Manitori

The note records the author's effort to learn Perfectoid geometry. However, due to the limitation of the author's capacity, there may be some mistakes in the note. For simplicity, some details may be omitted. And the readers are assumed to be as familiar as the author. Moreover, the order of the sections is irrelevant. The note was finished when the author was an undergraduate student.Sometimes $\mathbb{N}$ will stand for $\{1,2,3,...,n,...\}$ instead of $\{0,1,2,...,n,...\}$ depending on the context. And whether the categories are big categories or small categories relies on the context. However, the following categories are big categories:The category $\textbf{Sets}$ of sets.The category $\textbf{Groups}$ of groups and homomorphisms.The category $\textbf{Ab}$ of abelian groups and homomorphisms.We will make use of Zermelo-Fraenkel system of axioms for set theory, so that every mathematical object is a set. Note that sometimes we will use "ring map" as a synonym for "ring homomorphism". If $p$ is a prime number, one may use $\mathbb{Z}(p)$ to distinguish the ring of integers module $p$ from the ring of $p$-adic integers $\mathbb{Z}_{p}$. And note that $\mathbb{Z}_{(p)}$ stands for the localization of $\mathbb{Z}$ at the prime ideal $(p)$.

In this section, we will introduce the notion of valuation, which is an important tool in algebraic number theory and algebraic geometry. It provides a measurement of elements of a field or a ring. We will first review some abstract algebra. Some materials can be found in [4], [52], [7], [18], and [35].Recall that an abelian totally ordered group or an ordered abelian group is an abelian group (written multiplicatively) $G$ endowed with a total order, such that $x\leq y$ implies $xz \leq yz$ for all $z\in G$. Since $1<x$ implies $1 < x < x^{2} < x^{3} < \cdot\cdot\cdot< x^{n}< \cdot\cdot\cdot$ (note that $1<x$ would imply $x^{-1}<1$), the abelian ordered group is torsion-free, i.e. no elements in it except the identity have finite order.And recall that a chain of subsets of a set $S$ is a family $(C_{j})_{j\in J}$ of subsets of $S$, such that for any pair $C_{i},C_{j}$ in it, we have $C_{i}\subseteq C_{j}$ or $C_{i}\supseteq C_{j}$. The length of a chain is the number of the inequalities. For example, a singleton is a chain of length zero. And in the sequel, the maximal chain of prime ideals in the valuation ring of the non-archimedean absolute value has length one.Definition 2.1. An order-preserving map $\varphi : (G,\leqq) \rightarrow (H,\leqq)$ of partially ordered sets is a map $G\rightarrow H$ such that for $x \leqq y$ in $G$, we have $\varphi(x)\leqq \varphi(y)$ in $H$. A morphism of ordered abelian groups is an order-preserving homomorphism of abelian groups. An isomorphism of ordered abelian group is an order-preserving isomorphism of abelian groups.Example 2.2. The logarithmic functions with base $a>1$ from $\mathbb{R}_{>0}$ to $\mathbb{R}$ is an isomorphism of ordered abelian groups, i.e. $\log_{a}: (\mathbb{R}_{>0},\cdot,\leq)\xrightarrow{\sim}(\mathbb{R},+,\leq)$ for some $a>1$.Example 2.3. $(\mathbb{R},+,\leq)$, $(\mathbb{R}_{>0},\cdot,\leq)$, and $(\mathbb{Q},+,\leq)$ are ordered abelian groups with the usual order. Note that $(\mathbb{R},+,\leq)\cong (\mathbb{R}_{>0},\cdot,\leq)$.$(\mathbb{R}^{n}_{>0},\cdot)$ with the lexicographic order ($(x_{1},...,x_{n})<(y_{1},...,y_{n})$ if and only if $x_{1}< y_{1}$ or $x_{k}< y_{k}$ when there exists $k\leq n$ such that $x_{i}=y_{i}$ for all $i<k$) is an ordered abelian group.Definition 2.4. Let $(\Gamma,\cdot,\leq)$ be an ordered abelian group. A subgroup $H$ of $\Gamma$ is a convex subgroup or an isolated subgroup if for any $x\in\Gamma,x'\in H$ such that $1\geq x\geq x'$, we have $x\in H$. The spectrum of $\Gamma$ is the set of all convex subgroups of $\Gamma$, denoted by Spec $\Gamma$.Proposition 2.5. The set Spec $\Gamma$ forms a well-ordered set under inclusion. The cardinality of non-trivial convex subgroups in Spec $\Gamma$ is called the height of $\Gamma$ which is denoted by ht$(\Gamma)$.Proof. Let $H,H'$ be two convex subgroups of $\Gamma$. Assume that $H\nsubseteq H',H'\nsubseteq H$, and $1\geq x\geq x'$ for $x\in H,x'\in H'$. Then since $H'$ is a convex subgroup, $x\in H'$, which is a contradiction! So $H'\subseteq H$ or $H\subseteq H'$.Definition 2.6. The convex rank of $\Gamma$, denoted by $$ c.rk(\Gamma)\in\mathbb{N}\cup\{\infty\}, $$ is the supremum over the lengths of the chains $1\subsetneq H_{1}\subsetneq \cdot\cdot\cdot\subsetneq H_{r}:=\Gamma$ of convex subgroups of $\Gamma$.Remark 2.7. Note that the convex rank of $\Gamma$ is equal to the height of $\Gamma$.Example 2.8. If $\textrm{c.rk}(\Gamma)=0$, then the only convex subgroup is the trivial group.If $\textrm{c.rk}(\Gamma)=1$, then the convex subgroups of $\Gamma$ are the trivial group $1$ and $\Gamma$. If $\Gamma$ is a non-trivial subgroup of $\mathbb{R}_{>0}$, then $\textrm{c.rk}(\Gamma)=1$.If $\Gamma=\mathbb{R}_{>0}^{n}$ (with lexicographic order), then $\textrm{c.rk}(\Gamma)=n$.Proposition 2.9. Let $(\Gamma,\cdot,\leq)$ be an ordered abelian group. Then the following are equivalent:${\rm{c.rk}}(\Gamma)=1$;for all $a,b\in\Gamma$ with $b<1,a\leq1$, there exists $n\geq0$ such that $b^{n}\leq a$;there exists an injective morphism of ordered abelian group $(\Gamma,\cdot,\leq)\rightarrow(\mathbb{R}_{>0},\cdot,\leq)$.Remark 2.10. Note that condition (2) is the multiplicative version of Archimedean property.Definition 2.11. The rational rank of $\Gamma$, denoted by $$rk(\Gamma)\in\mathbb{N}\cup\{\infty\},$$ is the dimension of the $\mathbb{Q}$-vector space $\Gamma\otimes_{\mathbb{Z}}\mathbb{Q}$.Remark 2.12. We define the $\mathbb{Q}$-vector space structure on $\Gamma\otimes_{\mathbb{Z}}\mathbb{Q}$ by $\lambda(a\otimes b)=a\otimes\lambda b$ for $a\in\Gamma;b,\lambda\in\mathbb{Q}$.Lemma 2.13. Let $K$ be a field and let $\left|\cdot\right|$ be an absolute value on $K$. Then the topology induced from $\left|\cdot\right|$ makes $K$ a topological field.Proof. Let $\mathbb{B}_{1}=\{x\mid\left|x-a\right|<r_{1}\}$ and $\mathbb{B}_{2}=\{y\mid\left|y-b\right|<r_{2}\}$ be two open balls in $K$. We first check the continuity of the addition. Since $\left|(x-y)-(a-b)\right|=\left|(x-a)+(-y+b)\right|$, we have $$\left|(x-y)-(a-b)\right|=\left|(x-a)+(-y+b)\right|\leq\left|x-a\right|+\left|y-b\right|<r_{1}+r_{2},$$ which implies that $$ \mathbb{B}_{1}-\mathbb{B}_{2}=\{x-y\mid\left|(x-y)-(a-b)\right|<r_{1}+r_{2}\}. $$Next, for multiplication, dividing both sides by $\left|a\right|$ or $\left|b\right|$, we have $\displaystyle\left|\frac{x}{a}\right|<\frac{r_{1}}{\left|a\right|}+1$ and $\displaystyle\left|\frac{y}{b}\right|<\frac{r_{2}}{\left|b\right|}+1$, which implies that $\begin{align*} \left|\frac{x}{a}\cdot\frac{y}{b}\right|=\frac{\left|xy\right|}{\left|ab\right|}&<(\frac{r_{1}}{\left|a\right|}+1)(\frac{r_{2}}{\left|b\right|}+1) \\ \left|xy\right|&<\left|ab\right|(\frac{r_{1}}{\left|a\right|}+1)(\frac{r_{2}}{\left|b\right|}+1) \\ &=r_{1}r_{2}+\left|b\right|r_{1}+\left|a\right|r_{2}+\left|ab\right|. \end{align*}$So the multiplication of two open balls is the open ball $$ \mathbb{B}_{1}\cdot\mathbb{B}_{2}=\{xy\mid\left|xy-ab\right|<r_{1}r_{2}+\left|b\right|r_{1}+\left|a\right|r_{2}\}.$$Finally, consider the inverse map. We have $\left|\left|y\right|-\left|b\right|\right|<r_{2}$, which implies that $-\left|y\right|+\left|b\right|<r_{2}$ or $\left|y\right|-\left|b\right|<r_{2}$. So we have $$ \frac{1}{r_{2}+\left|b\right|}<\frac{1}{\left|y\right|}<\frac{1}{\left|b\right|-r_{2}},$$ which shows that the inverse of an open ball is the open set $$ \mathbb{B}_{2}^{-1}=\{y^{-1}\mid\frac{1}{r_{2}+\left|b\right|}<\left|y^{-1}\right|<\frac{1}{\left|b\right|-r_{2}} \}. $$Consequently, for every neighborhood $U=\{z\mid \left|z-(a-b)\right|<r\}$ of $a-b$, there exist $a\in\mathbb{B}_{1}',b\in\mathbb{B}_{2}'$ such that $\mathbb{B}_{1}'-\mathbb{B}_{2}'\subset U$. And for every neighborhood $V=\{z\mid\left|z-ab\right|<r\}$ of $ab$, there exist $a\in\mathbb{B}_{1}',b\in\mathbb{B}_{2}'$ such that $\mathbb{B}_{1}'\mathbb{B}_{2}'\subset V$. For every neighborhood $W=\{z\mid\left|z-b^{-1}\right|<r\}$ of $b^{-1}$, there exists $b\in\mathbb{B}_{2}'$ such that $\mathbb{B}'^{-1}_{2}\subset W$.Definition 2.14. A (non-archimedean) absolute value on a field $K$ is a map $\left| \cdot \right| : K \rightarrow \mathbb{R}_{\geq0}$, such that for all $x,y\in K$ the following conditions are verified: $\left| x \right| = 0 \Leftrightarrow x=0$,$\left| xy \right| = \left| x \right| \left| y \right|$,$\left| x+y \right| \leq \max\{\left| x \right|, \left| y \right|\}$.Then we say that $(K,\left|\cdot\right|)$ is a valued field. Using absolute value, we can define Cauchy sequence in the usual way. And $K$ is complete if every Cauchy sequence converges in $K$.A field that is complete with respect to a non-archimedean absolute value is called a non-archimedean field. Similarly, a field that is complete under an archimedean absolute value is called an archimedean field.Remark 2.15. Note that, in Peter Scholze's thesis, [16], non-archimedean fields need not to be complete. And instead, a non-archimedean field is defined to be a topological field $k$ endowed with a non-trivial valuation of rank 1 inducing the topology.When a field is endowed with a non-archimedean absolute value, something amazing happens! We get some beautiful results that are distinct from the archimedean case. In this section, $K$ will be always equipped with a non-archimedean absolute value. We will show some peculiarities of $K$ when equipped with a non-archimedean absolute vale.Proposition 2.16. Let $a,b\in K$. If $a\neq b$, then $$ \left|a+b\right|=\max\{\left|a\right|,\left|b\right|\}.$$Proof. Assume that $\left| b \right|<\left| a \right|$, then we have $\left| a+b \right|\leq\left| a \right|$ by definition. We assume further $\left| a+b \right|\neq\left| a \right|$, then we have $\left| a+b \right|<\left| a \right|$, which implies $$ \left| a \right|=\left| (a+b)-b \right|\leq \max\{\left| a+b \right|,\left| b \right|\}<\left| a \right|, $$ which is contradictory! So we must have $\left| a+b \right|=\left| a \right|=\max\{\left|a\right|,\left|b\right|\}$.Lemma 2.17. A series $\sum_{v=0}^{\infty}a_{v}$ in $K$ is a Cauchy sequence if and only if the coefficients $a_{v}$ form a zero sequence, i.e. $\lim_{v\to\infty}\left|a_{v}\right|=0$.In particular, when $K$ is complete, the series $\sum_{v=0}^{\infty}a_{v}$ is convergent if and only if $\lim_{v\to\infty}\left|a_{v}\right|=0$.Proof. cf.([7], Lemma 3, p 10).Remark 2.18. Note that a series $\sum_{v=0}^{\infty}a_{v}$ can be viewed as a Cauchy sequence if we consider every partial sum $\sum_{v=0}^{n}a_{v}$ as a term. And $\lim_{v\to\infty}\left|a_{v}\right|=0$ is equivalent to $\lim_{v\to\infty}a_{v}=0$, since $\left|a_{v}-0\right|=\left|\left|a_{v}\right|-0\right|$ in this case.Clearly, the non-archimedean absolute value on $K$ induces a distance on $K$ by $d(a,b)=\left|a-b\right|$, so there is an associated topology on $K$. Consider the distance between points in $K$. Let $x,y,z\in K$, since $\left|y-z\right|=\left|(x-y)+(z-x)\right|$, the non-archimedean inequality implies $$d(y,z)\leq\max\{d(x,y),d(x,z)\}.$$By Proposition 2.16, if $d(x,y)\neq d(x,z)$, then $d(y,z)=\max\{d(x,y),d(x,z)\}$. In other words, consider a triangle in $K$, if two sides are not equal in length, then the longer one has the same length as the third side, which implies that any triangle in $K$ is isosceles.Next, we consider the disks in $K$ as an example.For a centre $a\in K$ and a radius $r\in \mathbb{R}_{> 0}$, we define the disk without boundary or open disk to be the set $$D^{-}(a,r) = \{ x \in K\mid d(x,a)<r \}. $$Similarly, we define the disk with boundary or closed disk to be the set $$ D^{+}(a,r) = \{ x \in K\mid d(x,a)\leq r\}. $$In addition, we can define the boundary or periphery of the disk $$ \partial D(a,r) = \{ x \in K\mid d(x,a)= r\}.