9 250422

Let $\nu$ be an exponential valuation of a field $k$. Fix a real number $\gamma$ with $0<\gamma <1$. Define $\phi :k\to \mathbb{R},a\mapsto {\gamma }^{\nu (a)}$. Then

• $\phi (a)⩾0$ and $\phi (a)=0$ yields $a=0$;
• $\phi (ab)=\phi (a)\phi (b)$
• $\phi (a+b)⩽\max \{\phi (a),\phi (b)\}$
• $\phi (a^{-1})=\phi (a{)}^{-1}$
• $\phi (a)<\phi (b)$ yields $\phi (a+b)=\phi (b)$.

Define the valuation ring $R=\{a\in k:\phi (a)⩽1\}$ and valuation ideal $P=\{a\in k:\phi (a)<1\}$.

Conversely, if we are given a function $\phi :k\to R$ satisfying

• $\phi (a)⩾0$ and $\phi (a)=0$ yields $a=0$;
• $\phi (ab)=\phi (a)\phi (b)$
• $\phi (a+b)⩽\max \{\phi (a),\phi (b)\}$

and define $\nu (a)={\log }_{\gamma }\phi (a)$, then $\nu$ is a valuation on $k$. $\phi$ is called multiplicative valuation of $k$. Two multiplicative valuation ${\phi }_1$ and ${\phi }_2$ are called equivalent if there exists $r\in {\mathbb{R}}_{⩾0}$ such that ${\phi }_2(a)={\phi }_1(a{)}^r$ for all $a\in k$.

Define $d(a,b)=\phi (a-b)$, which is a metric on $k$. Then $k$ is a Hausdorff space and

• $k\times k\to k,(a,b)\mapsto a+b$
• $k\times k\to k,(a,b)\mapsto ab$
• $k^{\times }\times k^{\times }\to k,(a,b)\mapsto a{b}^{-1}$

are continuous.

If every Cauchy sequence in $k$ converges, $k$ is called complete, or a complete field. The associated valuation ring is said to be a complete valuation ring.

Given a valuation field $k$, let $C$ denote the set of Cauchy sequence in $k$. Define $(a_n)+(b_n)=(a_n+b_n)$ and $(a_n)(b_n)=(a_n{b}_n)$. Then $C$ is a commutative ring. If $I$ denotes the subset of $C$ consisting of the sequences converge to $0$, then $I$ is a maximal ideal. Hence $\tilde{k}=C/I$ is a field.

If $(a_n)$ is a Cauchy sequence, then $|\phi (a_m)-\phi (a_n)|⩽\phi (a_m-a_n)$. Hence $(\phi (a_n))$ is a Cauchy sequence in $\mathbb{R}$ and ${\lim }_{n\to \infty }\phi (a_n)$ exists.

If $(a_n)-(b_n)\in I$, then ${\lim }_{n\to \infty }\phi (a_n)={\lim }_{n\to \infty }\phi (b_n)$. Define

$$\tilde{\phi }:\tilde{k}\to R,\widetilde{(a_n)}=\underset{n\to \infty }{\lim }\phi (a_n)$$

for $\widetilde{(a_n)}=(a_n)+I$. It is well-defined by ^147dxw. Furthermore, we can check $\tilde{\phi }$ is a multiplicative valuation of $\tilde{k}$.

For $a\in k$, the sequence $(a)=(a,a,a,\cdots )$ is a Cauchy sequence and $\widetilde{(a)}\ne \widetilde{(b)}$ if $a\ne b$. Thus we can regard $k$ as a subfield of $\tilde{k}$ by identifying $a$ with $\widetilde{(a)}$. So $\tilde{\phi }$ is an extension of $\phi$ and $k$ is dense in $\tilde{k}$ by definition.

Lemma

$\tilde{k}$ is a complete field.

Theorem

Let $\tilde{R}$ and $\tilde{P}$ be the valuation ring and valuation ideal of $\tilde{\phi }$ respectively. Then $\tilde{R}/\tilde{P}\simeq R/P$.

\begin{proof} See ^nlam3t. \end{proof}

Let $\nu$ be a normalized discrete valuation of $k$ with valuation ring $R$ and valuation ideal $P=(\pi )$. Then $\{P,P^2,\cdots \}$ is a fundamental neighborhood system of $0$ by ${\cap }_{i=1}^{\infty }({\pi }^i)=0$.

