If $I$ is locally nilpotent, then $R_{perf}\simeq (R/I{)}_{perf}$.
2.	If ${\varphi }^n(I)=0$ for some $n$, then $R^{perf}\simeq (R/I{)}^{perf}$.

证明. colimit perfection: As filtered colimit is exact, clearly $R_{perf}\to (R/I{)}_{perf}$ is surjective. If $r\in R_{perf}$ coming from $R_i$ is $0$ in $(R/I{)}_{perf}$, then $r^{p^m}$ is $0$ in $R/I$ for some $m$ and $r^{p^m}\in I$. Thus $r^{p^{m}k}=0$, which implies $r=0$ in $R_{perf}$.

limit perfection: clearly ${\lim }_{\varphi }I=0$, and $I$ is Mittag – Leffler. Thus ${\lim }_{\varphi }R={\lim }_{\varphi }R/I$

$\square$

例 1.5. If $\varphi$ is injective, then $R^{perf}$ is the subring of perfect elements in $R$. Thus ${\mathbb{F}}_p[t{]}^{perf}={\mathbb{F}}_p$.

If $R$ is a domain, $K=\mathrm{Frac}R$, then $R_{perf}=R[r^{\frac{1}{p^{\infty }}}{]}_{r\in R}$.

If $\mathrm{char}R=p$, and $R$ is perfect, then $(R/f{)}^{perf}$ is the $f$-adic completion of $R$. In fact, we have the commutative diagram where each vertical arrow is an isomorphism.

定义 1.6 (Fontaine). For any ring $R$, set $R^{♭}:=(R/p{)}^{perf}$. Unless otherwise specified, this ring is endowed with the inverse limit topology.

引理 1.7. Let $R$ be a ring, and let $t\in R$ be an element such that $p\in (t)$. Given $a,b\in R$ with $a\equiv b\mathrm{mod}t$, we have $a^{p^n}\equiv b^{p^n}\mathrm{mod}{t}^{n+1}$ for all $n$.

命题 1.8. Assume $R$ is $p$-adically complete. The projection map $R\to R/p$ induces a bijection $$\underset{x\mapsto x^p}{\lim }R\to \underset{\varphi }{\lim }R/p=:R^{♭}$$of multiplicative monoids.

证明. Injectivity: Suppose $a_n\equiv b_n\mathrm{mod}p$ for $n\ge 0$. As $a_{n+k}^{p^k}=a_n$ for $k\ge 0$, from the lemma we have $a_n\equiv b_n\mathrm{mod}{p}^k$. As $R$ is $p$-adically separated, $a_n=b_n$.

Surjectivity: For $(\overline{a_n})\in {\lim }_{\varphi }R/p$, take (arbitrary) lift $(a_n)\in R^{\mathbb{N}}$. Then $a_{n+k+1}^p\equiv a_{n+k}\mathrm{mod}p$ for all $n,k$. Thus $k\mapsto a_{n+k}^{p^k}$ is Cauchy for the $p$-adic topology and thus has a limit $b_n$. One checks $b_n=b_{n+1}^p$ and $b_n$ lifts $\overline{a_n}$.

$\square$

注 1.9. If one topologizes $R$ with the $p$-adic topology and $R/p$ with the discrete topology, then the bijection is a homeomorphism. In fact, the map $R\to R/p$ is continuous, and thus ${\lim }_{x\mapsto x^p}R\to R^{♭}$ is continous. It suffices to show it is open. (to be completed)

注 1.10. Via projection to the last term, we get multiplicative map $$♯:R^{♭}\to R$$denoted $f\mapsto f^{♯}$. For $R={\mathbb{Z}}_p$ this is the Teichmüller lifting.

引理 1.11. If a $p$-adically complete ring $R$ is a domain (resp. a valuation ring), the same is true for its tilt $R^{♭}$. In fact, if $|\cdot |:R\to \Gamma \cup \{0\}$ is the valuation on $R$, then the map $R^{♭}\stackrel{♯}{\to }R\stackrel{|\cdot |}{\to }\Gamma \cup \{0\}$ gives the valuation on $R^{♭}$. In particular, the rank of $R^{♭}$ is bounded by the rank of $R$.

证明. For $a,b\in R^{♭}$, let $(a_n),(b_n)\in {\lim }_{x\mapsto x^p}R$ be the corresponding lift. Then one checks $a\mid b$ precisely when $a_0\mid b_0$, or equivalently $|a_0|\ge |b_0|$.

For the last statement, notice that a subgroup ${\Gamma }^{\prime }$ of a totally ordered abelian group $\Gamma$ has smaller rank. In fact, for any convex subgroup $H^{\prime }$ of ${\Gamma }^{\prime }$, let $H$ be the subset of $\Gamma$ defined by $$\{\gamma \in \Gamma \mid h_1^{\prime }\le \gamma \le h_2^{\prime }\text{for some}{h}_1^{\prime },h_2^{\prime }\in H^{\prime }\},$$then $H$ is a convex subgroup of $\Gamma$, and different $H^{\prime }$ corresponds to different $H$.