$$Without any loss of generality, we consider only the open disk. Let $b\in D^{-}(a,r)$. Since we have $\left|x-a\right|<r$, then $$ \left|x-b\right|=\left|(x-a)+(a-b)\right|\leq\max\{\left|x-a\right|,\left|a-b\right|\}<r. $$This indicates that every point in a disk is a center. Moreover, if the intersection of two disks is not empty, then the two disks are concentric.The following proposition shows another peculiarity related to the topology of $K$.Proposition 2.19. The topology of $K$ is totally disconnected, i.e. the maximal connected components in $K$ are singletons.Proof. cf. [7], Proposition 4, p11.Next, we introduce the so-called valuations, which were first published by Krull in 1932. They generalize non-archimedean absolute values and have great applications in algebraic geometry. Their values are not constrained in real numbers and they can have more general values than non-archimedean absolute values.There are two kinds of valuations, each of which is in different notations. We define valuations in additive notation initially.Definition 2.20. Let $\Gamma$ be an abelian totally ordered group, we extend $\Gamma$ to $\Gamma \cup \{\infty\}$, such that $\alpha < \infty$  and $\alpha + \infty = \infty + \alpha = \infty$ for all $\alpha \in \Gamma$. A valuation in additive notation (or an additive valuation) on a field $K$ is a map $\upsilon: K \rightarrow \Gamma \cup \{\infty\}$ such that for all $x,y\in K$ the following hold:$\upsilon(x) = \infty \Leftrightarrow x = 0$$\upsilon(xy) = \upsilon(x) + \upsilon(y)$$\upsilon(x+y) \geq \min\{\upsilon(x),\upsilon(y)\}$Also we can define valuations in multiplicative notation.Definition 2.21. Let $\Gamma$ be an abelian totally ordered group, we extend $\Gamma$ to $\Gamma \cup \{0\}$, such that $0 < \alpha$  and $\alpha \cdot 0 = 0 \cdot \alpha = 0$ for all $\alpha \in \Gamma$. A valuation in multiplicative notation (or a multiplicative valuation) on a field $K$ is a map $\upsilon: K \rightarrow \Gamma \cup \{0\}$ such that for all $x,y\in K$ the following hold:$\upsilon(x) = 0 \Leftrightarrow x = 0$$\upsilon(xy) = \upsilon(x)\upsilon(y)$$\upsilon(x+y) \leq \max\{\upsilon(x),\upsilon(y)\}$It is clear that non-archimedean absolute value is an example of multiplicative valuation when taking $(\Gamma, \cdot,\leq)$ as $(\mathbb{R}_{>0}, \cdot,\leq)$. In fact, we can show that, in this case, non-archimedean absolute value and additive valuation are equivalent. If we write $\upsilon_{0}$ for an additive valuation and $\upsilon_{1}$ for an absolute value. Then just let $\upsilon_{0}(x) = - \ln\upsilon_{1}(x)$ for all $x \in \textit{K}$, so we can pass from additive valuation back to multiplicative valuation by setting $\upsilon_{1}(x) = e^{-\upsilon_{0}(x)}$. This sets up a one-to-one correspondence or an isomorphism. In terms of this correspondence, we will make no difference between non-archimedean absolute value and real additive valuation.In the sequel, we will use multiplicative valuation, and simply call it valuation. A real valuation will mean a valuation with $\Gamma\subseteq \mathbb{R}_{>0}$.A valuation is trivial if $\upsilon(x) = 1$ for all $x \neq 0$.Definition 2.22. Let $v: K\rightarrow \Gamma\cup\{0\}$ be a valuation. The value group of $v$ is $\Gamma_{v}:=\{v(x)\in \Gamma\mid x\in K^{\times}\}$.Definition 2.23. Let $v: K\rightarrow\Gamma\cup\{0\}$ be a valuation. The rank of $v$ is the convex rank or height of the value group $\Gamma_{v}$.Remark 2.24. A non-trivial valuation $\upsilon$ on $K$ has rank 1 if $\Gamma \subset \mathbb{R}_{>0}$ as abelian totally ordered group, and it has higher rank if it is not of rank 1.Remark 2.25. $K^{\times}$ is the multiplicative group of any field $K$. It can be observed that the valuation $\upsilon$ of rank 1 on $K$ is the same as the (non-archimedean) absolute value on $K$. Consequently, we can make no difference between rank-1 valuation and absolute value.Definition 2.26. A field with a valuation is called a valuation field or a valued field.Remark 2.27. The valued field in Definition 2.14 coincides with this.Definition 2.28 ([4]). The valuation ring of a valuation $v$ on a field $K$ is $$ \mathscr{O}_{K}:= \{x\in K\mid v(x)\leq 1 \}.$$Remark 2.29 ([4]). The valuation ring $\mathscr{O}_{K}$ is a subring of the field $K$. It has a unique maximal ideal $\mathfrak{m}: = \{ x\in K\mid v(x)<1\}$ and a group of units $\{x\in K\mid v(x) = 1\}$. The field $K$ is the fraction field of $\mathscr{O}_{K}$, that is we have $\textrm{Frac}(\mathscr{O}_{K}) \cong K$, so the valuation on $K$ is determined by its valuation ring.Proposition 2.30. Let $R$ be a subring of the field $K$. The following conditions are equivalent :$R$ is the valuation ring of some valuation on $K$.$Q(R)$ = $K$ and ideals of $R$ form a chain.For all $x\in K\backslash\{0\}$, $x\in R$ or $x^{-1}\in R$.Proof. cf.([4], Chapter VI, Proposition 6.2).Proposition 2.31. Let $\left|\cdot\right|_{v}$ and $\left|\cdot\right|_{w}$ be two absolute values on a field $K$. The following conditions are equivalent:$\left|\cdot\right|_{v}$ and $\left|\cdot\right|_{w}$ induce the same topology on $K$;$\left|x\right|_{v}<1$ if and only if $\left|x\right|_{w}<1$ for all $x\in K$;there exist $c>0$ such that $\left|x\right|_{w} = (\left|x\right|_{v})^{c}$ for all $x\in K$.Proof. cf.([4], Proposition 3.1, p240).Definition 2.32. Two absolute values on a field are equivalent if they satisfy one of the conditions in Proposition 2.31.Definition 2.33. Two valuations $v,w : K\rightarrow \Gamma\cup\{0\}$ are equivalent if there exists an order-preserving isomorphism $\alpha$ between $\Gamma_{v}$ and $\Gamma_{w}$ such that $w(x)=\alpha(v(x))$ for all $x\in K^{\times}$.Remark 2.34. For each $c>0$, there exists an automorphism $x\mapsto x^{c}$ of $\mathbb{R}_{>0}$ which is order-preserving. So non-archimedean absolute values are equivalent as absolute values if and only if they are equivalent as valuations.Definition 2.35 ([7], p154). The length of the maximal chain of prime ideals in a valuation ring $R$ is called the height or Krull dimension of $R$.Example 2.36. Consider the non-archimedean absolute value $\left|\cdot\right|$ on a field $K$, the valuation ring $R$ of $\left|\cdot\right|$ is $\{x\in K\mid\left|x\right| \leq1\}$. The prime ideals in $R$ are the maximal ideal and the zero ideal. So the height of $R$ is 1.Remark 2.37. A valuation ring $R$ has height 1 if and only if $R$ has convex rank 1.

