We still fix these notations, $L$ is a non-archimedean local field, $k:=k_L$ is a residue field of $L$, $\mathscr{O}:=\mathscr{O}_L$ is its ring of integers with $\pi:=\pi_L$ is a fixed uniformizer, $q=\#k_L$, $B$ is a perfect topological $k_L$-algebra, and $W(B)$ is the ring of Witt vectors with coefficients in $L$. For simplicity, we will denote the addition and multiplication on $W(B)$ as usual, instead of $\boxplus, \boxdot$.
For any open ideal $\mathfrak{a}$ of $B$, we define
$$V_{\mathfrak{a}, m}:=\ker(W(B)\xrightarrow{pr}W_m(B)\xrightarrow{W(pr)}W_m(B/\mathfrak{a}))=$$
$$=\{(b_0,...,b_{m-1})\in W(B)|b_0,...,b_{m-1}\in \mathfrak{a}\}$$
We can see that $V_{\mathfrak{a},m}$ is an ideal of $W(B)$, and
$$V_{\mathfrak{a}\cap \mathfrak{b}, \max{m,n}}\subset V_{\mathfrak{a},m}\cap V_{\mathfrak{b},n}$$
For any open ideal $\mathfrak{b}$ of $B$. And hence, by Proposition 1.6, there exists a unique topological structure on $W(B)$ such that $W(B)$ is a topological ring and that such $V_{\mathfrak{a},m}$ become a fundamental system of open neighborhoods around $0$. If we consider
$$W_m:=\pi^mW(B)=\{(0,...,0,b_m,...)\in W(B)|b_m,b_{m+1},...\in B\}$$
then for any $V_{\mathfrak{a},m}$, we always have $W_m\subset V_{\mathfrak{a},m}$. So, the topology on $W(B)$ we have equipped is weaker than the $\pi$-adic topology on $W(B)$. We call it the weak topology on $W(B)$.
Lemma 3.1. For any $a=(a_0,a_1,...)\in W(B)$, we have
$$a + W_{\mathfrak{a},m}=\{(b_0,b_1,...)\in W(B)|b_i\equiv a_i\mod \mathfrak{a}, 0\le i\le m-1\}$$
Hence, the weak topology on $W(B)$ coincides with the product topology on $B\times B\times ...$
Proof. Take any $(c_0,c_1,...)\in V_{\mathfrak{a},m}$, i.e. $c_0,...,c_{m-1}\in \mathfrak{a}$, we have
$$(a_0,a_1,...) + (c_0,c_1,...)=(a_0+c_0,...)=:(b_0,b_1,...)$$
We can see that $b_0=a_0+c_0\equiv a_0\mod \mathfrak{a}$. Assume that $b_i\equiv a_i\mod \mathfrak{a}$ holds to $n-1$ where $1\le n\le m-1$, we will prove that this holds for $n$. By the addition formula for Witt vectors, we have
$$\Phi_n(a_0,...,a_n)+\Phi_n(c_0,...,c_n)=\Phi_n(b_0,...,b_n)$$
Assume that $b_i=a_i+d_i$ for $d_i\in \mathfrak{a}$, $0\le i\le n-1$, we deduce from the definition of Witt polynomials that
$$\Phi_{n-1}(a_0^q,...,a_{n-1}^q)+\Phi_{n-1}(c_0^q,...,c_{n-1}^q)+\pi^n(a_n+c_n)=\Phi_{n-1}(b_0^q,...,b_{n-1}^q)+\pi^nb_n$$
And this yields
$$\frac{\Phi_{n-1}(a_0^1,...,a_{n-1}^q)+\Phi_{n-1}(c_0^q,...,c_{n-1}^q)-\Phi_{n-1}(b_0^q,...,b_{n-1}^q)}{\pi^n}+a_n+c_n=b_n (*)$$
And we know that
$$\Phi_{n-1}(b_0^q,...,b_{n-1}^q)=\Phi_{n-1}((a_0+d_0)^q,...,(a_{n-1}+d_{n-1})^q)=$$
