[Galois cohomology III] Cohomology Groups as Derived Functors

Let $M$ be a $G$-module, and $M,N$ $G$-modules. One can consider $M$ as an $\mathbb{Z}[G]$-module by an induced action from $G$, and vice versa. Also, any $G$-module homomorphism  between $M$, $N$ can be considered as a $\mathbb{Z}[G]$-module homomorphism, and vice versa. Say another words, one can identify $Mod_G$ to $Mod_{\mathbb{Z}[G]}$. Hence, $Mod_G$ is an abelian category. In this section, we will use the theory of derived functors to deduce some important results: Shapiro's lemma and the description of cohomology groups in terms of co-chains.

We denote $M^G:=\{m\in m|gm=m,\forall g\in G\}$, then $M^G$ is an abelian group, and this define a functor $(.)^G:Mod_G\to Ab$ defined by $M\mapsto M^G$. Let

$$0\to M\to N\to P\to 0$$

be an exact sequence of $G$-modules, we then apply the functor $(.)^G$ to obtain the following exact sequence of abelian groups

$$0\to M^G\to N^G\to P^G$$

And this yields $(.)^G$ is a left exact functor.

Proposition 3.1. $Mod_G$ has enough injectives.

Proof. We recall that an abelian category $C$ is said to e enough injective if for all object $M$ in $C$, there exists a monomorphism $M\to I$ where $I$ is an injective object in $C$, i.e. the functor $Hom_C(_,I)$ is exact. 

Let $M$ be a $G$-modules, and $M_0$ its underlying abelian group structure. We know that the category of abelian group has enough injectives, so there exists an injective object $I$ in $Ab$ such that $M_0\hookrightarrow I$, and since $Ind^G$ is an exact functor, we have $Ind^G(M_0)\hookrightarrow Ind^G(I)$, which is also an injective object in $Mod_G$, since $Ind^G$ is exact. And by Proposition 2.6, we can embed $M$ into $Ind^G(M_0)$. And this yields $Mod_G$ has enough injectives. (Q.E.D)

Proposition 3.2. Let $M$ be in $Mod_G$, then there exists an injective resolution of $M$, i.e. there exists an exact sequence of $G$-modules 

$$0\to M\to I^0\to I^1\to...$$

of $G$-modules, and $I^i$ is injective object in $G$-modules, for all $i$.

Proof. Due to Proposition 3.1, there exists $I^0$: an injective object in $Mod_G$ such that $f_0: M\hookrightarrow I^0$. Let $B^0$ be the cokernel of the map $f_0$, i.e. $B^0=I^0/Im f^0$, then we can embed $B^0\hookrightarrow I^1$, where $I^1$ is an injective object in $Mod_G$. And the composition $I^0\to B^0\to I^1$ has the kernel exactly $Im(f_0)$. Continuing this process, we obtain the desired exact sequence

$$0\to M\to I^0\to I^1\to...$$

(Q.E.D)

Fon Proposition 3.2, assume that 

$$0\to M\to I^0\to I^1\to...$$

is an injective resolution of a $G$-module $M$. If we apply the functor $(.)^G$, we then obtain the following complex of abelian groups

$$0\xrightarrow{d^{-1}}(I^0)^G\xrightarrow{d^0}(I^1)^G\xrightarrow{d^1}...$$

And we can define $H^i(G,M):=\ker d^i/Im (d^{i-1})$, which is called the $i$-th cohomology group. The theory of derived functors implies that the $i$-th cohomology group $H^i(G,M)$ does not depend on the choice of injective resolution of $M$. And it also follows that if we begin with a short exact sequence

$$0\to M\to N\to P\to 0$$

of $G$-modules, then we will get a long exact sequence of abelian groups

$$0\to H^0(G,M)\to H^0(G,N)\to H^0(G,P)\to H^1(G,M)\to ...$$

Proposition 3.3. $H^0(G,M)=M^G$ for any $G$-module $M$.

Proof. We have $H^0(G,M)=\ker d^0$, and since $(.)^G$ is a left exact functor, we have

$$0\to M^G\to (I^0)^G\to (I^1)^G$$

is exact, and this yields $\ker d^0=Im(M\to (I^0)^G)=M^G$. (Q.E.D)

We can consider $\mathbb{Z}$ as a $G$-module with trivial action from $G$. Let $M$ be any $G$-module , then any $G$-module homomorphism $f$ from $\mathbb{Z}$ to $M$ is uniquely determined by $f(1)$, and since $f(1)=f(g1)=gf(1)$, we have $f(1)\in M^G$. And it follows that $Hom_G(\mathbb{Z},M)=M^G$. Using this, one can prove the Shapiro's lemma

Proposition 3.4. Let $H$ be a subgroup of $G$, and $N$ an $H$-module, then there is a canonical isomorphism $H^r(G, Ind^G_H(N))\cong H^r(H,N)$, for all $r\ge 0$.

Proof. When $r=0$, we have

$$H^0(H,N)=N^H=Hom_H(\mathbb{Z},N)=Hom_G(\mathbb{Z},Ind^G_H(N))=Ind^G_H(N)^G=H^0(G,Ind^G(N))$$

where the third identity follows from the Proposition 2.3. Let

$$0\to N\to I^0\to I^1\to...$$

be an injective resolution of $N$ in $Mod_H$, then since $Ind^G_H$ is an exact functor, which preserves injectives, we have

$$0\to Ind^G_H(N)\to Ind^G_H(I_0)\to Ind^G_H(I_1)\to ...$$

is an injective resolution of $Ind^G_H(N)$, and

$$H^r(G, Ind^G_H(N))=H^r(Ind^G_H(I^\bullet)^G)\cong H^r((I^\bullet)^H)=H^r(H,N)$$

where the second identity follows from the previous argument that $Ind^G_H(I^j)^G=(I^j)^H$. (Q.E.D)

As a corollary, we get

Corollary 3.5. If $M$ is an induced $G$-module, then $H^r(G,M)=0$, for all $r>0$.