Therefore, $(a_n)$ is a Cauchy sequence iff for any $\ell \in {\mathbb{Z}}_{>0}$, there exists $N$ such that $a_m-a_n\in P^{\ell }$ for all $m,n⩾N$. Also ${\lim }_{n\to \infty }{a}_n=a$ iff given $\ell$, there exists $N$ such that $a_n-a\in P^{\ell }$ for all $n⩾N$.

Lemma

Let ${\phi }_1,{\phi }_2$ be discrete multiplicative valuations of $k$. TFAE:

• ${\phi }_1$ and ${\phi }_2$ are equivalent;
• $\{a\in k:{\phi }_1(a)<1\}=\{a\in k:{\phi }_2(a)<1\}$;
• the topologies on $k$ induced by ${\phi }_1$ and ${\phi }_2$ are same.

\begin{proof} See ^9b52ya. \end{proof}

Definition

Let $\phi$ be a multiplicative valuation of $k$. Let $V$ be a vector space over $k$. A real-valued function $||\cdot ||$ defined on $V$ is said to be a norm on $V$ if

• $||v||⩾0$ and $||v||=0⟺v=0$.
• $||av||=\phi (a)||v||$ for any $a\in k$ and $v\in V$.
• $||u+v||⩽\max \{||u||,||v||\}$.

If we set $d(u,v)=||u-v||$, then $V$ is a metric space. Note that $V\times V\to V,(u,v)\mapsto u+v$ and $k\times V\to V,(a,v)\mapsto av$ are continuous.

Lemma

Let $u_1,\cdots ,u_n$ be a basis of a finite-dimensional $k$-space $V$. For $v=a_1{u}_1+\cdots +a_n{u}_n\in V$, define $||v|{|}_0=\max \{\phi (a_i):1⩽i⩽n\}$. Then $||\cdot |{|}_0$ is a norm on $V$. The following holds for the metric indeed by $||\cdot |{|}_0$.

• Let ${\nu }_r=a_{r_1}{u}_1+\cdots +a_{r_n}{u}_n$ be given for $r=1,2,\cdots$. Then the sequence $\{v_r{\}}_r$ is a Cauchy sequence iff the sequence $(a_{r_i}{)}_r$ is a Cauchy sequence in $k$ for all $i=1,\cdots ,n$. Also ${\lim }_{r\to \infty }{v}_r={\sum }_{i=1}^n{a}_i{v}_i$ iff ${\lim }_{r\to \infty }{a}_{r_i}=a_i$ for all $i$.
• If $k$ is complete, then so is $V$.

\begin{proof} See homework. \end{proof}

Lemma

Suppose that $k$ is complete. Then for any norm $||v||$ on $V$, there exists constant $\lambda ,\mu$ such that $||v||⩽\lambda ||v|{|}_0$, $||v|{|}_0⩽\mu ||v||$ for all $v\in V$. Thus the topological induces by these two norms are the same.

\begin{proof} See ^zs3z31. \end{proof}

Let $K,L$ be discrete valuation fields with multiplicative valuation $\phi$ and $\Phi$, respectively. If $K\subseteq L$ and $\Phi$ is an extension of $\phi$, then we say $(L,\Phi )$ is an extension of $(K,\phi )$, and write $(K,\phi )\subseteq (L,\Phi )$. Consider $L$ as an $K$-space. Let $||x||=\Phi (x)$ for $x\in L$.

Theorem

Let $\phi$ be a complete discrete multiplicative valuation of $K$ and $L$ be a finite extension of $K$. If there exists an extension $(L,\Phi )$ of $(K,\phi )$, then $\Phi$ is unique up to equivalent. Furthermore, $(L,\Phi )$ is complete.