$$=\Phi_{n-1}(a_0^q+d_0^q,...,a_{n-1}^q+d_{n-1}^q)=(a_0^q+d_0^q)^{q^{n-1}}+...+\pi^{n-1}(a_{n-1}^q+d_{n-1}^q)$$
$$=\Phi_{n-1}(a_0^q,...,a_{n-1}^q)+\Phi_{n-1}(d_0^q,...,d_{n-1}^q)$$
And from $(*)$, we get
$$d+a_n+c_n=b$$
for some $d\in \mathfrak{a}$, and this yields $b_n\equiv a_n\mod \mathfrak{a}$, since $c_n$ is also in $\mathfrak{a}$. We then get
$$a + V_{\mathfrak{a},m}\subset \{(b_0,...,b_{m-1},...)|b_i\equiv a_i\mod \mathfrak{a}, 0\le i\le m-1\} $$
For the converse direction, with the same argument, we deduce that for $b_i\equiv a_i\mod \mathfrak{a}$, for all $0\le i\le m-1$,
$$(b_0,...,b_{m-1},...) - (a_0,...,a_{m-1},...)=(c_0,...,c_{m-1},...)$$
with $c_i\in \mathfrak{a}$, for $0\le i\le m-1$. For the second statement, we can see by the first statement that the set
$$a+V_{\mathfrak{a},m}=\{(b_0,...,b_{m-1})\in W(B)|b_i\equiv a_i\mod \mathfrak{a}, 0\le i\le m-1\}$$
forms a fundamental system of open neighborhoods around $a$. And this follows directly that the weak topology on $W(B)$ is the same as the product topology $B\times B\times ...$. (Q.E.D)
Via this lemma, we can prove
Proposition 3.2. If $B$ is Hausdorff (complete), then $W(B)$ is Hausdorff (complete, resp.).
Proof. It follows easily that if $B$ is Hausdorff then the product topology $B\times B\times...$ is also Hausdorff. Now, assume that $B$ is complete. In this case, the canonical map
$$\phi: B\to \varprojlim_{\mathfrak{a}} B/\mathfrak{a} $$
is surjective. Let $\mathfrak{c}$ be its kernel, we have $B/\mathfrak{c}\cong \varprojlim_{\mathfrak{a}} B/\mathfrak{a}$. And this yields
$$W_m(B/\mathfrak{c})\cong W_m(\varprojlim_{\mathfrak{a}} B/\mathfrak{a})\cong \varprojlim_{\mathfrak{a}} W_m(B/\mathfrak{a})=\varprojlim_{\mathfrak{a}}W(B)/V_{\mathfrak{a},m}$$
where the second isomorphism comes from the functorial properties of Witt vectors, and the last isomorphism follows from the fact that the map $W(B)\xrightarrow{W(pr)} W(B/\mathfrak{a})\xrightarrow{pr} W_m(B/\mathfrak{a})$ has kernel $V_{\mathfrak{a},m}$.
Now, it follows from our (W.4) in the chapter about Witt vectors that
$$W(B/\mathfrak{c})\cong \varprojlim_{m}W_m(B/\mathfrak{c})\cong \varprojlim_m\varprojlim_{\mathfrak{a}}W(B)/V_{\mathfrak{a},m}$$
And so, we obtain the following surjective map
$$W(B)\xrightarrow{W(pr)} W(B/\mathfrak{c})\cong \varprojlim_m\varprojlim_{\mathfrak{a}}W(B)/V_{\mathfrak{a},m}$$
And this yields $W(B)$ is complete. (Q.E.D)
Remark 3.3. In the case $B$ is complete and Hausdorff, we can equip the topological structure on the ring $W_m(B)$, with $m$ is fixed, by the method in the proof, such that $W_m(B)$ is Hausdorff and complete.