Proof. We can write $M=Ind^G(N)$, for some abelian group $N$. According to Proposition 3.4, we have

$$H^r(G, Ind^G(N))\cong H^r({1},N)=0\forall r>0$$

(Q.E.D)

To make things more explicit, we need to describe cohomology groups in terms of co-chains. And we will need some more about projective resolution. Because we can identify $Mod_G$ and $Mod_{\mathbb{Z}[G]}$, it follows that if $P$ is a free $\mathbb{Z}[G]$-module, then $P$ is a projective object in $Mod_G$. We recall that $P$ is a projective object in an abelian category $\mathscr{C}$ if $Hom_{\mathscr{C}}(P,_)$ is an exact functor. Equivalently, any morphism from $P$ to a quotient object $M/N$ can be lifted to a morphism from $P$ to $M$.

Proposition 3.6. $Mod_G$ has enough projectives, i.e. for all $M$ in $Mod_G$, there exists $P$: projective in $Mod_G$, such that $P\to M\to 0$ is exact.

Proof. Let $(m_i)_{i\in I}$ be the set of generators of $M$ as $G$-modules, i.e. for all $n\in M$, there exists $(g_i)_{i\in I}$, such that $m=\sum_{i}g_im_i$. Consider the $G$-module $\mathbb{Z}[G]^I$, which is a direct sum of $\mathbb{Z}[G]$, with indexes run on $I$. Then the map $\mathbb{Z}[G]^I\to M$ defined by $\sum_i \gamma_i\mapsto \sum_i\gamma_im_i$, this map is clearly a surjective $G$-module homomorphism. And $\mathbb{Z}[G]^I$ is projective $G$-module. (Q.E.D)

Proposition 3.7. Any $G$-module $M$ has a projective resolution, i.e. there exists $P_0,P_1,...$: projective $G$-modules, such that

$$...\to P_2\to P_1\to P_0\to M\to 0$$

is an exact sequence of $G$-modules.

Proof. From Proposition 3.6, there exists a projective object $P_0$ in $Mod_G$ such that

$$P_0\xrightarrow{f_0} M\to 0$$

is exact. Let $B_0:=\ker (P_0\to M)$, there exists a projective object $P_1$ in $Mod_G$ such that $P_1\to B_0\to 0$ is exact. And it is clear that $P_1\to P_0\to M\to 0$ is exact. Continuing this process, we get that $M$ admits a projective resolution. (Q.E.D)

Now, let $A,B$ be two $G$-modules, let $B\to I^\bullet$ be an injective resolution of $B$, then since $Hom_G(A,_)$ is a left exact functor, we get 

$$0\to Hom_G(A,I^0)\to Hom_G(A,I^1)\to... $$

is a complex in $Ab$, and let us denote $Ext^r(A,B):=H^r(Hom_G(A,I^bullet))$

Now, let $A\leftarrow P_\bullet$ be a projective resolution of $A$, then since $Hom_G(_,B)$ is a contravariant left exact functor, we have:

$$0\to Hom_G(A,B)\to Hom_G(P_0,B)\to Hom_G(P_1,B)\to...$$

is a complex in $Ab$, and we denote $Ext'^r(A,B):=H^r(Hom_G(P_\bullet,B))$. And an interesting thing is that $Ext^r(A,B)=Ext'^r(A,B)$ [Milne ...]. Using this, we can deduce an interesting result

Proposition 3.8. Let $M$ be a $G$-module, and $\mathbb{Z}$ with the trivial action from $G$. Let $\mathbb{Z}\leftarrow P_\bullet$ be a projective resolution of $\mathbb{Z}$, then

$$H^r(Hom_G(P_\bullet,M))\cong H^r(G,M)$$

Proof. We have $H^r(Hom_G(P_\bullet,M))=Ext^r(\mathbb{Z},M)$, and $Ext^r$ is the $r$-th derived functor of $Hom_G(\mathbb{Z},-)$, but it follows that $Hom_G(\mathbb{Z},-)\equiv (.)^G$, so their $r$-th derived functor are the same. (Q.E.D)

We can use Proposition 8 to write explicitly elements in a cohomology groups. Let $G$ be a group, we denote $P_r$ the free $\mathbb{Z}$-module generated by tuples $(g_0,...,g_{r})$, where $g_i\in G$. We define a map $d_r: P_r\to P_{r-1}$ defined by $(g_0,...,g_r)\mapsto \sum_{i=1}^r(-1)^i (g_0,...,\hat{g_i},...,g_r)$. And it can be seen that

$$...\xrightarrow{d_{r+1}}P_r\xrightarrow{d_r} P_{r-1}\xrightarrow{d_{r-1}}...\xrightarrow{d_2}P_1\xrightarrow{d_1}P_0\xrightarrow{d_0} \mathbb{Z}\to 0 (*)$$

is a complex of $G$-modules, where $G$ acts trivially on $\mathbb{Z}$.

Lemma 3.9. The complex $(*)$ is exact.

