# 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.