# perfectoid spaces-6

In this post, we will say something on Witt vectors, which will be used in Scholze’s paper. All the materials in this post can be found in Serre’s book, ‘Corps Locaux’.

We shall not discuss the original motivation for Witt to introduce the Witt vector. One interesting motivation is the following simple result:

Suppose that $A$ is a complete discrete valuation ring, with fraction field $K$ and residue field $\kappa$. Suppose that $S$ is a set of representatives of $\kappa$ in $A$, and $\pi$ is a uniformizer of $A$. Then each element $a$ in $A$ is a ‘power series’ of $\pi$ with coefficients in $S$, that is, $a$ can be written uniquely as $a=\sum_{n=0}^{\infty}s_n\pi^n$ with $s_n\in S$.

Here a uniformizer of $A$ is just a generator for the maximal ideal of $A$. Since $A$ is a discrete valuation ring, this means that $\pi$ always exists. The essential part in the proof to this proposition is that, the ideals $m=(\pi)\supset m^2\supset m^3\supset...$ form a system of open neighborhood of $0$. And of course, elements in the fraction field $K$ can also be written as Laurent power series as above.

Now note that $A$ is a ring, which has addition, subtraction, and multiplication operations. What will happen if we use the above ‘expansion’ expression for these operations? That is to say, if $a=\sum_n a_n\pi^n,b=\sum_n b_n\pi^n,c=a+b=\sum_n c_n\pi^n$,then what are the relations between $a_n,b_n,c_n$? Is there some simple algebraic expression for this? If not, can we choose $S$ to be as simple as possible such that the expression desired is very simple?

The above observation can be seen as a motivation for Witt vectors. Note that in the above proposition, we can show that as sets, we have, by construction, that $A=\kappa^{\mathbb{N}}$. So, the questions raised at the end of the last paragraph is just to find a ring structure on $\kappa^{\mathbb{N}}$ such that this ring is isomorphic to $A$. Of course, conditions could be posed on $A$.

One general construction is as follows:

Fix a prime number $p$. For a countable set of indeterminates $(X_0,X_1,X_2,...)$ and we set $W_0=X_0,W_1=X_0^p+pX_1,...,W_k=X_0^{p^k}+pX_1^{p^{k-1}}+p^2X_2^{p^{k-2}}+...+p^kX_k$. Then it is easy to see that

For any $F\in\mathbb{Z}[X,Y]$, there is a unique sequence $(f_0,f_1,...)$ in $\mathbb{Z}[X_0,X_1,...,Y_0,Y_1,...]$ such that $W_n(f_0,f_1,...)=F(W_n(X_0,X_1,...),W_n(Y_0,Y_1,...))$ for all $n=0,1,...$

I will not prove this result. One method is just to use induction on $n$ and some calculations. The interesting thing is that, when we express $X_i$ in terms of $W_k$, the coefficients lie in $\mathbb{Z}[1/p]$, yet this proposition tell us that these $f_j$ have integer coefficients. This is due to the assumption that $p$ is a prime number.

We shall apply this result to two polynomials, that is $F=X+Y$ and $G=XY$, and we denote the corresponding sequence as $(f_0,f_1,...),(g_0,g_1,...)$. Then a surprising result is:

Suppose that $R$ is a commutative ring, and for any $a=(a_0,a_1,...),b=(b_0,b_1,...)\in R^{\mathbb{N}}$, if we set $a+b=(f_0(a,b),f_1(a,b),...),a*b=(g_0(a,b),g_1(a,b),...)$, then with these two operations(addition and multiplication, as the symbols suggest), $(R^{\mathbb{N}},+,*)$ is a commutative ring.

Here we simply write $g_i(a,b)=g_i(a_0,a_1,...,b_0,b_1,...)$ and the same for others.

One little remark, in some sense, this result is part of the theory of formal groups.

We shall denote this ring as $W(R)$, and $a\in W(R)$ is called a Witt vector with coefficients in $R$. Note that since $f_i$ contains only addition, subtraction and multiplication operations and divide $p$ operation, so if this proposition is valid for a ring $R$, then it is also valid for subrings and quotient rings of $R$. As any ring is a quotient ring of some $R=\mathbb{Z}^I$, so it suffices to prove this proposition for these kinds of rings. If $R=\mathbb{Z}^I$, then we construct a map $W:W(R)\rightarrow R^{\mathbb{N}}$(the latter is equipped with the product ring structure), $a\mapsto (W_0(a),W_1(a),...)$. This is a bijection, which is obvious. Now, it is rather clear that $W$ respects the addition and multiplication operations of these two rings, thus proving that $F$ is an isomorphism. Moreover, it is easy to see, $(1,0,0,...)$ is the unit element in $W(R)$.

Now we can put the first result and the last result above together to see what happens: from a discrete valuation ring $A$ we get a $\pi$-expansion for each of its elements, and thus a bijection $A\simeq \kappa^{\mathbb{N}}$. And conversely, from $\kappa$, we construct a ring $W(\kappa)\simeq\kappa^{\mathbb{N}}$. Then what is the relation between $A$ and $W(\kappa)$? In fact, they are isomorphic in some cases:

If $\kappa$ is a perfect field of characteristic $p$, then $A\approx W(A/m)$.

To prove this result, we need Teichmüller representatives:

Suppose that $\kappa$ is a perfect field of characteristic $p$, then there exists a unique map $f:\kappa\rightarrow A$ such that:(1) $f(a^p)=f(a)^p$;(2)for $a\in A$, $a\in f(\kappa)$ if and only if $a=b^p$ for some $b\in A$;(3)$f(ab)=f(a)f(b)$ for any $a,b\in \kappa$;(4) if $A$ is of characteristic $p$, then $f(a+b)=f(a)+f(b)$ for any $a,b\in \kappa$.

The construction used in the proof is the same flavor as the one we have seen in Scholze’s paper, or in Fontaine-Wintenberg’s theorem. For any $a\in\kappa$, we consider $(a_0,a_1,a_2,...)\in \kappa^{\mathbb{N}}$ such that $a_{i+1}^p=a_i$. And for each $a_i$ we find a lift $b_i\in A$, and we set $f(a)=\lim_n b_n^{p^n}$. And the first three points are automatically satisfied(the uniqueness is a consequence of the fact that $f$ commutes with the $p$-power operation). As for the last point, it is also very clear.

This map is called a Teichmüller map, and $f(\kappa)$ are called the Teichmüller representatives of $\kappa$ in $A$.

Now we can construct a map from $W(\kappa)$ to $A$ as follows: for any $a\in W(\kappa)$, we send it to $T(a)=\sum_n f(a_n)^{1/p^n}p^n$, for some special rings, we can show that this is an isomorphism:

If $A$ is a strict $p$-adic ring such that $p$ is invertible in $A$, then $T:W(\kappa)\rightarrow A$ is an isomorphism.

Here a strict $p-$adic ring $A$ is a discrete valuation ring such that the $(p)\supset (p^2)\supset (p^3)\supset...$ forms a system of open neighborhoods of $0$ in $A$. Note that, under this assumption, the very first $\pi$-expansion is can be rephrased for $p$-expansion. And we can in the before last proposition replace the residue field $\kappa$ by the quotient ring $A/pA$, everything is still correct since the essential point is of characteristic $p$. And then this proposition can be proved by calculations.