Proof. Fix $g\in G$, let us define the map $e_r: P_r\to P_{r+1}$ defined by $(g_0,...,g_r)\mapsto (g,g_0,...,g_r)$, then it can be seen easily that $d_{r+1}e_r+e_{r-1}d_r=Id_{P_r}$. And so, $\alpha\in \ker d_r$ implies that $d_{r+1}\circ e_r(\alpha)=\alpha$, i.e. $\alpha\in Im(d_{r+1})$. So, the sequence above is exact. (Q.E.D)

From this lemma, we can see that, in fact, $P_\bullet\to \mathbb{Z}$ is a projective resolution of $\mathbb{Z}$. And from Proposition 3.8, we have $H^r(Hom_G(P_\bullet,M))\cong H^r(G,M)$, where the first group begins with the following complex

$$0\to Hom_G(P_0,M)\to Hom_G(P_1,M)\to...$$

Basically, $\tilde{\varphi}\in Hom_G(P_r,M)$ iff $g\tilde{\varphi}(g_0,...,g_r)=\tilde{\varphi}(gg_0,...,gg_r)$. And the map induced from $Hom_G(P_r,M)$ to $Hom_G(P_{r+1},M)$ is defined to be ${d^r}\tilde{\varphi}:=\tilde{\varphi}\circ d_r$, and

$${d^r}\tilde{\varphi}(g_0,...,g_{r+1})=(\tilde{\varphi}\circ d_r)(g_0,...,g_{r+1})=\sum_{i=1}^{r+1}(-1)^i\tilde{\varphi}(g_0,...,\hat{g_i},...,g_{r+1})$$

And we denote 

$$\widetilde{C^r}(G,M):=Hom_{G}(P_r,M)=\{\tilde{\varphi}:G^{r+1}\to M|g\tilde{\varphi}(g_0,...,g_r)=\tilde{\varphi}(gg_0,...,gg_r)\}$$

And the map $d^r: \widetilde{C^r}(G,M)\to \widetilde{C^{r+1}}(G,M)$ is defined to be $\widetilde{\varphi}\mapsto d^r\circ \widetilde{\varphi}$. On the other hand, we will prove that

Lemma 3.10. There are a bijective map between $\widetilde{C^r}(G,M)$ and $C^r(G,M):=\{G^r\to M\}$ defined by $\widetilde{\varphi}(1,g_1,g_1g_2,...,g_1...g_n)=:\varphi(g_1,...,g_n)$, for all $g_1,...,g_r\in G$.

Proof. Assume that $\widetilde{\varphi_1}(1,g_1,...,g_1...g_r)=\varphi(g_1,...,g_r)=\widetilde{\varphi_2}(1,g_1,...,g_1...g_r)$, for all $g_i\in G$, for all $g_i\in G$. If we let $h_i:=g_{i-1}^{-1}g_i$, then

$$\widetilde{\varphi_1}(g_0,...,g_r)=g_0\widetilde{\varphi_1}(1,g_0^{-1}g_1,...,g_0^{-1}g_r)=g_0\widetilde{\varphi_1}(1,h_1,...,h_1...h_r)=$$

$$=g_0\varphi(g_0^{-1}g_1,g_1^{-1}g_2,...,g_{r-1}^{-1}g_r)=g_0\widetilde{\varphi_2}(1, g_0^{-1}g_1,...,g_{r-1}^{-1}g_r)=\widetilde{\varphi_2}(g_0,...,g_r)$$

So we get $\widetilde{\varphi_1}=\widetilde{\varphi_2}$. Now, take any $\varphi: G\to M$, we have to define $\widetilde{\varphi}\in \widetilde{C^r}(G,M)$, such that $\widetilde{\varphi}(1,g_1,...,g_1...g_r)=\varphi(g_1,...,g_r)$. We define 

$$\widetilde{\varphi}(g_0,...,g_r):=g_0\widetilde{\varphi}(g_0^{-1}g_1,...,g_{r-1}^{-1}g_r)$$

And it is easy to check that $\widetilde{\varphi}(1,g_1,g_1g_2,...,g_1...g_r)=\varphi(g_1,...,g_r)$. And also,

$$g\widetilde{\varphi}(g_0,...,g_r)=gg_0\widetilde{\varphi}(g_0^{-1}g_1,...,g_{r-1}^{-1}g_r)$$

And $\widetilde{\varphi}(gg_0,...,gg_r)=gg_0\varphi(g_0^{-1}g_1,...,g_{r-1}^{-1}g_r)$. And hence, $\widetilde{\varphi}\in \widetilde{C^{r+1}}(G,M)$. (Q.E.D)

We can see that the map in Lemma 3.10 defines an isomorphism between $\widetilde{C^r}$ and $C^r$. And we can build a map $d^r: C^r(G,M)\to C^{r+1}(G,M)$ such that it is compatible with the map $d^r: \widetilde{C^r}(G,M)\to \widetilde{C^{r+1}}(G,M)$. 

For the case $r=0$, for any $\widetilde{\varphi}\in \widetilde{C^0}(G,M)$, we have in this case $\widetilde{\varphi}(1)=\varphi(1)$. And

$$(d^0\widetilde{\varphi})(g_0,g_1)=\widetilde{\varphi}(g_1)-\widetilde{\varphi}(g_0)=g_1\widetilde{\varphi}(1)-g_0\widetilde{\varphi}(1)=g_1\varphi(1)-g_0\varphi(1)$$

And

$$(d^0\widetilde{\varphi})(g_0,g_1)=g_0(d^0\widetilde{\varphi})(1,g_0^{-1}g_1)=g_0(d^0\varphi)(g_0^{-1}g_1)$$

Hence, $g_0(d^0\varphi)(g_0^{-1}g_1)=g_1\varphi(1)-g_0\varphi(1)$, i.e. $(d^0\varphi)(g_0^{-1}g_1)=g_0^{-1}g_1\varphi(1)-\varphi(1)$. If we denote $h:=g_0^{-1}g_1$, we then get $(d^0\varphi)(h):=h\varphi(1)-\varphi(1)$.