Proof. In this case, the canonical map $B\to \varprojlim_{\mathfrak{a}}B/\mathfrak{a}$ is bijective, and this yields by the previous proof that
$$W(B)/V_m(B)=W_m(B)\cong \varprojlim_{\mathfrak{a}}W_m(B/\mathfrak{a})\cong$$
$$\cong \varprojlim_{\mathfrak{a}} W(B)/V_{\mathfrak{a},m}\cong \varprojlim_{\mathfrak{a}}(W(B)/V_m(B))/(V_{\mathfrak{a},m}/V_m(B))$$
From this, there exists a unique topological structure on $W_m(B)$, such that $\{V_{\mathfrak{a},m}/V_m(B)|\mathfrak{a}\subset B: \text{ open ideal }\}$ becomes a fundamental system of open neighborhood around $0$. And $W_m(B)$ is also Hausdorff, and complete. (Q.E.D)
We will be mainly interested in the case $B:=\mathscr{O}_F$, where $F$ is a complete, non-archimedean, perfect field containing $k$. In this case, we get $W(B)$ is Hausdorff, complete, and is a subring of $W(F)$.
Lemma 3.4. Let $\mathscr{O}_F$ be as above, then an ideal $\mathfrak{a}$ of $\mathscr{O}_F$ is open iff $\mathfrak{a}$ is non-zero.
Proof. Assume that $\mathfrak{a}$ is open, then it is obvious that $\mathfrak{a}$ is non-zero. Now, let $\mathfrak{a}\subset \mathscr{O}_F$ be any non-zero ideal. Take $0\ne x\in \mathfrak{a}$, it is sufficient to prove that $(x)$-the ideal generated by $x$ is open in $\mathscr{O}_F$. We can see that
$$(x) =\{y\in \mathscr{O}_F||y|\le |x|)\}$$
Let us take any $z\in \mathscr{O}_F$, such that $|y-z|<|x-y|$. This yields $|y-z|< \max\{x,y\}\le |x|$. From this, we have $|z|\le |x|$, and $z\in (x)$. This yields $(x)$ is open, and hence, $\mathfrak{a}$ is open. (Q.E.D)
We can define for any open ideal $\mathfrak{a}$ of $\mathscr{O}_F$, and any $m\ge 1$ an $\mathscr{O}_F$-submodule
$$U_{\mathfrak{a},m}:=V_{\mathfrak{a},m} + \pi^mW(F):=\{(b_0,...,b_{m-1},...)\in W(F)|b_0,...,b_{m-1}\in \mathfrak{a}\}$$
We note that $U_{\mathfrak{a},m}$ are not ideals of $W(F)$, and we again have
$$U_{\mathfrak{a}\cap \mathfrak{b},\max\{m,n\}}\subset U_{\mathfrak{a},m}\cap U_{\mathfrak{b},n}$$
And these subgroups satisfy the conditions of Proposition 1.3. So, there exists a unique topology on $W(F)$, such that $W(F)$ is a topological group, and $U_{\mathfrak{a},m}$ forms a fundamental system of neighborhoods around $0$. We recall that because $F$ is an extension of $k$, and perfect, $W(F)$ is a D.V.R, with maximal ideal generated by $\pi$. Again, the topology we have equipped for $W(F)$ is weaker that the $\pi$-adic topology. We can call it the weak topology on $W(F)$. We actually want to prove that this topology actually defines a topological ring on $W(F)$, and that when $\mathscr{O}_F$ admits a filtered fundamental system, then $W(F)$ is complete.
We will need the multiplicative property of Teichmuller's representative.
Lemma 3.5.
(i) Let $a_1,...,a_r\in W(F)$, then there exists $0\ne \alpha\in \mathscr{O}_F$, such that
$$\tau(\alpha)a_1,...,\tau(\alpha)a_r\in U_{\mathscr{O}_F,m}$$
(ii) Let $\mathfrak{a}$ be an open ideal of $\mathscr{O}_F$, then for any $0\ne \alpha\in \mathscr{O}_F$, and $m\ge 1$, we have
$$\tau(\alpha^{-1})U_{\alpha^{q^{m-1}}\mathfrak{a},m}\subset U_{\mathfrak{a},m}$$
Proof.
(i) By Proposition 4.6 in the section about Witt vectors, we can represent
$$a_i=\sum_{j\ge 0}\tau(a_{i,j})\pi^j$$
And from this
$$\tau(\alpha)a_i=\sum_{j\ge 0}\tau(\alpha a_{i,j})\pi^j=(\alpha a_{i,0},\alpha a_{i,1},...)$$
And we can choose $\alpha$ such that $\alpha a_{i,j}\in \mathscr{O}_F$, for all $1\le i\le r, 0\le j\le m-1$.