Almost-mathematics is a tool studying $p$-adic Hodge theory. In this section, we use the book of Gabber and Ramero, [20], as our main reference like Peter Scholze in [16].Unless otherwise stated, all rings are commutative with identity. We fixed a base ring $V$ that contains an ideal $\mathfrak{m}$ satisfying $\mathfrak{m}^{2} = \mathfrak{m}$. And we also assume that $\widetilde{\mathfrak{m}} := \mathfrak{m}\otimes_{V}\mathfrak{m}$ is a flat $V$-module. For the definition of tensor product, see §11.Question 3.1. Let $(V,\left| \cdot \right|)$ be a non-discrete valuation ring of rank 1, and $\mathfrak{m}$ is an ideal of $V$. If we have $\mathfrak{m}^{2} = \mathfrak{m}$, can we get that $\mathfrak{m}$ is the maximal ideal of $V$?The basic idea of almost-mathematics is that one neglects $\mathfrak{m}$-torsion everywhere.Definition 3.2. Let $M$ be a $V$-module. We say that $M$ is almost zero if $\mathfrak{m}M$ = 0. A map of $V$-modules $\phi: M\rightarrow N$ is an almost isomorphism, if Ker $\phi$ and Coker $\phi$ are both almost zero $V$-modules. Then we say that $M$ is almost isomorphic to $N$.Remark 3.3. A map of $V$-modules $\phi: M\rightarrow N$ is almost injective if $\textrm{Ker }\phi$ is almost zero. And $\phi$ is almost surjective if $\textrm{Coker }\phi$ is almost zero.Proposition 3.4. A $V$-module $M$ is almost zero if and only if $\mathfrak{m}\otimes_{V} M = 0$.Proof. Just consider the natural isomorphism $V\otimes_{V}M\cong M$.Proposition 3.5. A map $\varphi:M\rightarrow N$ of $V$-modules is an almost isomorphism if and only if the induced map $\widetilde{\mathfrak{m}}\otimes_{V}M\rightarrow \widetilde{\mathfrak{m}}\otimes_{V}N$ is an isomorphism.Proof. First we assume that $\varphi:M\rightarrow N$ is almost isomorphic. Consider the exact sequence $$ \textrm{Ker }\varphi\rightarrow M\rightarrow N\rightarrow \textrm{Coker }\varphi; $$since $\widetilde{\mathfrak{m}}$ is flat, we have the induced exact sequence $$ \widetilde{\mathfrak{m}}\otimes_{V}\textrm{Ker }\varphi\rightarrow \widetilde{\mathfrak{m}}\otimes_{V}M\rightarrow\widetilde{\mathfrak{m}}\otimes_{V}N\rightarrow\widetilde{\mathfrak{m}}\otimes_{V}\textrm{Coker }\varphi. $$Since Ker $\phi$ and Coker $\phi$ are both almost zero $V$-modules, we have the following exact sequence $$ 0\rightarrow \widetilde{\mathfrak{m}}\otimes_{V}M\rightarrow\widetilde{\mathfrak{m}}\otimes_{V}N\rightarrow0, $$which implies that $\widetilde{\mathfrak{m}}\otimes_{V}M\rightarrow\widetilde{\mathfrak{m}}\otimes_{V}N$ is an isomorphism.Conversely, assume that $\widetilde{\mathfrak{m}}\otimes_{V}M\rightarrow\widetilde{\mathfrak{m}}\otimes_{V}N$ is an isomorphism. We consider the exact sequence again $$ \textrm{Ker }\varphi\rightarrow M\rightarrow N\rightarrow \textrm{Coker }\varphi. $$Then similarly, we have the induced exact sequence $$ \widetilde{\mathfrak{m}}\otimes_{V}\textrm{Ker }\varphi\rightarrow \widetilde{\mathfrak{m}}\otimes_{V}M\rightarrow\widetilde{\mathfrak{m}}\otimes_{V}N\rightarrow\widetilde{\mathfrak{m}}\otimes_{V}\textrm{Coker }\varphi.$$Since $\widetilde{\mathfrak{m}}\otimes_{V}M\rightarrow\widetilde{\mathfrak{m}}\otimes_{V}N$ is an isomorphism, we conclude that $\widetilde{\mathfrak{m}}\otimes_{V}\textrm{Ker }\varphi=\widetilde{\mathfrak{m}}\otimes_{V}\textrm{Coker }\varphi=0$. So we get $\mathfrak{m}\otimes_{V}\textrm{Ker }\varphi=\mathfrak{m}\otimes_{V}\textrm{CoKer }\varphi=0$ as desired.Lemma 3.6. If $\mathfrak{m}$ is flat, then we have an isomorphism $\widetilde{\mathfrak{m}}\cong\mathfrak{m}$ of $V$-modules.Example 3.7. If $V$ is a non-discrete valuation ring of rank one, then its maximal ideal $\mathfrak{m}$ is flat as a $V$-module by ([11], Tag0539). Therefore, we have $\widetilde{\mathfrak{m}}\cong\mathfrak{m}$.Proposition 3.8. The set of all almost isomorphisms forms a multiplicative system.Proof. First, any identity of $V$-module is almost isomorphic. And the composition of almost isomorphic maps is almost isomorphic. Given a map of $V$-modules $g:X\rightarrow Y$, an almost isomorphic map of $V$-modules $t: X\rightarrow Z$, and a $V$-module $W$. Then we have an induced solid diagramwhich is made to be commutative with $\widetilde{\mathfrak{m}}\otimes s:\widetilde{\mathfrak{m}}\otimes_{V}Y\cong \widetilde{\mathfrak{m}}\otimes_{V}W$ and $\widetilde{\mathfrak{m}}\otimes f:\widetilde{\mathfrak{m}}\otimes_{V}Z\rightarrow\widetilde{\mathfrak{m}}\otimes_{V}W$. So the solid diagramcan be made to be commutative with a map of $V$-modules $f: Z\rightarrow W$ and an almost isomorphic map of $V$-modules $s: Y\rightarrow W$.Given a pair of maps of $V$-modules $f,g: X\rightarrow Y$ and an almost isomorphic map of $V$-modules $t: Z\rightarrow X$ such that $f\circ t=g\circ t$. We have the induced maps $f',g':\widetilde{\mathfrak{m}}\otimes_{V}X\rightarrow\widetilde{\mathfrak{m}}\otimes_{V}Y, t':\widetilde{\mathfrak{m}}\otimes_{V}Z\cong\widetilde{\mathfrak{m}}\otimes_{V}X$ such that $f'\circ t'=g'\circ t'$. Then there exists an almost isomorphic map of $V$-modules $s: Y\rightarrow I$ such that $s':\widetilde{\mathfrak{m}}\otimes_{V}Y\cong\widetilde{\mathfrak{m}}\otimes_{V}I$ satisfies $s'\circ f'=s'\circ g'$ by virtue of $f'\circ t'=g'\circ t'$. Consequently, we have $s\circ f=s\circ g$.We have just proved the left multiplicative case. However, the right multiplicative case is similar.