When $r\ge 1$, by similar argument, for any $\varphi\in C^r(G,M)$, we have

$$(d^r\varphi)(g_1,...,g_{r+1})=g_1\varphi(g_2,...,g_{r+1})-\sum_{i=1}^r(-1)^i\varphi(g_1,...g_{i}g_{i+1},...,g_r)+(-1)^{r+1}\varphi(g_1,...,g_r)$$

And this yields $H^r(G,M)=H^r(\widetilde{C^\bullet}(G,M))=H^r(C^\bullet(G,M))=\ker d^r/Im (d^{r-1})$. We can now write things explicitly to $H^2$. For $H^0(G,M)$, we already know that it is $M^G$. And for $H^1$, take any $\varphi: G\to M$, we have $\varphi\in \ker d^1$ iff $\varphi(g_1g_2)=g_1\varphi(g_2)+\varphi(g_1)$, and such $\varphi$ is said to be a crossed homomorphism. And $\varphi\in Im(d^0)$ iff $\varphi(g)=g\varphi(1)-\varphi(1)$, i.e. $\varphi(g)=gm-m$, for some $m\in M$, and such $\varphi$ is said to be a principal homomorphism. So, we get

$$H^1(G,M)=\frac{\text{crossed homomorphisms from G to M}}{\text{principal homomorphisms from G to M}}$$

For $H^2$, we have $\varphi\in \ker d^2$ iff $\varphi:G\times G\to M$ satisfying

$$\varphi(g_1g_2,g_3)+\varphi(g_1,g_2)=g_1\varphi(g_2,g_3)+\varphi(g_1,g_2g_3)$$

And $\varphi\in Im(d^1)$ iff $\varphi(g_1,g_2)=h_1\varphi'(h_2)-\varphi'(h_1h_2)+\varphi'(h_1)$, for some $\varphi': G\to M$. And the quotient of them yields the group $H^2(G,M)$.

[Galois Cohomology II] Induced Modules

For basic definition of group cohomology, one can read it the book of Lorenz Falco (Algebra II). We will now treat group (co)homology in terms of derived functors, as presented in the lecture notes of Milne. In the first section, we will introduce about induced modules. Roughly speaking, it defines an exact functor from the category of $H$-modules to the category of $G$-modules, where $H$ is a subgroup of $G$. Especially, when $H=\{1\}$, then we obtain an exact functor from the category of abelian groups to the category of $G$-modules.

Definition. Let $G$ be a group, an abelian group $M$ is said to be a $G$-module if there exists an action from $G$ to $M$, and $g(m_1+m_2)=gm_1+gm_2$, for all $g\in G, m_1,m_2\in M$. Let $M,N$ be $G$-modules, a map $\varphi: M\to N$ is said to be a $G$-module homomorphism if $\varphi$ is a homomorphism of abelian groups, and $\varphi(gm)=g\varphi(m)$, for all $g\in G,m\in M$. The set of all $G$-module homomorphisms between $M$ and $N$ is denoted by $Hom_G(M,N)$. We also denote $Mod_G$ the category with objects consisting $G$-modules, and morphisms are $G$-modules homomorphisms.

Let $G$ be a group and $H$ a subgroup of $G$, and $M$ is an $H$-module, we denote:

$$Ind^G_H(M):=\{\varphi: G\to M|\varphi(hg)=h\varphi(g), \forall h\in H,g\in G\}$$

One can equip $Ind^G_H(M)$ a structure of $G$-module by defining $(\varphi_1+\varphi_2)(g)=\varphi_1(g)+\varphi_2(g)$, and $(g_1\varphi)(g)=\varphi(gg_1)$.

Lemma 2.1. The map $\phi_M: Ind^G_H(M)\to M$ defined by $\varphi\mapsto \varphi(1_G)$ is an $H$-module homomorphism.

Proof. It can be seen that $\phi$ is a homomorphism of abelian groups, and

$$\phi_M(h\varphi)=(h\varphi)(1_G)=\varphi(1_Gh)=\varphi(h1_G)=h\varphi(1_G)=h\phi_M(\varphi)$$

(Q.E.D)

Lemma 2.2. With the assumption in Lemma 2.1, when $H=G$, then $\phi_M$ is an isomorphism.

Proof. We have $Ind^G_G(M)=\{\varphi:G\to M|\varphi(g_1g_2)=g_1\varphi(g_2),\forall g_1,g_2\in G\}$, and any $\varphi$ in this set is completely determined by $\varphi(1)$. And the map $\phi_M$ above is injective and surjective. (Q.E.D)

Proposition 2.3. Let $G$ be a group, $H$ a subgroup of $G$, $M$ a $G$-module, and $N$ an $H$-module, then there exists a canonical isomorphism $Hom_G(M, Ind^G_H(N))\cong Hom_H(M,N)$.

Proof. Let $\phi$ be a $G$-module homomorphism from $M$ to $Ind^G_H(M)$, then the map $\phi':=\phi_N\circ \phi: M\to N$, where $\phi_N$ is defined in Lemma 2.1 is an $H$-module homomorphism.