So, in this way, we find an algebraic expression for $(A/pA)^{\mathbb{N}}$ such that the corresponding ring structure coincides with that of $A$.

One important remark: for any $n>0$, we can consider $W_n(R)=\{(r_0,r_1,...,r_{n-1},0,0,...)\in W(R)\}$. Note that this $W_n(R)$ is a quotient of $W(R)$ by $W'_n(R)=\{(0,0,...,0,r_n,r_{n+1},...)\in W(R)\}$. Indeed, $W'_n(R)$ is easily seen to be an ideal of $W(R)$, thus $W_n(R)=W(R)/W'_n(R)$ is a quotient ring. And it is not hard to show that $W(R)=\varprojlim_nW_n(R)$.

Another remark is about the topology on $W(R)$. As above, we have defined some ideals $W'_n(R)$ of $W(R)$, and it is easy to see that $W'_n(R)W'_m(R)\subset W'_n(R)$ if $n\geq m$. So, we can use these ideals to define a topology on $W(R)$.

# learning Italian

## Articles

There are several forms of articles in italian. By this, I mean that the article will change its form depending on the noun following it. There are several cases to consider. The first is of course the gender of the noun, masculine or feminine. The second case is the starting sound of the noun. And the last case is the number of the noun, singular or plural. The first and the third cases are, in some sense, very natural, at least for me, if someone has learned some other languages, like French or German. The problem is the second case. Yet, if we think more, we should find that this is the point which shows that Italian is a logical language: it always tries to make pronunciation easier.

More precisely, if we only consider the first and the third cases, we have that, before a noun singular and masculine, we should use il(definite article) or un(indefinite article). Before a noun singular and feminine, we should use la or una; before a noun plural and masculine, we should use i or dei; before a noun plural and feminine, we should use le or delle.

Yet Italian considers also the first sound of the noun. If this sound is a vowel, then we should avoid two vowels appearing at the same time, and if this sound is something like ‘st,sc,gn,ps,z’, then we should avoid too many consonants appearing together. For the first situation, we change la to l’, una to un’, i to gli, and dei to degli. And for the second situation, we change change il to lo, un to uno, i to gli, dei to degli. For example, il/un treno, lo/uno studente, l’/un amico, la/una ragazza, l’/un’amica, i/dei quadri, gli/degli italiani, le/delle città.

Note that the sound gli appears in both these two situations. This sound is a little particular in Italian. It serves both as a vowel and as a consonant. In gli, the sound l is stronger than the sound i after it, and this makes it sounds like a consonant.

# Huber space-9

In this post, we will study more closely the generizations of a valuation.

Before all, we should make an important remark: suppose that $v:A\rightarrow\Gamma\bigcup\{0\}$ is a valuation, $\pi:A\rightarrow K(v)$ is the canonical morphism, $H$ is a convex subgroup of $\Gamma$ and $p\subset A(v)$ the prime ideal $p=\{a\in A(v)|v(a). Then

(1)$c\Gamma_v\subset H$ if and only if $\pi(A)\subset A(v)_p$.

Proof: if $c\Gamma_v\subset H$, then for any $a\in \pi(A)-A(v)$, we have that $1/a\in A(v)$. This shows that $v(a)\geq 1$, thus $v(a)\in c\Gamma_v\subset H$. It is worth pointing out that in the case $c\Gamma_v\subset H$, we have that for any $a\in A$, either $v(a)\in H$, either $v(a)(just consider the two cases: (1)$v(a)\geq1$, then $v(a)\in c\Gamma_v\in H$;(2)$v(a)<1$, then if $v(a)\not\leq H$, say $h\leq v(a)<1$ for some $h\in H$, since $H$ is convex, thus $v(a)\in H$). Thus $v^{-1}(H)=A-\pi^{-1}(p)$, so $\pi(a)\in A(v)_p$. This shows that $\pi(A)\subset A(v)_p$. Conversely, if $\pi(A)\subset A(v)_p$, then for any $v(a)>1$, since $\pi(a)\in A(v)_p$, thus $\pi(a)=a'/a''$ with $a'\in A(v),a''\in A(v)-p$. So we have that $v(a)v(a'')=v(a')$(we use the same letter for the valuations on $A$ and the induced valuation on $K(v)$). And since $v(a'')\geq h$ for some $h\in H$ and $v(a)>1$, so $v(a')>h$. So we have that $a',a''\not\in p$, which implies that $\pi(a)\not\in p$ since $p$ is a prime ideal. So, $v(a)>v(p)$, and use the correspondence of the set of prime ideals of $A(v)$ and the set of convex subgroups of $\Gamma$, we have that $v(a)\in H$, which shows that $c\Gamma_v\subset H$.

Again this proof is not illuminating. One way to understand this proof is to note that, $c\Gamma_v\subset H$ means that $H$ contains all the elements of $\Gamma$ that are $\geq1$. Thus for any $a\in A$, if $v(a)\leq 1$, then $v(a)\in A(v)$ automatically, if $v(a)>1$, then $v(a)\in H$, and use the fact that $A(v)$ is a valuation ring, $1/\pi(a)\in A(v)$, so we must have that $\pi(a)\in(v^{-1}(H))^{-1}A(v)$. Yet this multiplicative set is just the complement of $p$ in $A(v)$ provided that $c\Gamma_v\subset H$.

Note that we have transformed problems on $A$ to problems on $A(v)$, latter of which is a subring of $K(v)=Frac(A/supp(v))$, this is because we want to use the correspondence of prime ideals of a ring and the convex subgroups of a totally ordered group, and this fact is established only for valuation rings. That is why we do this in the above.

(2)We assume that $c\Gamma_v\subset H$. Then it is easy to see that $supp(v|_H)=\pi^{-1}(pA(v)_p)$. This is just the definition. Moreover, $supp(v)\subset supp(v|_H)$. And it is clear that $supp(v)=supp(v|_H)$ if and only if $H=\Gamma$, if and only if $p=0$.

(3)We assume that $c\Gamma_v\subset H$. Then consider the local ring $A(v)_p$. Its residue field is $A(v)_p/p$, with a valuation ring $A(v)/p$. Note that we can not extend directly the valuation $v$ to $A(v)_p/p$, we have to kill all the elements $v(p)$. This reminds us to use $v|_H$ to do this extension. Indeed, $v|_H:A(v)_p/p\rightarrow H\bigcup\{0\}$ defines well a valuation. And it is easy to see that $A(v|_H)=A(v)/p$. So, here we get an explicit description of $v|_H$, horizontal generizations of $v$.

In the following, we shall prove a series of proposition to get the result that any generization of $v$ is the composite of vertical generization and horizontal generization(similar results for specializations), as promised at the end of the precedent post.

The first proposition below gives a description of horizontal specializations of $v$. Firstly, we need a concept:

Definition: if $v:A\rightarrow\Gamma\bigcup\{0\}$ is a valuation, a subset $T\subset A$ is call $v$-convex if for any $v(t)\leq v(a)\leq v(t')$ in $A$ with $t,t\in T$ implies that $a\in T$.

Now another description of horizontal specializations of a given valuation:

Suppose $v:A\rightarrow\Gamma\bigcup\{0\}$ is a valuation, and $C_v$ is the set of $v$-convex prime ideals, equipped with an order of inclusion, and $S_h$ is the set of horizontal specializations of $v$, equipped with an order of being specialization($v'\leq v''$ if $v'$ is a specialization of $v''$), then $C_v$ and $S_h$ are totally ordered sets and the map $F:S_h\rightarrow C_v,v\mapsto supp(v)$ is an order-preserving bijection.