In this section, we will introduce the so-called complex analytic spaces using ([46]) as our reference.Definition 4.1. Let $K$ be a ring. A $K$-locally ringed space is a locally ringed space $(X,\mathscr{O}_{X})$ equipped with a $K$-algebra structure $p$ in $\mathscr{O}_{X}$. A $K$-algebra structure $p$ in $\mathscr{O}_{X}$ is a collection of ring homomorphisms $p_{U}:K\rightarrow\mathscr{O}_{X}(U)$, one for each open $U\subset X$, such that for every $U_{1}\supset U_{2}$ the following diagram commutes:If $z\in\mathscr{O}_{X}(U)$ and $a\in K$, we usually write $az$ instead of $p_{U}(a)z$.Let $(X',\mathscr{O}_{X'})$ be another $K$-locally ringed space. A morphism of $K$-locally ringed space or simply a $K$-morphism from $(X,\mathscr{O}_{X})$ to $(X',\mathscr{O}_{X'})$ is a morphism $(f,f^{*})$ of locally ringed spaces such that $$ f^{*}_{U'}\circ p'_{U'}=p_{f^{-1}(U')} $$ for every open set $U'\subset X'$.Example 4.2. The complex number space $$ \mathbb{C}^{n}=\{(\xi_{1},...,\xi_{n})\mid\xi_{i}\in\mathbb{C}\textrm{ for }1\leq i\leq n\}$$ will be viewed as a $\mathbb{C}$-locally ringed space with the natural structure as follows:The underlying topological space is $\mathbb{C}^{n}$ equipped with the standard topology. And the structure sheaf $\mathscr{O}_{\mathbb{C}^{n}}$ is the sheaf of homological functions. Namely, for each open set $U\subset\mathbb{C}^{n}$, $\mathscr{O}_{\mathbb{C}^{n}}(U)$ is the $\mathbb{C}$-algebra of all holomorphic functions on $U$, where the elements of $\mathbb{C}$ are identified with constant functions. Moreover, any open subset $V$ of $\mathbb{C}^{n}$ will be also viewed as a $\mathbb{C}$-locally ringed space with the structure sheaf $\mathscr{O}_{V}$ obtained as the restriction of $\mathscr{O}_{\mathbb{C}^{n}}$ to $V$.Example 4.3. Let $V$ be an open subset in $\mathbb{C}^{n}$ and let $(h_{1},...,h_{m})$ be a finite system of holomorphic functions in $V$. Let $$S=\{\xi\in V\mid h_{i}(\xi)=0,1\leq i\leq m\},$$ with the induced topology from $\mathbb{C}^{n}$. The quotient sheaf of $\mathscr{O}_{V}$ by the ideal generated by $(h_{1},...,h_{m})$ is concentrated on $S$. We will denote by $\mathscr{O}_{S}$ the restriction of this sheaf to $S$. Then we obtain a $\mathbb{C}$-locally ringed space $(S,\mathscr{O}_{S})$ with the $\mathbb{C}$-algebra structure $\rho$ which is the identification of constants with constant functions.Definition 4.4. A complex analytic space or simply a $\mathbb{C}$-space is a locally ringed space $(X,\mathscr{O}_{X})$ such that every point in $X$ admits an open neighborhood $U$ such that $(U,\mathscr{O}_{X|U})$ is $\mathbb{C}$-isomorphic to some $(S,\mathscr{O}_{S})$ described in Example 4.3.

Lemma 5.1. Let $X={\rm{Sp}}\ A$ be an affinoid $K$-space and $f_{0}, ..., f_{r}\in A$. Then $f_{0}, ..., f_{r}$ have no common zeros on $X$ if and only if $f_{0}, ..., f_{r}$ generate $A$ as the unit ideal.Proof. If $f_{0}, ..., f_{r}$ generate $A$ as the unit ideal, assume that $f_{0}, ..., f_{r}$ have some common zero $x_{0}$ on $X$. Then for any $f\in A\backslash x_{0}$,$$f=\sum_{i=0}^{r}a_{i}f_{i},\ a_{i}\in A,$$but then $f(x_{0})=0$ implies that $f\in x_{0}$, which is clearly a contradiction! So $f_{0}, ..., f_{r}$ have no common zeros on $X$.Conversely, if $f_{0}, ..., f_{r}$ have no common zeros on $X$, assume that $f_{0}, ..., f_{r}$ generate an ideal $I\varsubsetneq A$. Since there exists some maximal ideal $x\in X$ such that $I\subseteq x$, we have$$f_{0}(x)=\cdot\cdot\cdot=f_{r}(x)=0,$$which shows that $f_{0}, ..., f_{r}$ have a common zero. So by contradiction, we conclude that $f_{0}, ..., f_{r}$ generate the unit ideal in $A$.$\Box$Definition 5.2. Let $X=\textrm{Sp }A$ be an affinoid $K$-space.(1)  A subset of $X$ of type $$X(f_{1}, ..., f_{r}):=\{x\in X\mid \left|f_{i}(x)\right|\leq1\}$$ for $f_{1},...,f_{r}\in A,1\leq i\leq r$ is called a Weierstrass domain in $X$.(2)  A subset of $X$ of type $$ X(f_{1},...,f_{r},g_{1}^{-1},...,g_{s}^{-1}):=\{x\in X\mid\left|f_{i}(x)\right|\leq1, \left|g_{j}(x)\right|\geq1 \} $$ for $f_{1},...,f_{r},g_{1},...,g_{s}\in A,1\leq i\leq r,1\leq j\leq s$ is called a Laurent domain in $X$.(3)  A subset of $X$ of type $$ X(\frac{f_{1}}{f_{0}},...,\frac{f_{r}}{f_{0}}):=\{x\in X\mid\left|f_{i}(x)\right|\leq\left|f_{0}(x)\right|\} $$ for $f_{0},...,f_{r}\in A,1\leq i\leq r$ without common zeros is called a rational domain in $X$.Remark 5.3. The domains in Definition 5.2 are special cases of affinoid subdomains.