Conversely, if we begin with $\phi':M\to N$ an $H$-module homomorphism, then for any $\alpha\in Ind^G_G(M)$, we have $\phi'\circ \alpha\in Ind^G_H(M)$, since for all $h\in H,g\in G$, we have

$$(\phi'\circ \alpha)(hg)=\phi'(h\alpha(g))=h(\phi'\circ \alpha(g))$$

And we hence obtain a map $\phi: Ind^G_G(M)\to Ind^G_H(N)$ defined by $\alpha\mapsto \phi'\circ \alpha$, and it can be seen that for all $g,g_1\in M$, we have

$$\phi(g\alpha)(g_1)=\phi'\circ (g\alpha)(g_1)=\phi'\circ\alpha(g_1g)=(g(\phi'\circ\alpha))(g_1)=g\phi(\alpha)(g_1)$$

Hence, $\phi$ is a homomorphism of $G$-modules. Due to Lemma 2.2, we have $M\cong Ind^G_G(M)$, and hence, one obtains $Hom_G(M,Ind^G_H(N))\cong Hom_H(M,N)$. (Q.E.D)

Via this theorem, we obtain the universal property of induced modules.

Corollary 2.4. With the assumption in Proposition 2.3, then for any $H$-module homomorphism $\phi'$ from $M\to N$, there exists one and only one $G$-module homomorphism $\phi$ from $G$ to $Ind^G_H(N)$, such that $\phi_N\circ \phi=\phi'$.

Via the proof of Proposition 2.3, if $M,N$ are $H$-modules, and $\phi'\in Hom_H(M,N)$, then one can construct $\phi: Ind^G_H(M)\to Ind^G_H(N)$ defined by $\alpha\mapsto \phi'\circ\alpha$, and $\phi$ is a $G$-module homomorphism. We hence obtain a functor from $Mod_H$ to $Mod_G$ defined by $M\mapsto Ind^G_H(M)$.

Proposition 2.5. The functor $Ind^G_H$ defined above is exact.

Proof. Let

$$0\to M\xrightarrow{i} N\xrightarrow{p} P\to 0$$

be an exact sequence of $H$-modules, then via the functor $Ind^G_H$, the induced map from $Ind^G_H(M)$ to $Ind^G_H(N)$ is defined to be $\alpha\mapsto i\circ \alpha$, and the induced map $Ind^G_H(N)\to Ind^G_H(P)$ is defined to be $\beta\mapsto p\circ\beta$, and this yields the following complex of $G$-modules

$$0\to Ind^G_H(M)\to Ind^G_H(N)\to Ind^G_H(P)\to 0$$

The exactness at $Ind^G_H(M)$ is clear. We will prove next the exactness at $Ind^G_H(P)$. Taking any $\gamma\in Ind^G_H(P)$, one can represent $G=\bigcup_{i\in I}Hs_i$, where $\{s_i\}_{i\in I}$ are representatives of right coset of $H$ in $G$. One has $\gamma(hs_i)=h\gamma(s_i)$. Let $n_i\in N$, such that $p(n_i)=\gamma(s_i)$. If we defined $\beta(hs_i):=hn_i$, then the map $\beta: G\to N$ is well-defined, and $p\circ \beta=\gamma$. Moreover, for any $g\in G$ there exists a unique $s_i$ such that $g\in Hs_i$, and hence, for all $h\in H$, there exists $h'$ such that $hg=h's_i$, i.e. $g=h^{-1}h's_i$, and

$$\beta(g)=\beta(h^{-1}h's_i)=h^{-1}(h'n_i)=h^{-1}\beta(h's_i)=h^{-1}\beta(hg)$$

And this yields $h\beta(g)=\beta(hg)$, and $\beta\in Ind^G_H(N)$. This yields the surjectivity of the map $Ind^G_H(N)\to Ind^G_H(P)$.

To complete the proof, we have to show that this complex is also exact at $Ind^G_H(N)$. Let $\beta\in Ind^G_H(N)$, such that $p\circ \beta=0$. From the original exact sequence, for any $g\in G$, there exists a unique $m_g\in M$ such that $\beta(g)=i(m_g)$. And one can define the map $\alpha: G\to M$ defined by $g\mapsto m_g$, and $\alpha(hg)=m_{hg}$, and $i(m_{hg})=\beta(hg)=h\beta(g)=hi(m_g)=i(hm_g)$. And since $i$ is injective, we have $m_{hg}=hm_g$, and this yields $\alpha(hg)=h\alpha(g)$, for all $h\in H,g\in G$. From this, $\alpha\in Ind^G_H(M)$ with $i\circ \alpha=\beta$. And it follows that $Ind^G_H$ is an exact functor. (Q.E.D)

When $H=\{1\}$, an $H$-module is just an abelian group. Let $M_0$ be an abelian group, we have $Ind^G(M_0):=Ind^G_{\{1\}}(M_0)=\{\alpha:G\to M_0 \text{ as sets}\}$.  We said that a $G$-module $M$ is induced if there exists an abelian group $M_0$ such that $M\cong Ind^G(M_0)$ as $G$-modules.

If we denote

$$\mathbb{Z}[G]:=\{\sum_{g\in G}n_gg|n_g\in \mathbb{Z}\text{ and }n_g\ne 0\text{ for finitely many } g\in G\}$$

 then $\mathbb{Z}[G]$ becomes a $G$-module under the action $g'(\sum_{g\in G}n_gg):=\sum_{g\in G}n_g(g'g)$. And this follows easily that $Ind^G(M_0)=Hom_{Ab}(\mathbb{Z}[G],M_0)$. And so, when $G$ is finite, we get

$$Ind^G(M_0)\cong Hom_{Ab}(\mathbb{Z}[G],M)\cong \mathbb{Z}[G]\otimes_\mathbb{Z}M_0$$
Proposition 2.6. Let $G$ be a group, and $M$ is a $G$-module.

