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.