We will introduce the notion of Berkovich's analytic spaces, following \cite{Berkovich1}. Berkovich's analytic space is one of the nonarchimedean analogues of the complex analytic space.§6.1. Seminorms.Let $(K,\left|\cdot\right|)$ be a valued field. Recall that a \textit{seminorm} on a $K$-vector space $V$ is a map $\|\cdot\|:V\rightarrow\mathbb{R}_{\geq0}$ such that $\|0\|=0$, $\|\lambda x\|=\left|\lambda\right|\|x\|$, and $\|x+y\|\leq\|x\|+\|y\|$ for $x,y\in V$ and $\lambda\in K$.We first generalize this notion to the case of abelian groups. Let $A$ be an abelian group.Definition 6.1. A seminorm on $A$ is a map $\|\cdot\| : A\rightarrow \mathbb{R}_{\geq0}$ such that $\|0\| = 0$, $\|-f\|=\|f\|$ , and $\|f+g\|\leq \|f\| + \|g\|$ for all $f,g\in A$. A seminorm is non-archimedean if it satisfies $\|f+g\|\leq \max\{\|f\| , \|g\|\}$.Remark 6.2. The conditions $\|-f\|=\|f\|$ , and $\|f+g\|\leq \|f\| + \|g\|$ can be equivalently written as $\|f-g\|\leq \|f\|+\|g\|$ for all $f,g\in A$.Each seminorm $\|\cdot\|$ induces a topology on $A$. Since we have a notion of open balls, a subset $U\subset A$ is open if for every point in $U$, there exists some open ball that contained in $U$.This topology is Hausdorff if and only if $\|\cdot\|$ is a norm, i.e., $\|f\| = 0 \Leftrightarrow f=0$. In fact, if $\|\cdot\|$ is not a norm, then there exists some $a\in A$ such that $\|a\| = \|a-0\| = 0$, which implies that the points $0\neq a$ do not have disjoint neighborhood.In terms of the seminorm, we can construct completion of $A$ in the usual sense. A sequence $(x_{n})_{n\in\mathbb N}$ in $A$ is a \textit{Cauchy sequence} if for every $\varepsilon>0$, $\|x_{n}-x_{m}\|<\varepsilon$ when $n,m$ sufficiently large. We define the addition of two Cauchy sequences to be termwise addition. For any two Cauchy sequences $x=(x_{n})_{n\in\mathbb N}$ and $y=(y_{n})_{n\in\mathbb N}$ in $A$, we define a seminorm as follows$$\|x-y\|_{0}:=\lim_{n\to\infty}\|x_{n}-y_{n}\|.$$Then we say that two Cauchy sequences $x=(x_{n})_{n\in\mathbb N}$ and $y=(y_{n})_{n\in\mathbb N}$ in $A$ are \textit{equivalent} if $\|x-y\|_{0}=0$. We denote the set of equivalence classes of Cauchy sequences by $\widehat{A}$. And we define the seminorm of an equivalence class $x$ represented by $(x_{n})_{n\in\mathbb N}$ of Cauchy sequences by$$ \|x\|_{1}:=\lim_{n\to\infty}\|x_{n}\|. $$The limit exists since $(\|x_{n}\|)_{n\in\mathbb N}$ forms a Cauchy sequence in $\mathbb{R}$. Let $x,y\in\widehat{A}$ which are represented by $x=(x_{n})_{n\in\mathbb N}$ and $y=(y_{n})_{n\in\mathbb N}$ respectively. We define the \textit{addition} of $x$ and $y$ to be the equivalence class $x+y$ represented by $(x_{n}+y_{n})_{n\in\mathbb N}$. So $\widehat{A}$ forms a group under the addition defined before.Next, we show that $\widehat{A}$ is complete. Let $(x_{n})_{n\in\mathbb N}$ be a Cauchy sequence in $\widehat{A}$. Assume that $(x_{n})_{n\in\mathbb N}$ has no limit in $\widehat{A}$. Then for every $a\in\widehat{A}$ and every integer $N>0$, there exists $\varepsilon>0$ such that $\|x_{v}-a\|_{2}\geq\varepsilon$ for all $v>N$. Choose representative Cauchy sequences $(x_{v,i})_{i\in\mathbb N}$ and $(a_{i})_{i\in\mathbb N}$, then we have $\lim_{i\to\infty}\|x_{v,i}-a_{i}\|\geq\varepsilon$. Since$$ \|x_{n,i}-a_{i}\|-\|x_{n,i'}-a_{i'}\|\leq\|x_{n,i}-x_{n,i'}+a_{i'}-a_{i}\|\leq\|x_{n,i}-x_{n,i'}\|+\|a_{i}-a_{i'}\|, $$the limit $\lim_{i\to\infty}\|x_{v,i}-a_{i}\|$ exists. This constructs a contradiction, i.e. the limit is not necessarily greater than $\varepsilon$. So there exists some $a\in\widehat{A}$ such that for every $\varepsilon>0$, there exists integer $N>0$ such that $\|x_{v}-a\|_{2}<\varepsilon$ for all $v>N$.Consequently, the group $\widehat{A}$ of equivalence classes of Cauchy sequences in $A$ is complete. We call it the completion of $A$.Definition 6.3. Two seminorms $\|\cdot\|$ and $\|\cdot\|'$ on $A$ are equivalent if there exist $C,C'\in\mathbb{R}_{>0}$ such that $C\|f\|\leq\|f\|'\leq C'\|f\|$ for all $f\in M$.Remark 6.4. Note that $C\|f\|\leq\|f\|'\leq C'\|f\|$ implies $\frac{1}{C'}\|f\|'\leq\|f\|\leq\frac{1}{C}\|f\|'$. And if there is a seminorm $\|\cdot\|''$ on $A$ such that $\|\cdot\|'$ and $\|\cdot\|''$ are equivalent, i.e. $C_{0}\|f\|'\leq\|f\|''\leq C_{0}'\|f\|'$ for some $C_{0},C_{0}'\in\mathbb{R}_{>0}$, then we have $CC_{0}\|f\|\leq\|f\|''\leq C'C_{0}'\|f\|$. These show that we have an equivalence relation.Let $M$ be a seminormed group and $N$ be a subgroup of $M$. Then one can define the residue seminorm on $M/N$ by $\|f\|:=\inf\{\|g\|\mid g\in\pi^{-1}(f)\}$, where $\pi$ denotes the canonical homomorphism $M\rightarrow M/N$.Lemma 6.5. Consider the situation above. The residue seminorm on $M/N$ is a norm if and only if $N$ is closed. If $M$ is a normed group whose norm is multiplicative and $N$ is closed, the residue norm on $M/N$ is multiplicative.Proof. If $g\in f$, then $f-g\in N\Rightarrow g\in f+N$ for $f,g\in M$, we can write $g=f+n$ for some $n\in N$. Then $\|f\|=\inf_{g\in f}\|g\|=\inf_{n\in N}\|f+n\|=\inf_{n\in N}\|f-n\|=d(f,N)$. Since $\|f\|=0\Leftrightarrow d(f,N)=0\Leftrightarrow f\in\overline{N}$, $f=0$ if and only if $\overline{N}=N$.If $f,g\in M/N$, then we have$$ \begin{align*} \|fg\|=\inf\{\|g_{1}\|\mid g_{1}\in\pi^{-1}(fg)\}&\leq \inf\{\|g_{0}'\|\|g_{0}\|\mid g_{0}'g_{0}\in\pi^{-1}(f)\pi^{-1}(g)\} \\ &=\|f\|\|g\| \\ &=\inf\{\|g_{0}'g_{0}\|\mid g_{0}'g_{0}\in\pi^{-1}(fg)\} \\ &\leq \|fg\|, \end{align*} $$which shows that $\|fg\|=\|f\|\|g\|$.