(i) Let $M_0$ be the underlying abelian group structure of $M$, then there exists an embedding of $G$-modules $\phi: M\to Ind^G(M_0)$ defined by $m\mapsto \varphi_m$, where $\varphi_m(g)=gm$, for all $g\in G$.

(ii) When $G$ is finite, $M$ is a quotient of an induced module.

Proof.

(i) The map $\phi$ is obvious injective, and $\phi(g'm)=\varphi_{g'm}$, where $\varphi_{g'm}(g)=gg'm$, and $(g'\phi(m))(g)=(g'\varphi_m)(g)=\varphi_m(gg')=gg'm$. Hence, $\phi(g'm)=g'\phi(m)$, for all $g'\in G,m\in M$. This yields $\phi$ is an embedding of $G$-modules.

(ii) Let $M_0$ be the underlying abelian group structure of $M$, we can have a map $\phi$ from $Ind^G(M_0)$ to $M$ defined by $\varphi\mapsto \sum_{g\in G}g\varphi(g^{-1})$. It can be seen that $\phi$ is a surjective homomorphism of abelian groups. Moreover, we have

$$\phi(g_1\varphi)=\sum_{g\in G}g(g_1\varphi)(g^{-1})=\sum_{g\in G}g\varphi(g^{-1}g_1)$$

And $g_1\phi(\varphi)=\sum_{g\in G}g_1g\varphi(g^{-1})$. Let $g'^{-1}=g^{-1}g_1$, then $g'=g_1^{-1}g$, and $g_1g'=g$. And this yields

$$\sum_{g\in G}g\varphi(g^{-1}g_1)=\sum_{g'\in G}(g_1g')\varphi(g'^{-1})=\sum_{g\in G}g_1g\varphi(g^{-1})$$

So $\phi(g_1\varphi)=g_1\phi(\varphi)$. Hence, the map $\phi$ above is a $G$-module homomorphism. (Q.E.D)

[Tilts and Field of Norms I] Perfectoid Fields and Tilts

Let us fix these notations, $L/\mathbb{Q}_P$ a finite extension, with $\mathscr{O}:=\mathscr{O}_L$ is its ring of integers, $\pi:=\pi_L$ is a uniformizer, with $q:=\#k_L$, where $k_L$ is the residue field of $L$. $\overline{\mathbb{Q}_p}$ denotes the algebraic closure of $\mathbb{Q}_p$, and $\mathbb{C}_p$ its completion. Let $L_\infty$ be the totally ramified  extension of $L$ as in the theory of Lubin-Tate.

When $L:=\mathbb{Q}_p$, $L_\infty=\mathbb{Q}_p(\zeta_{p^\infty})$, and a theorem of Fontaine yields $Gal(\overline{\mathbb{Q}_p}/\mathbb{Q}_p(\zeta_{p^\infty}))$ is isomorphic (as topological group) to $Gal(\mathbb{F}_p((t))^{\text{sep}}/\mathbb{F}_p((t)))$. This helps us classify all $p$-adic Galois representations, by the classification of Galois representations in char. $p$. Fontaine dealt with this by the theory of field of norms. It was generalized by Peter Scholze by the notions of perfectoid fields and their tilts. The goal of this part is to understand the construction of Peter Scholze.

Definition. Let $L\subset K\subset \mathbb{C}_p$ be an immediate field, then $K$ is said to be a perfectoid field if

i) $K$ is complete.
(ii) $|K^\times|$ is dense in $\mathbb{R}^\times_{>0}$.
(iii) The map $\mathscr{O}_K/p\mathscr{O}_K\to \mathscr{O}_K/p\mathscr{O}_K$ defined by $x\mapsto x^p$ is surjective.

Lemma 1.1. Let $K$ be a perfectoid field and $a\in K^\times$, then there exists $b\in K^\times$, such that $|a|=|b|^p$.

Proof. Because $K^\times$ is dense in $\mathbb{R}^\times_{>0}$, there exists some $\omega\in K^\times$ such that $|p|<|\omega|\le 1$, and some $m\in \mathbb{Z}$, such that $|\omega|^{m+1}\le |a|\le |\omega|^m$. From this, $|\omega|\le |a\omega^{-m}|\le 1$, and hence $|p|<|a\omega^{-m}|\le 1$, and $|a|=|\omega^m||a\omega^{-m}|$. And it is sufficient to prove that whenever $|p|<|a|\le 1$, there exists $b\in K^\times$, such that $|a|=|b|^p$. The condition (iii) of the definition above yields there exists some $b\in K^\times$, such that $|a-b^p|<|p|$. If $|a|\ne |b^p|$, then $|a-b^p|=\max\{|a|,|b^p|\}\ge |a|>|p|$, a contradiction. Hence, $|a|=|b|^p$. (Q.E.D)

Let us fix some $\omega\in K^\times$, where $K$ is a perfectoid fields, such that $|\omega|\ge |\pi|$ (so that $\omega \mathscr{O}_K\supset \pi \mathscr{O}_K\supset p\mathscr{O}_K$. We consider the following projective limit

$$\mathscr{O}_{K^\flat}:=\varprojlim (...\xrightarrow{(.)^q}\mathscr{O}_K/\omega \mathscr{O}_K\xrightarrow{(.)^q}\mathscr{O}_K/\omega \mathscr{O}_K\xrightarrow{(.)^q}\mathscr{O}_K/\omega\mathscr{O}_K)=$$

$$=\{(...,\alpha_i,...,\alpha_1,\alpha_0)|\alpha_i\in \mathscr{O}_K/\omega\mathscr{O}_K,\alpha_{i+1}^q=\alpha_i\}$$

Lemma 1.2. $\mathscr{O}_{K^\flat}$ is a perfect $k_L$-algebra.