(ii) Take $a=(a_0,a_1,...)\in \alpha^{q^{m-1}}\mathfrak{a}$, we can represent
$$(\alpha_0,\alpha_1,...)=\sum_{i\ge 0}\tau(a_i^{1/q^i})\pi^i$$
Hence
$$\tau(\alpha^{-1})a=\sum_{i\ge 0}\tau(\alpha^{-1}a_i^{1/q^i})\pi^i=\sum_{i\ge 0}(\alpha^{-q^i}a_i)$$
And hence, $\alpha^{-q^i}a_i\in U_{\mathfrak{a},m}$, for all $0\le i\le m-1$. (Q.E.D)
We are now ready for the main result of this section.
Proposition 3.7.
(i) $W(F)$ is a Hausdorff topological ring.
(ii) If $\mathscr{O}_F$ admits a filtered fundamental system, then so is $W(F)$, and in this case, $W(F)$ is complete.
Proof.
(i) We will prove that the multiplication map
$$W(F)\times W(F)\to W(F)$$
is continuous. Take any $a,b\in W(F)$, and an open neighborhood of $ab+U_{\mathfrak{a},m}$, for some open ideal $\mathfrak{a}$ of $\mathscr{O}_F$, and $m\ge 1$. By Lemma 3.6, one can find $0\ne \alpha\in \mathscr{O}_F$ such that $\tau(\alpha)a,\tau(\alpha)b\in U_{\mathscr{O}_F,m}$, which is equivalent to $a,b\in \tau(\alpha^{-1})U_{\mathscr{O}_F,m}$. By Lemma 3.5 again, we have
$$(a+U_{\alpha^{q^{m-1}}\mathfrak{a},m})(b+U_{\alpha^{q^{m-1}}\mathfrak{a},m})\subset ab+U_{\mathscr{O}_F,m}U_{\mathfrak{a},m}+U_{\mathfrak{a},m}\subset ab+ U_{\mathfrak{a},m}$$
And by Lemma 3.4, $U_{\alpha^{q^{m-1}}\mathfrak{a},m}$ is open. Hence, $W(F)$ is a topological ring. Moreover, one get easily that
$$\cap_{\mathfrak{a},m}U_{\mathfrak{a},m}=\cap_{\mathfrak{a},m}\{(b_0,...,b_{m-1},...)\in W(F)|b_i\in \mathfrak{a}\}=0$$
since $\mathscr{O}_F$ is Hausdorff, and the intersection of all open ideals is just 0.
(ii) Let $I_1\supset I_2\supset ...$ be a filtered fundamental system of $\mathscr{O}_F$. Let
$$U_{n,m}:=\{(b_0,...,b_{m-1},...)\in W(F)|b_0,...,b_{m-1}\in I_n\}$$
And $U_m:=U_{m,m}$. Then it can be seen easily that $\{U_{n,m}\}$ forms a fundamental system of open neighborhood around $0$ in $W(F)$, because for any open ideal $\mathfrak{a}\in \mathscr{O}_F$, there exists $n$ such that $I_n\subset \mathfrak{a}$. And one has $U_{n,m}\subset U_{\min{n,m}}, U_m=U_{m,m}$. So $\{U_m\}$ also forms a fundamental system around $0$ in $W(F)$. And this yields $W(F)$ admits a filtration.
Now, in the case $\mathscr{O}_F$ is complete, and admits a filtered fundamental system, it is sufficient to prove any Cauchy sequence converges in $W(F)$. The main ideal of the proof is that we will use Lemma 3.5 to reduce the induced Cauchy sequence to $W_m(\mathscr{O}_F)$, which is complete, by Remark 3.3. And then, with the completeness of the $\pi$-adic topology on $W(F)$, we will prove that our sequence converges in $W(F)$.