Proof: (1)Note that for a $v$-convex subset $T$ of $A$ and an element $a\in A$, either $v(a), or $v(a)>v(T)$, or $a\in T$. Moreover, if $0\in T$, then $v(a)\leq v(t)$ with $t\in T$, we have $s\in T$. So, for two $v$-convex prime ideal $p,p'$ of $A$, if $a\in p-p'$, then we should have that $v(a)>v(p')$. This means that $p'\subset p$. So, any two elements in $C_v$ are comparable, thus $C_v$ is a totally ordered set. For the present we shall not show directly that $S_h$ is a totally ordered set, we will show that $F$ is an order-preserving bijection, which implies that $S_h$ is totally ordered.

(2)It is easy to show that $F$ is well-defined. Indeed, for any $v|_H$ horizontal specialization of $v$, $supp(v|_H)=Ker(A\rightarrow A(v)_p/p)$, thus showing that $supp(v|_H)$ is indeed a $v$-convex prime ideal. Moreover, if $H\subset H'$ two convex subgroups of $\Gamma$, then $v|_{H'}$ is a horizontal specialization of $v|_H$, that is $v|_{H'}\leq v|_H$. It is also easy to see that $Ker(A\rightarrow A(v)_p/p)\supset Ker(A\rightarrow A(v)_{p'}/p')$ where $p,p'$ are the corresponding prime ideals for $H,H'$. Thus $F$ preserves the order.

(3)It remains to see that $F$ is a bijection. It suffices to construct the inverse of $F$. For any $p\in C_v$, we define $H=\{h\in \Gamma|h>v(p)\}$. Then $H$ is a convex subgroup. Indeed, if $v(a),v(a')\in H$, that is $v(a),v(a')>v(p)$, then $a,a'\not\in p$, thus $a a'\not\in p$, that is $v(a a')>v(p)$. Moreover, note that $1\not\in p$, that is $1\in H$. So, if $v(a)\in H$, and if $v(a)^{-1}\in v(p)$, that is $v(a k)=1$ for some $k\in p$, then $a k\in p$, thus $1\in v(p)$, contradiction, showing that $H$ is indeed a subgroup. In addition, if $h in $\Gamma$ with $h,h'\in H$, then $s\not\in v(p)$, otherwise $h\in v(p)$. So, this shows that $H$ is a convex subset. Now we can define $G:C_v\rightarrow S_h,p\mapsto H$, and it is not difficult to show that $F,G$ are inverse one to the other.

The next proposition is very important for later use:

Suppose $w$ is a horizontal specialization of $v$, that is $w=v|_H$ for some $H$, then(1) for any $v'$ a vertical specialization of $v$($v'=v/H'$), there exists a unique vertical specialization $w'$ of $w$($w'=w/L'$) such that $w'$ is a horizontal specialization of $v'$($w'=v'|_L$);(2) if $w'$ is a vertical specialization of $w$($w'=w/L$), then there exists a vertical specialization $v'$ of $v$($v'=v/H'$) such that $v'$ is a horizontal specialization of $v'$($w'=v'|_{L'}$).

In other words, for (1), if $w=v|_H$ and $v'=v/H'$, then there exists a unique $w'$ such that $w'=w/L$ and $w'=v'|_{L'}$; for (2), if $w'=(v|_H)/L$, then $w'=(v/H')|_{L'}$.

The essential point for this proposition is a simple commutative diagram, for $\Gamma\rightarrow H,\Gamma\rightarrow \Gamma/H'$, we have that $H\rightarrow H/(H\bigcap H'),\Gamma/H'\rightarrow H/(H\bigcap H')$. The uniqueness of the first point also follows. And the non-uniqueness in the second point comes from the fact that $w'$ can be a trivial valuation, that is $H\subset H'$ in the commutative diagram. If it is not trivial, then the uniqueness is guaranteed.

A corollary of this proposition is

If $p$ is a prime ideal such that $p\subset supp(v)$, then there exists a horizontal generization $w$ of $v$($v=w|_H$) such that $p=supp(w)$.

Proof: the proof is a bit tricky. Let $\pi:A\rightarrow A/p$ be the canonical projection, $R=(A/p)_{\pi(supp(v))}$, then $R$ is a local ring, a subring of $Frac(A/p)$. Thus there exists a valuation ring $B$ of $Frac(A/p)$ dominating $R$. Suppose then $w$ is the valuation on $B$, thus on $A$, with $supp(w)=p$. Then it is easy to find a convex subgroup $H$ such that $w|_H=v$.

Now we give the main result about specializations of valuations:

Let $v:A\rightarrow\Gamma\bigcup\{0\}$ be a valuation, $w$ be a specialization of $v$, (1)then $w$ is a horizontal specialization of a vertical specialization of $v$;(2)$w$ is also a vertical specialization of $w'$ which is in turn a horizontal specialization of $v$ or $c\Gamma_v=1$ and $w'$ is a trivial valuation whose support contains $supp(v|_1)$.

In other words, if $w$ is a specialization of $v$, then $w=(v/L)|_H$ for some convex subgroup $L\subset \Gamma$ and some convex subgroup $H\subset \Gamma/L$ such that $H\supset c\Gamma_{v/L}$. It is easier to first consider the second point. We shall give only a sketch of the proof. We first show that $supp(w)$ is $v$-convex, and thus according to a precedent result, there exists a horizontal specialization $u$ of $v$ with $supp(u)=supp(w)$. Then it remains to see that $u$ is a specialization of $w$.

So, until now, we finished the part on valuation theory.

In the next several posts, we shall focus on spectral spaces, which are very important for proving some properties of affinoid spaces to be defined later.

# Huber space-8

In this post, we will consider the space of all the valuations on a ring. We have seen that this set is, intuitively a little larger than the spectrum of the ring(the set of prime ideals). Still we shall give this space a topology, to mimic the spectrum space.

We fix a ring $A$, and we set $Spv(A)=\{v|v\text{ valuation on } A\}/\approx$. It is called the valuation spectrum of $A$. Then what topology should we consider on $A$? This is a fundamental question. Recall what topology we pose on $Spec(A)$. The open sets are generated by sets of the form $Spec(A)(f)=\{p|p\text{ prime ideal of }A, f\not\in p\}$. If for a prime ideal $p$, there is a valuation $v$ on $A$ such that $supp(v)=p$, then $f\not\in p$ means that $v(f)\neq 0$. With this, perhaps we can try to give the topology on $Spv(A)$, it is generated by subsets of the form $Spv(A)(f/s)=\{v\in Spv(A)|v(f)\leq v(s)\neq 0\}$ where $f,s\in A$.