$\square$

Fix a prime number $p$.

定义 1.12. A perfectoid field $K$ is a NA field with residue characteristic $p$ such that

•	The value group $|K^{\ast }|\subset {\mathbb{R}}_{>0}$ is not discrete.
•	$K^{\circ }/p$ is semiperfect.

例 1.13.

1.	$K=\hat{{\mathbb{Q}}_p(p^{\frac{1}{p^{\infty }}})}$.
2.	$K=\widehat{{\mathbb{Q}}_p({\mu }_{p^{\infty }})}$.
3.	$K={\mathbb{C}}_p$.
4.	If $\mathrm{char}K=p$, then $K$ is perfectoid if and only if it is perfect.

引理 1.14. Let $K$ be a perfectoid field.

1.	The value group $|K^{\ast }|$ is $p$-divisible.
2.	We have $(K^{\circ \circ }{)}^2=K^{\circ \circ }$. Moreover, $K^{\circ \circ }$ is flat.
3.	The ring $K^{\circ }$ is not Noetherian.

For the rest of section, we fix a pseudouniformizer $p\in K^{\circ }$ with $|p|\le |\pi |<1$, so $p\in (\pi )$. If $\mathrm{char}K=p$, then $(K^{\circ }/\pi {)}^{perf}$ is the $\pi$-adic completion of $K^{\circ }$, which is agian $K^{\circ }$ since we assumed $K$ is complete. If $\mathrm{char}K=0$, we still have $(K^{\circ }/p{)}^{perf}\simeq (K^{\circ }/\pi {)}^{perf}$. To summurize, we have the commutative diagram We want to show that $K^{\circ ♭}$ is the valuation ring of a perfectoid field $K^{♭}$. To this end, we first find a pseudouniformizer:

引理 1.15. There exists some $t\in K^{\circ ♭}$ such that $|t^{♯}|=|\pi |$. Moreover, $t$ maps to $0$ in $K^{\circ }/\pi$, and this gives an isomorphism $K^{\circ ♭}/t\simeq K^{\circ }/\pi$.

证明. By $p$-divisibility of the value group, we may choose some $f\in K^{\circ }$ such that $|f{|}^p=|\pi |$, and hence $|f|>|\pi |$. Choose $g\in K^{\circ ♭}$ lifting $f$ under the map $K^{\circ ♭}\to K^{\circ }/\pi$. Then $g^{♯}=f\mathrm{mod}\pi$ from the diagram. By the NA property, $|g^{♯}|=|f|$. Setting $t=g^p$ we are done.

The kernel of the surjection $K^{\circ ♭}\to K^{\circ }/\pi$ is precisely $g\in K^{\circ ♭}$ such that $g^{♯}\in (\pi )=(t^{♯})$, or equivalently $t^{♯}\mid g^{♯}$. But this is equivalent to $t\mid g$. Thus $K^{\circ ♭}/t\simeq K^{\circ }/\pi$.

$\square$

推论 1.16. With $t$ as above, $K^{\circ ♭}$ is $t$-adically complete, and that the $t$-adic topology coincides with the given topology.

证明. $$K^{\circ ♭}\simeq (K^{\circ }/\pi {)}^{perf}\simeq (K^{\circ ♭}/t{)}^{perf}\simeq \widehat{K^{\circ ♭}}.$$

$\square$

命题 1.17. Fix an element $t$ as in the lemma above.

1.	The ring $K^{\circ ♭}$ is a valuation ring, and the ring $K^{♭}:=K^{\circ ♭}[\frac{1}{t}]$ is a perfect field.
2.	The ideal $(t^{\frac{1}{p^{\infty }}})$ is maximal, and the Krull dimension of $K^{\circ ♭}$ is $1$.
3.	The valuation topology on $K^{♭}$ comming from (1) coincides with the one induced by the $t$-adic topology on $K^{\circ ♭}$. In this topology, $K^{♭}$ is a perfectoid field, and $K^{♭,\circ }=K^{\circ ♭}$.
4.	The value groups and residue fields of $K$ and $K^{♭}$ are canonically identified.

证明. We know that $K^{\circ ♭}$ is a valuation ring of rank $\le 1$. As $|t{|}^{♭}=|t^{♯}|=|\pi |$ is not trivial, the rank is exactly $1$. Thus $K^{\circ ♭}$ is microbial with $t$ as as a pseudouniformizer. Thus $K^{♭}=K^{\circ ♭}[\frac{1}{t}]$, which is perfect as $K^{\circ ♭}$ is.

Since $(t^{\frac{1}{p^{\infty }}})$ is radical, it is prime and thus maximal.

$\square$