$\Box$Definition 6.6. Let $\varphi: (M,\|\cdot\|)\rightarrow (N,\|\cdot\|')$ be a homomorphism of seminormed groups. We say that $\varphi$ is bounded if there exists $C>0$ such that $\|\varphi(f)\|'\leq C\|f\|$ for all $f\in M$. We say that $\varphi$ is admissible if identifying Im$(\varphi)$ with $M/\ker(\varphi)$ via the canonical isomorphism, the residue norm on $M/\ker(\varphi)$ is equivalent to the restriction to Im$(\varphi)$ of the seminorm of $N$.Remark 6.7. Note that a homomorphism of seminormed groups is simply a homomorphism of groups which are seminormed.Next, we consider the case of seminorms on rings. In the sequel, we let $A$ be a ring with unity.Definition 6.8. A seminorm (resp. norm) on $A$ is a seminorm (resp. norm) on the additive group of $A$ such that $\|1\|=1$ and $\|ab\| \leq \|a\|\|b\|$ for all $a,b\in A$. A seminorm is power-multiplicative if $\|f^{n}\|=\|f\|^{n}$ for each integer $n>0$. A seminorm is multiplicative if it satisfies $\|ab\| = \|a\|\|b\|$ for $a,b\in A$.Remark 6.9. Note that Definition 6.8 above is slightly different to that in [24], which says that a seminorm is multiplicative if it satisfies $\|ab\| = \|a\|\|b\|$ and $\|1\|=1$ for all $a,b\in A$.Definition 6.10. A multiplicative norm is called a valuation, and a multiplicative seminorm is called a semivaluation.Remark 6.11. Note that the notion of valuation in Definition 6.10 is different from what we have defined before (compared with the Definition 2.14, 2.20, and 2.21).Definition 6.12. A ring endowed with a seminorm (resp. norm) is called seminormed (resp. normed) ring. A normed ring $\mathscr{A}$ is a Banach ring if $\mathscr{A}$ is complete with respect to its norm.Definition 6.13. Let $\mathscr{A}$ be a normed ring. A seminormed (resp. normed) $\mathscr{A}$-module is an $\mathscr{A}$-module $M$ equipped with a seminorm (resp. norm) $\|\cdot\|$ such that $\|fm\|\leq C\|f\|\|m\|$ for some $C>0$ and for all $f\in\mathscr{A}$, $m\in M$. A Banach $\mathscr{A}$-module is a complete normed $\mathscr{A}$-module.Definition 6.14. Let $(\mathscr{A},\|\cdot\|)$ be a Banach ring. A seminorm $\left|\cdot\right|$ on $\mathscr{A}$ is bounded if there exists $C>0$ such that $\left|f\right|\leq C\|f\|$ for all $f\in\mathscr{A}$.Remark 6.15. In the previous definition, if $\left|\cdot\right|$ is a power-multiplicative bounded seminorm, then we can set $C=1$. In fact, by $\left|f^{n}\right|\leq C\|f^{n}\|$, we have $\left|f\right|\leq \sqrt[n]{C}\|f\|$ for all $n\geq1$, which implies that $\left|f\right|\leq\|f\|$.Let $M,N$ be seminormed $\mathscr{A}$-module. We define the seminorm on $M\otimes_{\mathscr{A}}N$ by $\|f\|=\inf\sum_{i}\|m_{i}\|\cdot\|n_{i}\|$ where the infimum is taken over all representations $f=\sum_{i}m_{i}\otimes n_{i}$. We denote the completion of $M\otimes_{\mathscr{A}} N$ with respect to this seminorm by $M\widehat{\otimes}_{\mathscr{A}} N$. And we call $M\widehat{\otimes}_{\mathscr{A}} N$ the complete tensor product.§6.2. The Spectrum.Let $\mathscr{A}$ be a commutative Banach ring with identity. The spectrum of $\mathscr{A}$ denoted by $\mathscr{M}(\mathscr{A})$ is the set of all bounded multiplicative seminorms on $\mathscr{A}$ provided with the weakest topology $\tau$ such that all real-valued functions on $\mathscr{M}(\mathscr{A})$ of the form $\varphi_{f}:\mathscr{M}(\mathscr{A})\rightarrow\mathbb{R}_{\geq0},\left|\cdot\right|\mapsto\left|f\right|$, $f\in\mathscr{A}$, are continuous, i.e. $\tau=\{\varphi^{-1}_{f}(U)\mid U\subset\mathbb{R}_{\geq0}\textrm{ open},f\in\mathscr{A}\}$.Theorem 6.16. The spectrum $\mathscr{M}(\mathscr{A})$ is a non-empty and compact Hausdorff space.Proof. cf. [25, Theorem 1.2.1, p13]$\Box$§6.3. Analytic Spaces over a Commutative Banach Ring.Definition 6.17. Let $k$ be a commutative Banach ring. The $n$-dimensional affine space $A^{n}=A^{n}_{k}$ over $k$ is the set of all multiplicative seminorms on $k[T_{1}, T_{2}, ..., T_{n}]$ whose restriction to $k$ is bounded. The set is provided with the weakest topology such that all real-valued functions on $A^{n}$ of the form $\left|\cdot\right|\mapsto\left|f\right|$, $f\in k[T_{1}, T_{2}, ..., T_{n}]$, are continuous.