First we examine some examples to get a feeling of this topology.(1)Take $K=\mathbb{Q}$, the field of rational numbers. What are the possible valuations on $K$? Note that any valuation $v$ on $K$ gives a valuation ring $A(v)$. Yet, any valuation ring contains $1$, thus any of its multiples, that is $\mathbb{Z}\subset A(v)$. We can show that any ring $A$ of $K$ containing $\mathbb{Z}$ is of the form $A=\mathbb{Z}[(1/n_i)_{i\in I}]$ where $p_i\in\mathbb{N}-0$ are primes. Yet a valuation ring is a local ring, so this means that $A=\mathbb{Z}[1/p]$ for some prime. This gives all the possible valuation rings of $K$, so we get that $Spv(\mathbb{Q})=\{v_0,v_2,v_3,v_5,...\}$ of which the elements are indexed by the primes($v_p(p^k a/b)=p^{-k}$ where $a,b$ integers coprime one to another and both are prime to $p$, and $v_0$ is the trivial valuation). Moreover, the basic open sets are of the form $Spv(K)(n/m)=\{v_p|v_p(n)\leq v_p(m)\neq 0\}$. The condition $v_p(m)\neq0$ is almost trivial as long as we take $m\neq 0$. $v_p(n)\leq v_p(m)$ means that $v_p(n/m)\leq 1$. If we write $n/m=\pm\prod_k p_k^{e_k}$, then $Spv(K)(n/m)=\{p_k|e_k\geq0\}=Spec(\mathbb{Z})-\{p_k|e_k<0\}$. Yet the second set $\{p_k|e_k<0\}$ is always a finite set, and can be any finite subset of $Spec(\mathbb{Z})$ excluding $0$. So, this means that the topology on $Spv(K)$ is the same as $Spec(\mathbb{Z})$.(2) Next we take $A=\mathbb{Z}$, and consider $Spv(A)$. If $v\in Spv(A)$ has trivial support $supp(v)=0$, then $v(2)=v(1+1)\leq v(1)=1, ...,v(n)\leq 1$. And $v(1)=v(3-2)\leq \max(v(3),v(2))$, thus $v(2)=1$ or $v(3)=1$. If $v(2)<1$, then we can show that this valuation is just $v_2$. If $v(3)<1$, then $v=v_3$. If $v(2)=v(3)=1$, we can consider the pair $(3,4)$ in place of $(2,3)$, and continue the above argument, finally we can get all the valuations $v_p$ including the trivial valuation $v_0$. So, if $v$ has trivial support, then $v$ is the usual valuation. If $supp(v)=pA$, a prime ideal generated by a prime $p$, then $v$ factors by $v':A/pA=\mathbb{F}_p\rightarrow\Gamma\bigcup\{0\}$. Yet $\mathbb{F}_p^*$ is a finite group, thus the only valuation on $\mathbb{F}_p$ is the trivial valuation, which induces a valuation on $A$, we denote this valuation as $v_{p,0}$(more specifically, for any $a\in A$, if $p|a$, then $v_{p,0}(a)=0$, otherwise it is $1$). So, all in all, we get that $Spv(\mathbb{Z})=\{v_0,v_2,v_{2,0},...\}$. Now consider the topology on $Spv(A)$. The basic open subsets are of the form $Spv(A)(n/m)=\{v\in Spv(A)|v(n)\leq v(m)\neq 0\}$. First if $m=0,n=0$, then it is the whole set. If $m=0,n\neq 0$, then $v(n)=0$, writing $n=\pm\prod_k p_k^{e_k}$, then $Spv(A)(n/0)=\{v_{p_k,0}|e_k\neq 0\}\bigcup\{v_0\}$. If $m\neq 0$, then write $n/m=\prod_k p_k^{e_k}$, and $v(n)\leq v(m)$ means that $Spv(A)(n/m)=\{v_0\}\bigcup\{v_{p_k},v_{p_k,0}|e_k\geq 0\}$. Note that any open subset containing $v_{p,0}$ contains also $v_0,v_p$, this shows that $\overline{\{v_0\}}=Spv(A),\overline{\{v_p\}}=\{v_p,v_{p,0}\}$.

Now a little remark about morphism between valuation spectrums. Suppose that $f:A\rightarrow B$, then we have that $Spv(f):Spv(B)\rightarrow Spv(A)$ induced by composition. Note that we also have that $Spec(f):Spec(B)\rightarrow Spec(A)$. Moreover, we have a map $\pi_A: Spv(A)\rightarrow Spec(A), v\mapsto supp(v)$. It is worthwhile pointing out that $\pi_A$ is not always surjective. If $A$ is a valuation ring, then it is easy to see that $\pi_A$ is surjective. The converse is not true in general, as the above example $Spv(\mathbb{Z}), Spec(\mathbb{Z})$ shows. There are many interesting remarks to make concerning these maps $Spv(f),Spec(f),\pi_A$.

It is easy to see that $Spv(f), Spec(f)$ are always continuous. (1)If we take $S\subset A$ a multiplicative set, and set $B=S^{-1}A$, then for the canonical map $f:A\rightarrow B$, we have that $Spv(f)(v)=v\circ f=v'$. Since $v(1/s)v(s)=v(1)=1$, this shows that $v(s)\neq 0$, that is $supp(v')\bigcap S=\emptyset$. In fact, we can show that $Im(Spv(f))=\{v'\in Spv(A)|supp(v')\bigcap S=\emptyset\}$, and $Spv(f)$ is a homeomorphism of $Spv(B)$ onto its image. (2)If we take $I$ an ideal of $A$, and set $B=A/I$, then for the canonical map $f:A\rightarrow A/I$, this shows that for any $v\in Spv(B)$, $supp(Spv(f)(v))\supset I$, and similarly, we can show that $Spv(f)$ is a homeomorphism of $Spv(B)$ to the image $Im(Spv(f))=\{v\in Spv(A)|supp(v)\supset I\}$.(3) For the map $\pi_A:Spv(A)\rightarrow Spec(A)$, a basic open subset of $Spec(A)$ is $Spec(A)(f)=\{p\in Spec(A)|f\not\in p\}$. Then since $\pi_A(v)=supp(v)\in Spec(A)(f)$ means that $v(f)\neq 0$, so we get that $\pi_A^{-1}(Spec(A)(f))=Spv(A)(0/f)$, which shows that $\pi_A$ is continuous.(4) If $K$ is a field, then $Spv(K)$ admits the trivial valuation $v_0$ as a generic point. Indeed, any basic open set $Spv(K)(f/s)$ containing a valuation $v$ simply means that $v(f)\leq v(s)\neq 0$, this in turn shows that $s\neq 0$, and thus $v_0(f)\leq v_0(s)=1$, thus this basic open set contains $v_0$, too. So the trivial valuation is a generic point in $Spv(K)$. This means that $Spv(K)$ is an irreducible topological space.(5) If $v$ is a valuation on $A$, then for the canonical map $f:A\rightarrow K(v)$, we have that $Im(Spv(f))=\{v'\in Spv(A)|supp(v')=supp(v)\}$. It can be shown that $Spv(f)$ is a homeomorphism, and thus $Im(Spv(f))$ is an irreducible subspace of $Spv(A)$.

We have seen in the above that in $Spv(A)$ some points are more special than other points, just like in $Spec(A)$, that is the generic points. We give definitions concerning this situation:

Given two valuations $v,v'\in Spv(A)$, $v$ is said to be a specialization of $v'$ if $v\in \overline{\{v\}}$, and $v'$ is said to be a generization of $v$. $v$ is a specialization of $v'$ means that any basic open subset $Spv(A)(f/s)$ containing $v$ contains $v'$, too. That is to say, $v(f)\leq v(s)\neq 0$ implies that $v'(f)\leq v'(s)\neq 0$. If we take $f=0$, then this is: $v'(s)=0$ implies that $v(s)=0$, which shows that $supp(v')\subset supp(v)$. to memorize the relation, it is better to write this as $A-supp(v')\supset A-supp(v)$ if $v'$ is a generization of $v$.