Proof. There is a map from $(\mathscr{O}\mod \pi \mathscr{O})$ to $\mathscr{O}_{K^\flat}$ defined as

$$(a\mod \pi \mathscr{O}) \mapsto (...,a\mod \omega \mathscr{O}_K,...,a\mod \omega \mathscr{O}_K)$$

Because $a^q\equiv a \mod \pi \mathscr{O}$ for all $a\in \mathscr{O}$, we have $a^q\equiv a\mod \omega\mathscr{O}_K$, this yields a well-defined map from $k_L$ to $\mathscr{O}_K/\omega \mathscr{O}_K$. It is easy to check that this map is a ring homomorphism. Hence, $\mathscr{O}_{K^\flat}$ is a $k_L$-algebra.

Let us consider the map $\mathscr{O}_{K^\flat}\to \mathscr{O}_{K^\flat}$ defined as $\alpha \mapsto \alpha^q$. Assume that $\alpha^q:=(...,\alpha_i^q,...,\alpha_1^q,\alpha_0^q)=0$. The fact that $\alpha_{i+1}^q=\alpha_i$ yields $\alpha_i=0$, for all $i$, and hence, $\alpha=0$. Also, it is easy to see that $(...,\alpha_i,...,\alpha_1,\alpha_0)=(...,\alpha_i,...,\alpha_1)^q$. So, the map is also surjective. This yields $\mathscr{O}_{K^\flat}$ is perfect. (Q.E.D)

Now, for any $\alpha=(...,\alpha_i,...,\alpha_1,\alpha_0)$, we can lift $\alpha_i$ to $a_i\in \mathscr{O}_{K}$, such that $a_i\mod \omega \mathscr{O}_{K}=\alpha_i$, and we have $a_{i+1}^q=a_i\mod \omega \mathscr{O}_{K}$. And this yields

$$a_{i+1}^{q^{i+1}}\equiv a_i^{q^i}\mod \omega\mathscr{O}_{K}$$

so that the sequence $(a_i^{q^i})$ converges in $\mathscr{O}_{K}$. It can be checked easily that the limit of this sequence does not depend on the choice of $a_i$. And we denote this limit as $\alpha^\sharp$.

Lemma 1.3. The map

$$\varprojlim_{(.)^q}\mathscr{O}_K\xrightarrow{\psi} \mathscr{O}_{K^\flat}$$

defined by 

$$(..., a_i,...,a_1,a_0)\mapsto (...,a_i\mod \omega\mathscr{O}_{K},...,a_1\mod \omega \mathscr{O}_{K},a_0\mod \omega \mathscr{O}_{K})$$
and the map

$$\mathscr{O}_{K^\flat}\xrightarrow{\theta} \varprojlim_{(.)^q}\mathscr{O}_{K}$$

defined by $\alpha\mapsto (...,(\alpha^{1/q^i})^\sharp,...,(\alpha^{1/q})^\sharp,\alpha^\sharp)$ are multiplicative inverse of each other.

Proof. We can see that $\psi$ is well-defined. Also, if we denote $\alpha:=(...,\alpha_i,...,\alpha_1,\alpha_0)$, then it can be seen that $\alpha^{1/q^i}=(...,\alpha_{i+1},\alpha_i)$, and

$$(\alpha^{1/q^i})^\sharp=\lim_{j\to \infty}(a_{i+j}^{q^j})=(\lim_{k\to \infty}(a_k)^{q^k})^{1/q^i}=(\alpha^\sharp)^{1/q^i}$$

when we change variables $k=i+j$, and $a_i$ are lifts of $\alpha_j$. And this yields $\theta$ is also well-defined. Now, if we begin with $(...,a_i,...,a_1,...,a_0)\in \varprojlim_{(.)^q} \mathscr{O}_{K}$, then


$$\theta\psi(...,a_i,...,a_1,a_0)=\theta(...,a_i\mod \omega \mathscr{O}_{K},...,a_0\mod \omega \mathscr{O}_{K})=$$

$$=\theta (...,\alpha_i,...,\alpha_1,\alpha_0)=(...,(\alpha^{1/q^i})^\sharp,...,(\alpha^{1/q})^\sharp, \alpha^\sharp)$$

where $\alpha_i=a_i\mod \omega \mathscr{O}_{K}$, and $\alpha=(...,\alpha_i,...,\alpha_0)$. We have $(\alpha^{1/q^i})^\sharp=\lim_{j\to\infty}(a_{i+j}^{q^j})$. Because $a_{i+1}^q=a_i$, we have $a_j^{q^{i+j}}=a_i$. And hence $(\alpha^{1/q^i})^\sharp=a_i$. And hence, $\theta\circ \psi$ is just the identity map.

Now, if we begin with $\alpha=(...,\alpha_i,...,\alpha_1,\alpha_0)\in \mathscr{O}_{K^\flat}$, then we first note that $\alpha^\sharp\equiv \alpha_0\mod \omega \mathscr{O}_{K}$, hence

$$\psi\circ\theta(\alpha)=\psi(...,(\alpha^{1/q^i})^\sharp,...,(\alpha^{1/q})^\sharp,\alpha^\sharp)=\alpha$$

So, this yields $\psi$ and $\theta$ are inverse of each other. The multiplicative properties are easy to check. (Q.E.D)

Our net goal is to prove that in fact $\mathscr{O}_{K^\flat}$ is an integral domain of char. $p$, and that it is complete. We first introduce the following norm on $\mathscr{O}_{K^\flat}$

$$||_\flat: \mathscr{O}_{K^\flat} \to \mathbb{R}$$

defined as $|\alpha|_\flat:=|\alpha^\sharp|$.