§6.4. Affinoid Algebras.We fix a non-archimedean field $k$ with a valuation (multiplicative norm) $\left|\cdot\right|$ which may be trivial. We set $\sqrt{\left|k^{*}\right|}:=\{\alpha\in\mathbb{R}_{\geq0}\mid\alpha^{n}\in\left|k^{*}\right| \textrm{for some }n\geq1\}$.For $r_{1},r_{2},...,r_{n}>0$, we set:$$ k\{r_{1}^{-1}T_{1}, ..., r_{n}^{-1}T_{n}\}\overset{\textrm{def}}=\{f=\sum_{v=0}^{\infty}a_{v}T^{v}\mid a_{v}\in k\textrm{ and }\left|a_{v}\right|r^{v}\to0\textrm{ as }\left|v\right|\to\infty\} $$here $v=(v_{1},v_{2},...,v_{n})$, $\left|v\right|=v_{1}+v_{2}+\cdot\cdot\cdot+v_{n}$, $T^{v}=T^{v_{1}}_{1}T^{v_{2}}_{2}\cdot\cdot\cdot T^{v_{n}}_{n}$ and $r^{v}=r^{v_{1}}_{1}r^{v_{2}}_{2}\cdot\cdot\cdot r^{v_{n}}_{n}$. This is a commutative Banach $k$-algebra with respect to the multiplicative norm $\|f\|=\max_{v}\left|a_{v}\right|r^{v}$. And this algebra will be simply denoted by $k\{r^{-1}T\}$.The construction above can be extended to $\mathscr{A}\{r^{-1}T\}$ for any commutative Banach $k$-algebra $\mathscr{A}$.Example 6.18. Suppose that the valuation on $k$ is trivial. If $r_{i}\geq1$ for all $1\leq i\leq n$, then $k\{r^{-1}T\}$ becomes the ring of polynomials $k[T_{1},T_{2},...,T_{n}]$. If $r_{i}<1$ for all $1\leq i\leq n$, then $k\{r^{-1}T\}$ becomes the ring of formal power series $k[[T_{1},T_{2},...,T_{n}]]$.Definition 6.19. A commutative Banach $k$-algebra $\mathscr{A}$ is said to be $k$-affinoid if there exists an admissible epimorphism $k\{r^{-1}T\}\rightarrow\mathscr{A}$. If such an epimorphism can be found with $r=1$, $\mathscr{A}$ is said to be strictly $k$-affinoid. An affinoid $k$-algebra is a $K$-affinoid algebra for some non-archimedean field $K$ over $k$.Definition 6.20. A $k$-affinoid variety is a pair $\textrm{Sp }A=(\textrm{Max }A, A)$, where $A$ is a $k$-affinoid algebra and Max $A$ is the set of all maximal ideals of $A$. A morphism $\varphi: \textrm{Sp }A\rightarrow \textrm{Sp }B$ of affinoid varieties (also called a $k$-affinoid morphism or a $k$-affinoid map) is a pair $(^{a}\sigma,\sigma)$, where $\sigma: B\rightarrow A$ is a $k$-algebra homomorphism and $^{a}\sigma: \textrm{Max }A\rightarrow \textrm{Max }B$ is a map induced from $\sigma$.Remark 6.21. We will use Sp $A$ in the sense of Max $A$ in the sequel. For any affinoid variety Sp $A$, the elements in the affinoid algebra $A$ are called affinoid functions on Sp $A$.Definition 6.22. Let $\mathscr{A}$ be $k$-affinoid algebra. A Banach $\mathscr{A}$-module $M$ is finite if there exists an admissible epimorphism $\mathscr{A}^{n}\rightarrow M$. A commutative Banach $\mathscr{A}$-algebra $\mathscr{B}$ is finite if $\mathscr{B}$ is finite as a Banach $\mathscr{A}$-module.Proposition 6.23. If the valuation on $k$ is non-trivial, then for a strictly $k$-affinoid algebra $\mathscr{A}$ the canonical map ${\rm{Max}}(\mathscr{A})\rightarrow\mathscr{M}(\mathscr{A})$ induces a homeomorphism of ${\rm{Max}}(\mathscr{A})$ with an everywhere dense subset of $\mathscr{M}(\mathscr{A})$.Proof. cf. [25, Proposition 2.1.15, p26].$\Box$Remark 6.24. Everywhere dense is the same as dense.

A rigid-analytic space is an analogue of a complex analytic space over a non-archimedean field, which first made its appearance in John Tate's thesis \cite{Tate}. In this section, we will introduce the notion of rigid space, using \cite{Tate} as our reference. All rings are commutative with identity and $K$ is the fixed ground field, which is complete with respect to a non-trivial rank-1 valuation denoted by $\left|\cdot\right|$. The valuation ring of $K$ is denoted by $V$ and its maximal ideal is denoted by $\mathfrak{m}$. And $x$ is a fixed non-zero element in $\mathfrak{m}$, i.e. $0<\left|x\right|<1$.§7.1. Vector Spaces.Let $E$ be a vector space over $K$. Then $E$ is also a $V$-module if we restrict the scalar multiplication to $V\times E$. We let $E_{0}$ be a $V$-submodule of $E$ such that $E=KE_{0}$.Set $E_{n}:=x^{n}E_{0}$ for $n\in \mathbb{Z}$. We obtain a filtration$$ E\supset \cdot\cdot\cdot\supset E_{-n}\supset E_{-n+1}\supset\cdot\cdot\cdot\supset E_{0}\supset\cdot\cdot\cdot\supset E_{n}\supset E_{n+1}\supset \cdot\cdot\cdot\supset0 $$of $E$ (note that we have $x^{-\infty}E_{0}=E$ and $x^{\infty}E_{0}=0$). It is easy to see that using such filtration, we can form a fundamental system of neighborhoods of 0 in $E$.So one can define a topology on $E$ by saying a subset $U\subset E$ is open if for every $y\in U$, $y+x^{n}E_{0}\subset U$ for some $n\in\mathbb{Z}$. Such topology is said to be defined by $E_{0}$.Lemma 7.1. Let $F$ be another $K$-vector space whose topology is defined by $F_{0}$. A $K$-linear map $\varphi:E\rightarrow F$ is continuous if and only if there exists some integer $n$ such that $\varphi(E_{0})\subset x^{-n}F_{0}$.Proof. Let $p\in E$ be any point. If $\varphi$ is continuous, then for every neighborhood $\varphi(p)\in V$ in $F$, there exists a neighborhood $p\in U$ in $E$ such that $U\subset V$. This implies that for a neighborhood $F_{0}$ of 0 in $F$, there exists integer $n$ such that $\varphi(x^{n}E_{0})\subset F_{0}$. Hence we have $\varphi(E_{0})\subset x^{-n}F_{0}$.Conversely, if there is some integer $n$ such that $\varphi(E_{0})\subset x^{-n}F_{0}$, then we have $\varphi(x^{n}E_{0})\subset F_{0}$. Thus, for every $m\in\Z$, we have $\varphi(x^{n+m}E_{0})\subset x^{m}F_{0}$, which shows that for every open $V\subset F$ and $y\in\textrm{Im }\varphi\cap V$, we have $\varphi(y_{0}+x^{n+m}E_{0})\subset y+x^{m}F_{0}\subset V$ where $y=\varphi(y_{0})$.$\Box$