Here there are two cases to consider, the first is that $v$ is a specialization of $v'$ with $supp(v)=supp(v')$, in this case we will call $v$ to be a vertical specialization of $v'$(and of course, $v'$ is a vertical specialization of $v$), and the second is that $v$ is a specialization of $v'$ with $supp(v)\supset supp(v')$. To see another characterization of vertical specialization, we need:

Lemma:Let $K$ be a field, and $v,v'$ two valuations on $K$, then $v$ is a specialization of $v'$ if and only if $A(v)\subset A(v')$

This lemma is easy.

Then it is easy to see that, fixing a valuation on $A$, $v:A\rightarrow\Gamma\bigcup\{0\}$, we have that

$\{v'\in Spv(A)|v'\text{ vertical generization of }v\}=\{v'\in Spv(K(v))|v'\text{ generization of }v\}=\{v'\in Spv(K(v))|A(v')\supset A(v)\}=\{H|H\text{ convex subgroup of }\Gamma\}$

By the same reasoning, we see that

$\{v'\in Spv(A)|v'\text{ vertical specialization of }v\}=\{v'\in Spv(K(v))|A(v')\subset A(v)\}=Spv(\kappa(v))$.

From above we see that convex subgroups play an important role in the classification of valuations on a ring. Here we give some valuations induced from some convex subgroups:

Suppose that we have a valuation $v:A\rightarrow\Gamma\bigcup\{0\}$, and $H$ is a convex subgroup of $\Gamma$, then we define

$v/H=A\rightarrow\Gamma\bigcup\{0\}\rightarrow\Gamma/H\bigcup\{0\}$

to be the composition,

$v|_H=v*\chi_H$

where $\chi_H:\Gamma\bigcup\{0\}\rightarrow \{0,1\}$ is the characteristic function of $H$. It is easy to see that $v/H$ is again valuations on $A$, as long as $H$ is a convex subgroup. As for $v|_H$, more condition should be posed. Before this, we should introduce an important convex subgroup of $\Gamma$:

$c\Gamma_v$ is the convex subgroup of $\Gamma$ generated by all the elements $v(a)$ such that $v(a)\geq 1$. It is called the characteristic group of $v$.

Then we can state the condition:

$v|_H$ is a valuation on $A$ if and only if $H$ is a convex subgroup and contains $c\Gamma_v$.

These two conditions are not trivial at first sight. Suppose that $v'=v|_H$ is a valuation, then if $h,h'\leq 1$ and $h h'\in H$, writing $v(a)=h,v(a')=h'$, we have that $v'(a)v'(a')=v'(aa')=v(aa')=hh'\neq 0$, thus $v'(a),v'(a')\in H$, that is to say, $h,h'\in H$, which implies that $H$ is convex; note that $v'(1)\leq\max v'(a),v'(1-a)$ for any $a\in A$, this means that either $v(a)\in H$ or $v(1-a)\in H$. Yet if $v(a)>1$ and $v(a)\not\in H$, we should have $v'(a)=0$, thus $v'(1-a)\geq 1$. But $v(1-a)=\max{v(1),v(a)}=v(a)\not\in H$(it is easy to show that if $v(a)\neq v(b)$, then $v(a+b)=\max{v(a),v(b)}$), so $v'(1-a)=0$, this is a contradiction, which implies that $H$ contains all the elements of the form $v(a)\geq 1$ and thus contains the subgroup $c\Gamma_v$. Conversely, if $H$ is a convex subgroup and contains $c\Gamma_v$, then $v'=v|_H$ is a valuation. Indeed, it is clear that $v'(0)=0,v'(1)=1$. And since $H$ is a convex subgroup, $v(a),v(a')\in H$ if and only if $v(a a')\in H$, thus $v'(a)v'(a')=v'(a a')$ for any $a,a'\in A$. Further more, for any $a,a'\in A$, if $v(a),v(a')\in H$, then either $v(a+a')\in H$, thus $v'(a+a')=v(a+a')\leq \max{v(a),v(a')}=\max{v'(a),v'(a')}$, or $v(a+a')\not\in H$, thus $v'(a+a')=0\leq\max{v'(a),v'(a')}$. If $v(a)\not\in H,v(a')\in H$, then either $v(a), thus $v(a+a')=\max{v(a),v(a')}=v(a')\in H$, which implies  $v'(a+a')=v(a+a')=\max{v(a),v(a')}=\max{v'(a),v'(a')}$, or $v(a)>H$, which is impossible since $H\supset c\Gamma_v$, and $v(a)>H$ implies that $v(a)>1$, thus $v(a)\in c\Gamma_c\subset H$. If $v(a),v(a')\not\in H$, then $v(a),v(a'), thus $v(a+a'), showing that $v'(a+a')=0\leq \max{v'(a),v'(a')}$. All in all, we showed that $v'(a+a')\leq\max{v'(a),v'(a')}$.

The above proof is a bit long and not very illuminating. The essential point is that, to ensure that $v|_H$ is multiplicative, $H$ should be a convex subgroup(more precisely, $H$ is a subgroup guarantees that the multiplication of two elements of $H$ is still in $H$, and $H$ is a convex guarantees that if two elements do not lie in $H$ at the same time, then their product is not in $H$, this leads to the facts that $0h=h$ in $H\bigcup\{0\}$), and to ensure that $v|_H$ satisfies the triangle inequality, $H$ should contain $c\Gamma_v$.

We call $v|_H$ a horizontal specialization of $v$ and $v$horizontal generization of $v|_H$. It is easy to see that $v|_H$ is indeed a specialization of $v$, $v$ is indeed a generization of $v|_H$.

So, vertical generizations of $v$ correspond to valuations of the form $v/H$ where $H$ is a convex subgroup, and horizontal generizations of $v$ correspond to valuations of the form $v|_H$ where $H$ is a convex subgroup containing the characteristic subgroup.

In the next post, we shall see that all generization of $v$ is a composite of these two kind of generizations.

# Huber space-7

In this post, we will discuss valuations.

We suppose that $A$ is a ring and $\Gamma$ is a totally ordered group, then a valuation is a map $v:A\rightarrow \Gamma\bigcup\{0\}$ such that (1)$v(a+b)\leq \max(v(a),v(b))$ for any $a,b\in A$;(2)$v(ab)=v(a)v(b)$ for any $a,b\in A$; (3)$v(1)=1,v(0)=0$.

Note that this definition excludes immediately all the Euclidean metrics. A little remark on the morphism between $\Gamma\bigcup\{0\}$: we say that $F:\Gamma\bigcup\{0\}\rightarrow\Gamma'\bigcup\{0\}$ is a morphism if $Im(F)\neq 0$, $F$ preserves the orders, and $F(gg')=F(g)F(g')$ for any $g,g'\in\Gamma\bigcup\{0\}$.