Take any $(a_n)_n$ is a Cauchy sequence in $W(F)$. Fix an integer $m\ge 1$, then for any $\mathfrak{a}$: open ideal in $\mathscr{O}_F$, there exists an integer $n_{\mathfrak{a}}$ such that for all $n,n'\ge n_\mathfrak{a}$, $a_n-a_{n'}\in U_{\mathfrak{a},m}$. Then by Lemma 3.5, we can choose $0\ne \alpha\in \mathscr{O}_F$, such that
$$\tau(\alpha) a_1,...,\tau(\alpha) a_{n_\mathfrak{a}}\in U_{\mathscr{O}_F,m}$$
And hence, for all $n\ge n_\mathfrak{a}$, we have
$$\tau(\alpha)(a_n-a_{n_\mathfrak{a}})\in \tau(\alpha)U_{\mathscr{O}_F,m}\subset U_{\mathscr{O}_F,m}$$
And hence $(\tau(\alpha)a_n)_n\in U_{\mathscr{O}_F,m}$, and for $n,n'\ge n_{\mathfrak{a}}$, we have
$$\tau(\alpha)(a_n-a_{n'})\in \tau(\alpha)U_{\mathfrak{a},m}\subset U_{\mathfrak{a},m}$$
Take $(b_n)_n\in W(\mathscr{O}_F)$ such that $\tau(\alpha)-b_n\in \pi^mW(F)$. We then have
$$b_n-b_{n'}\in (\tau(\alpha)(a_m-a_n)+\pi^mW(F))\cap W(\mathscr{O}_F)\subset (U_{\mathfrak{a},m}+\pi^m W(F))\cap W(\mathscr{O}_F)=V_{\mathfrak{a},m}$$
for all $n,n'\ge n_{\alpha}$. This yields the sequence $(b_n\mod V_m(\mathscr{O}_F))_n$ is Cauchy, and hence, converges to some $b\mod V_m(\mathscr{O}_F)$ in $W_m(\mathscr{O}_F)$. Hence, for any open ideal $\mathfrak{b}\subset \mathscr{O}_F$, there exists $n_{\mathfrak{b}}$ such that for all $n\ge n_{\mathfrak{b}}$, we have $b-b_n\in V_{\alpha^{q^{m-1}}\mathfrak{b},m}$. Let us denote $a(m):=\tau(\alpha^{-1})b$, then
$$a(m)-a_n=\tau(\alpha^{-1})b-a_n= \tau(\alpha^{-1})(b-\tau(\alpha)a_n)=\tau(\alpha^{-1})(b-b_n+b_n-\tau(\alpha)a_n)$$
$$\subset \tau(\alpha^{-1})(V_{\alpha^{q^{m-1}}\mathfrak{b},m}+\pi^mW(F))\subset U_{\mathfrak{b},m}+\pi^mW(F)\subset U_{\mathfrak{a},m}$$
for all $n\ge n_{\mathfrak{a}}$. Now, if we vary $m$, then we will get
$$a(m+1)-a(m)=(a(m+1)-a_n)-(a(m)-a_n)\in U_{\mathfrak{b},m}$$
for $n$ is sufficiently large. That means
$$a_{m+1}-a_m\in \{(b_0,...,b_{m-1})\in W(B)|b_0,...,b_{m-1}\in \mathfrak{b}, \forall \mathfrak{b}\subset \mathscr{O}_F: \text{ open}\}$$
And this yields $a(m+1)-a(m)\in \pi^mW(F)$. Now, this yields $(a(m))_m$ is a Cauchy sequence with respect to the $\pi$-adic topology, and hence, a Cauchy sequence in the weak topology. Let $a$ be the convergent value of $(a(m))_m$ in the $\pi$-adic topology, we will prove that $a$ is also the convergent value of $(a_n)_n$. For any ideal $\mathfrak{a}\subset \mathscr{O}_F$: open, and any $m$, we have $\pi^m W(F)\subset U_{\mathfrak{a},m}$, and there exists some $n'$ such that $a-a(n')\in \pi^mW(F)$ and $a(n')-a_n\in U_{\mathfrak{a},m}$, for some $n\ge n_\mathfrak{a}$. Hence, $(a-a_n)\in U_{\mathfrak{a},m}$. This yields $a$ is the convergent value of $W(F)$, and $W(F)$ is complete. (Q.E.D)
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