HW10

1. Let $\phi$ be a multiplicative valuation of a field $K$. Then the following holds:

• (i) $\left(a_n\right)$ is a Cauchy sequence $\iff {\lim }_{n\to \infty }\phi \left(a_{n+1}-a_n\right)=0$.
• (ii) If ${\lim }_{n\to \infty }{a}_n=a\ne 0$, then $\phi \left(a_n\right)=\phi (a)$ for a sufficiently large $n$.

\begin{proof} i) If $(a_n)$ is a Cauchy sequence, then for any $k>0$ there is $N_k$ such that $d(a_n,a_m)=\phi (a_n-a_m)<1/k$ for any $n,m>N_k$. Then $\phi (a_{n+1}-a_n)\to 0$ as $n\to \infty$.

Conversely, if ${\lim }_{n\to \infty }\phi (a_{n+1}-a_n)=0$, then for any $k>0$, there exists $N_k$ such that $\phi (a_{n+1}-a_n)<1/k$ for all $n>N_k$. Then for any $m,n>N_k$, we have

$$\begin{array}{rl}& \phi (a_{n+k}-a_n)=\phi (a_{n+k}-a_{n+k-1}+\cdots +a_{n+1}-a_n)\\ ⩽& \max \{\phi (a_{n+k}-a_{n+k-1}),\cdots ,\phi (a_{n+1}-\phi (a_n))\}<1/k.\end{array}$$

Therefore, $\left(a_n\right)$ is a Cauchy sequence.

ii) If ${\lim }_{n\to \infty }{a}_n=a\ne 0$, then ${\lim }_{n\to \infty }d(a_n,a)=0$. Since

$$\phi (a)=\phi (a-a_n+a_n)⩽\max \{\phi (a-a_n),\phi (a_n)\}=\max \{d(a_n,a),\phi (a_n)\}=\phi (a_n)$$

for big enough $n_1$ and

$$\phi (a_n)=\phi (a_n-a+a)⩽\max \{d(a_n,a),\phi (a)\}=\phi (a)$$

for big enough $n_2$, we have $\phi \left(a_n\right)=\phi (a)$ for a sufficiently large $n$. \end{proof}

2. Let $\phi$ be a multiplicative valuation of a field $K$. Let $V$ be a finite dimensional $K$-space with basis $u_1,\dots ,u_n$. For $v=a_1{u}_1+\cdots +a_n{u}_n\in V$, define $\Vert v{\Vert }_0$ by

$$\Vert v{\Vert }_0=\max \left\{\phi \left(a_i\right):1\le i\le n\right\}.$$

Then this is a norm on $V$. Moreover, the following holds concerning the metric induced by this norm.

• (i) Let $v_r=a_{r1}{u}_1+\cdots +a_{rn}{u}_n\in V$ be given for $r=1,2,\dots$. Then the sequence $\left(v_r\right)$ is a Cauchy sequence if and only if the sequence ${\left(a_{ri}\right)}_r$ in $K$ is so for all $i$. Also ${\lim }_{r\to \infty }{v}_r={\sum }_{i=1}^n{a}_i{u}_i$ if and only if ${\lim }_{r\to \infty }{a}_{ri}=a_i$ for all $i$.
• (ii) If $K$ is complete, then so is $V$.

\begin{proof} i) If $(v_r)$ is a Cauchy sequence, then for any $ϵ>0$, there exists $N$ such that $||v_k-v_m|{|}_0<ϵ$ for all $m,k>N$. Note that

$$\begin{array}{rl}||v_k-v_m|{|}_0& =||(a_{k1}-a_{m1})u_1+\cdots +(a_{k1}-a_{m1})u_n|{|}_0\\ & =\max \{\phi (a_{ki}-a_{mi}):1⩽i⩽n\}<ϵ\end{array}$$

for any $m,k>N$, and it deduces that the sequence ${\left(a_{ri}\right)}_r$ in $K$ is Cauchy.

Conversely, if the sequence ${\left(a_{ri}\right)}_r$ in $K$ is Cauchy, then for any $ϵ>0$, there exists $N_1,\cdots ,N_n$ such that $\phi (a_{ki}-a_{mi})<ϵ$ for $i>N_i$. Then we define $N=\max \{N_1,\cdots ,N_n\}$ and there is $||v_k-v_m|{|}_0<ϵ$ for any $k,m>N$. Therefore, $(v_r)$ is Cauchy.