We should have heard valuation rings in commutative algebra. For example, $A$ is a field, and $v:A\rightarrow\Gamma\bigcup\{0\}$ is a valuation, then we say that $A^o=\{a\in A|v(a)\leq 1\}$ is the valuation ring of $A$(associated to this valuation), and this valuation ring is a local ring, with the maximal ideal $A^{o o}=\{a\in A|v(a)<1\}$. Note that, for any $0\neq a\in A$, either $a$ or $1/a$ lies in $A^o$. This is the most important property of a valuation ring. In fact, we can use this as the definition for valuation rings, that is:

A subring $A^o$ of a field $A$ is a valuation ring if and only if for any non-zero element $a\in A$, either $a$ or $1/a$ lies in $A^o$.

Proof: If $A^o$ satisfies the condition above, then in fact we can define a totally ordered group using $A,A^o$: we consider the group $A^*/A^{o *}$, and we say that $\overline{a}\leq \overline{a'}$ if $a=a'a_0$ where $a_0\in A^o$. The condition above implies that this order is a total order, and we define the valuation as the projection $\pi:A\rightarrow A^*/A^{o *}\bigcup\{0\}$. It is easy to show that the valuation ring of $A$ associated to this valuation is exactly $A^o$, thus finishing the proof.

There is an important notion concerning valuations of a ring. For a valuation $v:A\rightarrow\Gamma\bigcup\{0\}$, $supp(v)=\{a\in A|v(a)=0\}$, it is called the support of this valuation. This name is perhaps a little confusing at first, since it reminds us the support of a function, for example $f:\mathbb{R}\rightarrow \mathbb{R}$, then $supp(f)$ is just the closure of the set of the points such that their image is not zero. Still, we will use this name for a valuation as defined above.

This notion give some immediate consequences. In fact, we have that $supp(v)$ is a prime ideal of $A$. It is rather easy. Furthermore, this valuation factors by $v':A/supp(v)\rightarrow\Gamma\bigcup\{0\}$. And now $A/supp(v)$ is an integral domain, and thus we can extend this valuation $v'$ to its fraction field by $v':Frac(A/supp(v))\rightarrow\Gamma\bigcup\{0\},a/b\mapsto v(a)v(b)^{-1}$ for $0\neq b\in A/supp(v)$.

And then we can consider the valuation ring of this valuation on the field $Frac(A/supp(v))$. To simplify notions, we will write $K(v)=Frac(A/supp(v))$, and $A(v)$ is the valuation ring of $K(v)$ associated with $v'$, $m(v)$ is the maximal ideal of $A(v)$, and $\kappa(v)$ is the residue field of $A(v)$, that is $\kappa(v)=A(v)/m(v)$.

A last notion is an equivalence relation on the set of valuations on a fixed ring. We say two valuations $v_1:A\rightarrow\Gamma_{v_1}\bigcup\{0\},v_2:A\rightarrow\Gamma_{v_2}\bigcup\{0\}$, then $v_1,v_2$ are called equivalent if there is an isomorphism between $\Gamma_{v_1}\bigcup\{0\}$ and $\Gamma_{v_2}\bigcup\{0\}$. Equivalently, we can say that for any two $a,a'\in A$, $v_1(a)\leq v_1(a')$ if and only if $v_2(a)\leq v_2(a')$. Or $supp(v_1)=supp(v_2)$ and $A(v_1)=A(v_2)$.

Using this last characterization, we get a bijection between $\{v|v \text{ valuation on} A\}/\approx$ and the set $\{(p,A')|p \text{ prime ideal of }A, A' \text{ valuation ring of }Frac(A/p)\}$ where $\approx$ is the equivalence relation defined above. If we have that $K$ is a valuation field with valuation $v$ and $A=A(v)$ its valuation ring, then there is also a bijection between $\{B|B\text{ valuation ring of }\kappa(v)\}$ and $\{B'|B'\text{ valuation ring of }K\text{ contained in }A\}$, defined by sending $B$ to $\pi^{-1}$ where $\pi:A\rightarrow \kappa(v)$ is the canonical projection.

This last bijection is at first sight a little mysterious. For any $B'$ valuation of $K$ contained in $A$, it is clear that $\pi(B')$ is again a valuation ring in $\kappa(v)$, so $\pi(B')$ defines a valuation $v:\kappa\rightarrow\Gamma\bigcup\{0\}$. Now this valuation can be pulled back to $A$, and then to $K$, defining a valuation $v':K\rightarrow\Gamma\bigcup\{0\}$ such that $v'|_A=v\circ \pi$. Note that the valuation ring of $v$ is just $\pi(B')$ according to the definition of $v$, and the valuation ring of $v'$ is $B''=\pi^{-1}(\pi(B'))$ which obviously contains $B'$. We have to show that $B''=B'$. Suppose that $b\in B''-B'$, then since $b\in K^*$, this means that, according to the equivalent definition of valuation rings, $1/b\in B'$. This shows that $b\in A$ is a unit, and since $1/b\in B'$, thus $b\in B'$, this implies that $B''=B'$, thus finishing the proof.

Note that a valuation $v$ on a field $K$ always has trivial support, that is $supp(v)=0$ or $supp(v)=K$. In the last bijection, the essential assumption is that $A$ is a valuation ring.

In the above, we considered valuation rings contained in a valuation $A$, next we will consider the valuation rings containing the valuation ring $A$, that is(in the following the always assume that for a valuation $v:A\rightarrow\Gamma\bigcup\{0\}$, $v$ is surjective, this can always be possible since we can just take $v:A\rightarrow Im(v)$):

Let $K$ be a valuation field with valuation $v:K\rightarrow\Gamma\bigcup\{0\}$, and $A$ its valuation group. (1)If we set $R=\{B| B\text{ valuation ring of }K, A\subset B\}$, and $I=\{H|H\text{ convex subgroup of }\Gamma\}$, and we define two maps $i:I\rightarrow R,H\mapsto A(K\rightarrow \Gamma\rightarrow\Gamma/H)$, and $r:R\rightarrow I,B\mapsto v(B^*)$;(2) Let $Spec(A)$ be the set of prime ideals of $A$, then we define another two maps, $s:Spec(A)\rightarrow R,p\mapsto A_p$ and $r':R\rightarrow Spec(A), B\mapsto m_B$ where $m_B$ is the maximal ideal of $B$. Then $(i,r),(s,r')$ define two bijections among these sets.

Proof: (1) First we should show that $r$ is well-defined. For any $B\in R$, if $v(k),v(k')\leq 1$ and $v(kk')\in v(B^*)$, then we should show that $v(k),v(k')\in v(B^*)$. But this is automatic since $A\subset B$. First we show that $i(r(B))=B$ for any $B\in R$. Indeed, for any $b\in B$, either $b\in A$, thus $v(b)\leq 1$, thus $v(b)\leq v(B^*)$, or $b\not\in A$, which means that $1/b\in A$, this shows that $b\in B^*$, and thus again $v(b)\leq v(B^*)$, so this means that $B\subset i(r(B))$:Conversely, for any $a\in i(r(B))$, this means that $v(a)\leq v(B^*)$. Then either $v(a)\in v(B^*)$, this shows that $a\in B^*\subset B$, or $v(a), which implies that $v(a)<1$, and thus $a\in A\in B$, so finally we get that $i(r(B))=B$. Then we show that $r(i(H))=H$ for any $H\in I$. But this is rather straightforward.(2)These two maps are indeed well-defined. First we show that $r'(s(p))=p$ for any $p\in Spec(A)$. But $s(p)=A_p$, the localization of $A$ with respect to $p$, and $r'(A_p)=m_{A_p}=pA_p$ the maximal ideal of $A_p$. Then we have that $p=m_{A_p}\bigcap A$. Yet, in general, for any $B\in R$ and any $b\in m_B$, then $1/b\not\in B$, thus $1/b\not\in A$, so $b\in m_A$, this shows that $m_B\subset m_A$. So, $m_B=m_B\bigcap A$ is a prime ideal of $A$, so $r'(s(p))=p$;Conversely, we should show that $s(r'(B))=A_{m_B}=B$. Note that for any $a\in A-m_B$, we have that $a\in B-m_B$, since $B$ is a local ring, thus $a\in B^*$, so $a\in A^*$, and thus $A_{m_B}\subset B$. On the other hand, for any $x\in B$, if $x\in B-A$, so $1/x\in A, x\not\in m_B$, so $x\in A_{m_B}$, thus $A_{m_B}=B$, finishing the whole proof.