Let $R$ be a ring, and let $V$ be an $R$-module. For an ideal $I\subseteq R$, assume that ${\cap }_{i=1}^{\infty }{I}^{i}V=0$. Fix $\gamma \in \mathbb{R}$ with $0<r<1$, define $||\cdot |{|}_I$ on $V$ by

• $||0|{|}_I=0$
• $||v|{|}_I={\gamma }^i$ for $v\in I^{i}V\setminus I^{i+1}V$.

Define $d_I(u,v):=||u-v|{|}_I$. Check: $d_I(u,v)⩽\max \{d_I(u,w),d_I(w,v)\}$. So $d_I$ is a metric on $V$.

In this situation, we say $d_I$ is defined on $V$.

Note that

• $v_n\in V$, $\{v_n{\}}_n$ is a Cauchy sequence iff for any integer $m>0$, there exists $N>0$ such that $v_i-v_j\in I^{n}V$ for all $i,j⩾N$.
• $v={\lim }_{n\to \infty }{v}_n$ iff for any integer $m>0$, there exists $N>0$ such that $v_i-v\in I^{m}V$ for all $i⩾N$.

Lemma

The maps

• $V\times V\to V$, $(a,b)\mapsto a+b$
• $R\times V\to V$, $(r,a)\mapsto ra$
• $f\in {\mathrm{Hom}}_R(V,W)$, where for $W$ we define $d_I$

are continuous.

Theorem

Assume that $V$ is finitely generated. If $R$ is complete (w.r.t. $d_I$), then $V$ is complete.

\begin{proof} See ^ffpm0c. \end{proof}

Lemma

Suppose that $d_I$ is defined for all finitely generated $R$-modules. If $V$ is a finitely generated $R$-module, then every submodule of $V$ is closed.

\begin{proof} See ^cpwp8z. \end{proof}

Let $R$ be complete discrete valuation ring with valuation $\nu$, and let $P=(\pi )$ and $F=R/P$ be the valuation ideal and residue field of $\nu$ respectively.

Let $V$ be a finitely generated $R$-module. Then $V=R{u}_1\oplus \cdots \oplus R{u}_r$ is a direct sum of cyclic submodule. Since $({\pi }^n)V=({\pi }^n)u_1\oplus \cdots \oplus ({\pi }^n)u_r$, there is ${\cap }_{n=1}^{\infty }{\pi }^{n}V=0$. Then $d_P$ is defined for all finitely generated $R$-module. Since $R$ is complete, the module $V$ is also complete by ^ffpm0c.

Let $A$ be an $R$-algebra, which is finitely generated as an $R$-module.

Fact. $A\times A\to A,(a,b)\mapsto ab$ is continuous.

$A^{\ast }=A/\pi A$ is a finitely generated $F$-algebra.

Theorem

Let $A$ be finitely generated $R$-algebra. Then

• $A\pi \subseteq J(A)$;
• $J(A{)}^n\subseteq A\pi$ for some $n>0$.

\begin{proof} See ^nuyi3u. \end{proof}

Theorem

Let $A$ be a finitely generated $R$-algebra. Let $I$ be an ideal of $A$ with $I\subseteq J(A)$ and $\overline{A}=A/I$.

• Let $\overline{c}$ be an idempotent in $\overline{A}$, and let $\overline{c}={\overline{c}}_1+\cdots +{\overline{c}}_n$ be an idempotent decomposition in $\overline{A}$. Then $\overline{c}$ lifts to an idempotent $e$ of $A$ and there exists orthogonal idempotent $e_1,\cdots ,e_n$ of $A$ satisfying ${\overline{e}}_i={\overline{c}}_i$ and $e=e_1+\cdots +e_n$.
• An idempotent $e$ of $A$ is primitive iff $\overline{e}$ is primitive in $\overline{A}$.
• Let $e,f$ be primitive idempotent of $A$, the. $Ae\simeq Af$ iff $\overline{A}\overline{e}\simeq \overline{A}\overline{f}$.

\begin{proof} See ^4kwh24. \end{proof}

Remark. Let $I=A\pi$, then $\overline{A}=A^{\ast }$ where $A$ is an $R$-algebra, $A^{\ast }$ is an $F$-algebra and $k{\otimes }_{R}A$ is a $k$-algebra.