Furthermore, ${\lim }_{r\to \infty }{v}_r={\sum }_{i=1}^n{a}_i{u}_i$ yields that $||{\sum }_{i=1}^n(a_i-a_{ri})u_i|{|}_0\to 0$ as $r\to \infty$. Since

$$||\sum _{i=1}^n(a_i-a_{ri})u_i|{|}_0=\max \{\phi (a_i-a_{ri})\}\to 0,$$

we have ${\lim }_{r\to \infty }{a}_{ri}=a_i$ for all $i$. On the other hand, if ${\lim }_{r\to \infty }{a}_{ri}=a_i$ for all $i$, then by $||{\sum }_{i=1}^n(a_i-a_{ri})u_i|{|}_0=\max \{\phi (a_i-a_{ki})\}$ we can prove that ${\lim }_{r\to \infty }{v}_r={\sum }_{i=1}^n{a}_i{u}_i$.

ii) For any Cauchy sequence $(v_r)$ in $V$, the corresponding ${\left(a_{ri}\right)}_r$ in $K$ is Cauchy and so has limit $a_i$ for all $i$. Thus by i) $(v_r)$ has limit ${\sum }_{i=1}^n{a}_i{u}_i$. Therefore, $V$ is complete. \end{proof}

3. Let $R$ be a ring and $V$ be an $R$-module. Let $I$ be an ideal of $R$. Suppose that $d_I$ is defined on $V$ and on $R$. Then

• (i) Addition is continuous from $V\times V$ to $V$.
• (ii) Multiplication is continuous from $R\times V$ to $V$.
• (iii) If $d_I$ is defined on the $R$-module $W$ and $f\in {\mathrm{Hom}}_R(V,W)$, then $f$ is continuous.

\begin{proof} i) It suffices to show if $\lim a_n=a$ and $\lim b_n=b$, then $\lim (a_n+b_n)=a+b$ for any $(a_n,b_n)\subseteq V$. Since $\lim a_n=a$, we have $||a_n-a|{|}_I\to 0$ as $n\to \infty$. For any $ϵ>0$, there exists $N_1$ such that $||a_n-a|{|}_I<ϵ$ for all $n>N_1$, that is, $a_n-a\in I^{M_1}V$ with ${\gamma }^{M_1}<ϵ$ for all $n>N_1$. Similarly, there exists $N_2$ such that $b_n-b\in I^{M_2}V$ with ${\gamma }^{M_2}<ϵ$ for all $n>N_2$. Then define $N=\max \{N_1,N_2\}$ and we have $a_n+b_n-a-b\in I^{\min \{M_1,M_2\}}V$ and so $||(a_n+b_n)-(a+b)|{|}_I<ϵ$ for all $n>N$. Hence $\lim (a_n+b_n)=a+b$ and so addition is continuous.

ii) If suffices to show for any $\lim r_n=r$ and $\lim a_n=a$, we have $\lim r_n{a}_n=ra$. Since $\lim r_n=r$ and $\lim a_n=a$, for any $ϵ>0$ there exists large enough $N$ such that $r_n-r\in I^N$ and $a_n-a\in I^{N}V$ with ${\gamma }^N<ϵ$. It deduces that $r_n{a}_n-ra=(r_n-r)a_n+r(a_n-a)\in I^{N}V$ and so $d_I(r_n{a}_n,ra)<ϵ$. Therefore, $\lim r_n{a}_n=ra$ and so multiplication is continuous.

iii) It suffices to show for any $v_n\to v$, one have $f(v_n)\to f(v)$. Note that $f(I^{n}V)=I^{n}f(V)\subseteq I^{n}W$. For any $ϵ>0$, there exists $N$ such that $v_n-v\in I^{N}V$ with ${\gamma }^N<ϵ$. Then $f(v_n-v)\in I^{N}W$ for all $n>N$ and so $d_I(f(v_n),f(v))<ϵ$. Therefore, $\lim f(v_n)=f(v)$ and so $f$ is continuous. \end{proof}