So, in this way, we get a one-to-one correspondence between the set of convex subgroups of $\Gamma$ and the set of valuation rings of $K$ containing $A$. More specifically, for any $H\subset \Gamma$ a convex subgroup, we send it to the maximal ideal of the valuation ring $A(K\rightarrow\Gamma\rightarrow \Gamma/H)$, that is the set $\{a\in A|v(a) which can be shown to be a prime ideal, and conversely, for any prime ideal $p$ of $A$, we send it to the image $v(A_p^*)$, that is the greatest convex subgroup $H$ of $\Gamma$ such that $v(p. Note that this subgroup indeed exists.

This last description clarifies the relation between prime ideals and convex subgroups.

In the next post, we will talk about the all these two types of valuations together.

# Huber space-6

In the upcoming several posts in this series, we will say something on valuation theory. The main reference is Wedhorn’s note on adic spaces.

Valuation theory is very important in non-archimedean analysis. Of course, this theory has many other applications, just like the example given at the beginning of this series of posts, yet non-archimedean analysis provides, or, demands an environnement for valuation theory, simply because the fundamental things like metric, are expressed via valuation.

Before saying anything about valuation theory, we should perhaps first talk about totally ordered groups.

Suppose that $(G,*)$ is an abelian group(written multiplicatively). If $G$ has a total order $(G,\leq)$ such that if $g\leq g'$, then $g''g\leq g''g'$ for any $g,g',g''\in G$. That is to say, this total order respects the algebraic structure on $G$. We call such a group $(G,*,\leq)$ a totally ordered group. We can give $G$ the order topology(open sets are generated by subsets of the form $\{g\in G|g where $g_0$ runs through $G$), thus making $G$ into a topological space. It is rather straight forward to see that $(G,*)$ is a topological group. At last, a morphism between two totally ordered groups $G,G'$ is a group morphism $f:G\rightarrow G'$ such that if $g\leq g'$ in $G$, then $f(g)\leq f(g')$ in $G'$. In the above, we can replace ‘group’ with the word ‘monoid’, the definitions do not change. In this way, we get a category of totally ordered groups(monoids). And clearly the subgroup(sub-monoid) of a totally ordered group(monoid) is again in the category. First let’s give some easy examples. The easiest one is $(\mathbb{R},+,\leq)$, the real numbers with the usual order. Similarly, $(\mathbb{R}_{>0},*,<)$, the positive real numbers. And it is easy to see that these two totally ordered groups are isomorphic. Next, given two totally ordered groups $G_1,G_2$, we can give $G_1\times G_2$ the lexicographic order(that is to say, two elements $(g_1,g_2)\leq(g_1',g_2')$ if $g_1 or $g_1=g_1', g_2\leq g_2'$), then it is easy to see that with this total order $G_1\times G_2$ becomes a totally ordered group. Note that the projection to the second factor $p_2:G_1\times G_2\rightarrow G_2$ is, in general, not a morphism of totally ordered groups. In this way, for any finite number of totally ordered groups, we can give their Cartesian product a total order, making it into a totally ordered group. If a set of totally ordered groups is indexed by some well-ordered set, that is $G_i$ where $i\in I$ with $(I,\leq)$ a well ordered set, then we can still give $\prod_iG_i$ a total order, as follows: for any two elements $(g_i)_{i\in I},(g_i')_{i\in I}$, we set $I'=\{i\in I|g_i\neq g_i'\}$, then since $I$ is well ordered, $I'$ has a minimal element $i_0\in I'$ or $I'$ is an empty set, in the first case, we define $(g_i)<(g'_i)$ if $g_{i'}, and in the second case, we naturally set $(g_i)=(g'_i)$. More general, we can relax the condition that $I$ be well ordered to that $I$ be totally ordered, and we define the Hahn product $HG_i$ t be the subset of $\prod_iG_i$ consisting of elements $(g_i)$ such that $\{i\in I|g_i\neq 1\}$ is well ordered. Then it is not hard to verify that $HG_i$ is again a totally ordered group.

The importance of Hahn product is shown in the following theorem:

Theorem: Every totally ordered group is isomorphic to a subgroup of the Hahn product of copies of $(\mathbb{R},+,\leq)$ indexed by some totally ordered set.

This theorem is striking at first sight, because a priori for the present we know almost nothing about totally ordered groups except the definition. Let’s still try to figure out what this theorem means. Given a totally ordered group $(G,*,\leq)$, and choose a non-trivial element $1\neq g\in G$, then consider the convex subgroup generated this element $g$(by convex subgroup$H$, we mean that for any two elements $h,h'\in H$, if $h\leq g\leq h'$, then $g\in H$, and the convex subgroup generated by some subset is just the intersection of all the convex subgroups of $G$ all of which contain this subset). We denote this convex subgroup by $C_g$, then in some sense, we can create a morphism from $C_g$ to $(\mathbb{R}_{>0},*,\leq)$. How to construct this morphism? For any element $1\leq h\in C_g$(we assume that $g>1$), we consider all the pairs $(n,m)\in\mathbb{Z}^2$ such that $h^n\leq g^m$ and take $D=\min m/n$ for all these pairs. Similarly, consider all the pairs $(n',m')$ of integers such that $h^{n'}\geq g^{m'}$ and take $d=\max m'/n'$. Then take $s=(d+D)/2$, and we define $F:C_g\rightarrow \mathbb{R}_{>0}, h\mapsto e^s$ where we assume that $g>1$ and $e=2.718...$ just for convenience(for those $h<1$, we just consider the inverse of $s$ for $h^{-1}$). We should show that this is a morphism of totally ordered groups. That is, if $h in $C_g$, then $s\leq s'$. For any two pairs of integers $(n,m),(n',m')$ such that $h^n\leq g^m,h'^{n'}\leq g^{m'}$, then we rewrite $h^{nm'}\leq g^{mm'}, h'^{n'm}\leq g^{mm'}$. Then consider all the possibilities of these pairs, we should have that $D\leq D'$ and similarly $d\leq d'$, and thus $s\leq s'$. There is a little problem with this construction, that is, $F$ can be non-injective. Here comes some modification. If there are two elements $h,h'$ such that $s=s'$, this in some sense means that we can pick $h$ out and define a morphism $F_1: C_g\rightarrow \mathbb{R}_{>0}\times 2^{\mathbb{Z}}$. And we can continue this procedure until all such cases are examined and at last we should get some morphism $F_I:C_g\rightarrow \prod_{i\in I}\mathbb{R}_{>0}$ which is injective.

Next we study the morphism between two totally ordered groups, $F:G_1\rightarrow G_2$. Note that $Ker(F)$ is again a subgroup of $G_1$. In fact, $Ker(F)$ has more interesting properties. That is, for any $g\leq g'\leq g''$ in $G_1$ with $g,g''\in Ker(F)$, then $1=F(g)\leq F(g')\leq F(g'')=1$, so this means that $g'\in Ker(F)$. This interesting property deserves a definition:

A subgroup $H$ of a totally ordered group $G$ is called convex(or isolated) if for any $g\leq g'\leq g''$ in $G$ with $g,g''\in H$ implies that $g'\in H$.

There is another equivalent definition for convex subgroup, that is:

A subgroup $H$ of a totally ordered group $G$ is convex if and only if for any $g,g'\leq 1$ in $G$ with $gg'\in H$, we have that $g\in H$ and $g'\in H$.

Proof: suppose that the above condition is satisfied, then for any $g\leq g'\leq g''$ in $G$, we have that $gg'^{-1}\leq 1,g'g''^{-1}\leq 1$. Furthermore, $gg'^{-1g'g''^{-1}}=gg''^{-1}\in H$, so $gg'^{-1}\in H$ and $g'g''^{-1}\in H$, in either case, we have that $g'\in H$ since $H$ is a group; Conversely, if $H$ is convex, then for $g,g'\leq 1$, we have that $gg'\leq g\leq 1,gg'\leq g'\leq 1$. This means that $g,g'\in H$.

We have said that the kernel of a morphism between totally ordered groups give some convex subgroups. In fact, we can see that all convex subgroups arise this way. Indeed, for a convex subgroup $H\subset G$, we see that if $g\in G-H$, then $g or $g>H$(by $g we mean that $g$ is smaller than all the elements in $H$;similar for $g>H$). So, this enables us to define an order on $G/H$. For any $gH$, we say that $gH<1H=H$ if $g. Note that this is indeed a definition, that is, if there $gH=g'H$, that is $g=g'h$ for some $h\in H$, then $g=g'h if and only if $g', which is rather clear. So, in this way, we define an order on $G/H$, and it is straight forward to see that this order is a total order, moreover, the canonical projection $\pi:G\rightarrow G/H$ is a morphism or totally ordered groups. And it is easy to see that $Ker(\pi)=H$. We thus prove that all convex subgroups of $G$ arise as the kernels of some morphisms of totally ordered groups. This is just like the normal subgroups in group theory.

This equivalent definition reminds us of the definition of prime ideals. Indeed, we will see later that there exists some correspondence between the set of prime ideals and the set of convex subgroups.

Before we state this correspondence, we need to define valuations, and for this, we need to modify the category of totally ordered groups.

For any totally ordered group $(\Gamma,*,\leq)$, we add an element $0$ to $\Gamma$ and write it as $\Gamma\bigcup \{0\}$. For the order on $\Gamma\bigcup \{0\}$, we define it as: for two elements in $\Gamma$, they take the same order as in $\Gamma$, and we set $0< \gamma$ for any $\gamma\in \Gamma$. Furthermore, we set that $0\gamma=0$. This $\Gamma\bigcup \{0\}$ resembles to a ring, the difference is that, there is no addition operation, only the multiplication operation.

In the next post, we will say something about valuations, such as the definition and some simple properties.

# perfectoid spaces-4

Suppose that $K$ is a perfectoid field. In this post, we will introduce algebras over $K$ as follows:

A perfectoid $K$-algebra is a Banach $K$ algebra $R$ such that the subset $R^o$ of $R$ consisting of power-bounded elements(as defined in the previous post) is open and bounded, and the Frobenius morphism $F:R^o/\omega\rightarrow R^o/\omega$ is surjective(here the element $\omega$ is defined in the first post of this series). Morphisms between perfectoid $K$-algebras are the continuous morphism of $K$-algebras. A perfectoid $K^{o a}$-algebra is a $\omega$-adically complete flat $K^{o a}$-algebra $R$ on which the Frobenius morphism $F$ induces an isomorphism $F:R/\omega^{1/p}\rightarrow R/\omega$(here $\omega$-adically complete means that $R\simeq \varprojlim R/\omega^n$); morphisms between perfectoid $K^{o a}$-algebras are morphisms of $K^{o a}$-algebras. A perfectoid $K^{o a}/\omega$-algebra is a $\omega$-adically complete flat $K^{o a}$-algebra $R$ on which the Frobenius morphism induces an isomorphism $F:R/\omega^{1/p}\rightarrow R$; morphisms between perfectoid $K^{o a}/\omega$-algebras are morphisms of $K^{o a}/\omega$-algebras.

We will use Scholze’s notations, that is, we write $K-perf$ denotes the category of perfectoid $K$-algebras and $K^{o a}-perf$ denotes the category of perfectoid $K^{o a}$-algebras, and so on.

The following proposition shows that perfectoid $K$-algebras and perfectoid $K^{o a}$-algebras are closely related:

If $R$ is a perfectoid $K$-algebra, then the Frobenius morphism $F$ induces an isomorphism $R^o/\omega^{1/p}\simeq R^o/\omega$, and $R^{o a}$ is a perfectoid $K^{o a}$-algebra.

Proof: By definition, $F$ is surjective. Suppose that $x\in R$ is sent to $\omega R^o$, that is $x^p\in \omega R^o$, say $x^p=\omega y(y\in R^o)$. Since $\omega y$ is power bounded, so is $x$, so $x\in R^o$. Moreover, it is clear that $x\in \omega^{1/p}R^o$, so $F$ induces an isomorphism. Next we need to show that $R^o$ is $\omega$-adically complete. This is not difficult. We define $f:R^o\rightarrow \varprojlim R^o/\omega^n, r\mapsto (r+(\omega^n))$, and $g:\varprojlim R^o/\omega^n\rightarrow R^o,(x_n+(\omega^n))\mapsto \lim_n x_n$. It is necessary to verify that $g$ is well-defined. Indeed, note that, according to the definition, $x_n-x_{n+k}\in (\omega^n)$ for $k>0$, thus $(x_n)_n$ is a Cauchy sequence in $R^o$, as $R$ is complete, so is $R^o$, thus $\lim_n x_n$ exists. Moreover, if $\lim_n x_n=0$, this means that $x_n-0=\lim_k x_n-x_{n+k}\in (\omega^n)$ for all $n$, so $(x_n+(\omega^n))=(0+(\omega^n))$. So, $g$ is indeed well-defined. And it is not difficult to show that $f$ and $g$ are inverse one to the other, so we showed that $R^o$ is $\omega$-adically complete. One last thing is to show that $R^o$ is a flat $K^{o a}$-module. This is not difficult. To finish the proof, we need the following results:

If $N$ is a flat $K^o$-module, then $N^a$ is a flat $K^{o a}$-module.

Conversely, we have that

If $A$ is a $K^{o a}$-algebra, and let $R=A_*[1/\omega]$. If we give $R$ the Banach $K$-algebra structure making $A_*$ open and bounded in $R$, then $A_*=R^o$ is the set of power-bounded elements, and $R$ is perfectoid $K$-algebra, and the Frobenius morphism $F:A_*/\omega^{1/p}\simeq A_*/\omega$.