Proposition 1.4.

(i) The norm $|.|_\flat$ on $\mathscr{O}_{K^\flat}$ is non-archimedean. 

(ii) $|\mathscr{O}_{K^\flat}|=|\mathscr{O}_{K}|$

(iii) For $\alpha,\beta\in \mathscr{O}_{K^\flat}$, $\alpha\mathscr{O}_{K^\flat}\subset \beta \mathscr{O}_{K^\flat}$ iff $|\alpha|_\flat\le |\beta|_\flat$.

(iv) $\mathscr{O}_{K^\flat}$ is a local domain of char. $p$, with the unique maximal ideal $\mathfrak{m}_{K^\flat}=\{\alpha\in \mathscr{O}_{K^\flat}||\alpha|_\flat<1\}$.

(v) $\mathscr{O}_{K^\flat}/\mathfrak{m}_{K_\flat}\cong \mathscr{O}_{K}/\mathfrak{m}_K$.

(vi) Let $\omega^\flat\in \mathscr{O}_{K^\flat}$, such that $|\omega^\flat|_\flat=|\omega|$, then the map $\mathscr{O}_{K^\flat}/\omega^\flat\mathscr{O}_{K^\flat}\to \mathscr{O}_{K}/\omega \mathscr{O}_{K}$ defined as $\alpha\mapsto \alpha^\sharp \mod \omega \mathscr{O}_{K}$ is an isomorphism of rings.

Proof. We first fix $\alpha:=(...,\alpha_i,...,\alpha_1,\alpha_0), \beta:=(...,\beta_i,...,\beta_1,\beta_0)$ in $\mathscr{O}_{K^\flat}$, and $a_i:=(\alpha^{1/q^i})^\sharp, b_i:=(\beta^{1/q^i})^\sharp$, we know that $b_{i+1}^q=b_i, a_{i+1}^q=a_i$.

(i) We have

$$|\alpha+\beta|_\flat=|(\alpha+\beta)^\sharp|=|\lim_{i\to \infty}(a_i+b_i)^{q^i}|=\lim_{i\to \infty}|(a_i+b_i)^{q^i}|$$
$$\le \lim_{i\to \infty}\max\{|a^{q^i}|, |b^{q^i}|\}=\lim_{i\to \infty}\{|a_0|,b_0\}=\max\{|\alpha|^\sharp|,|\beta^\sharp|\}=\max\{|\alpha|_\flat,|\beta|_\flat\}$$

Also, assume that $|\alpha|_\flat=0$, this yields $\alpha^\sharp=a_0=0$, and this yields $\alpha=0$. The multiplicative property of $|.|_\flat$ is easy to check. So, it is a non-archimedean norm on $\mathscr{O}_{K^\flat}$.

(ii) From the definition, we have $|\mathscr{O}_{K^\flat}|_\flat\subseteq |\mathscr{O}_{K}|$. Take any $a\in \mathscr{O}_{K}$, we know that there exists some $b$, such that $|\omega|<|b|\le 1$, and $|a|=|b|^{q^m}$. We can find $\alpha\in \mathscr{O}_{K^\flat}$ such that $\alpha_0\equiv b\mod \omega \mathscr{O}_{K}$. This yields $\alpha^\sharp\equiv b\mod \omega \mathscr{O}_{K}$, and $|\beta^\sharp - b|\le |\omega|$. It follows that $|\beta^\sharp|=|b|$. So, we get $|\alpha|_\flat=|b|$, and $|a|=|\beta^{q^m}|_\flat$. So $\mathscr{O}_{K^\flat}=\mathscr{O}_{K}$.

(iii) Assume that $\alpha \mathscr{O}_{K^\flat}\subseteq \beta \mathscr{O}_{K^\flat}$. Then there exists some $\gamma\in \mathscr{O}_{K^\flat}$, such that $\alpha=\beta\gamma$, and this yields $|\alpha|_\flat\le |\beta|_\flat$. Conversely, assume $|\alpha|_\flat\le |\beta|_\flat$, which yields $|(\alpha^{1/q^i})^\sharp|\le |(\beta^{1/q^i})^\sharp|$, because $|\alpha^{1/q^i}|_\flat\le |\beta^{1/q^i}|_\flat$. And this yields $|a_i|\le |b_i|$, and there exists some $c_i\in \mathscr{O}_{K}$, such that $c_ia_i=b_i$. It follows directly that $c_{i+1}^q=c_i$. And hence, $\gamma:=(...,c_i\mod \omega \mathscr{O}_{K},...,c_1 \mod \omega \mathscr{O}_{K}, c_0\mod \omega \mathscr{O}_{K})$ defines an element in $\mathscr{O}_{K^\flat}$. And it is clear that $\alpha\gamma=\beta$, and $\alpha \mathscr{O}_{K^\flat}\subseteq \beta \mathscr{O}_{K^\flat}$.

(iv) Now, if we take any element $\gamma\in \mathscr{O}_{K^\flat}\setminus\mathfrak{m}_{K^\flat}$, then we can see by our recent argument that $\gamma \mathscr{O}_{K^\flat}=\mathscr{O}_{K^\flat}$, i.e. $\gamma$ is invertible. This yields $\mathscr{O}_{K^\flat}$ is local with maximal ideal $\mathfrak{m}_{K^\flat}$. Assume for now, $\alpha\beta=0$, this yields $|\alpha\beta|_\flat=|a_0b_0|=0$, and hence, $a_0=0$ or $b_0=0$. From this $\alpha=0$ or $\beta=0$. This implies $\mathscr{O}_{K^\flat}$ is a domain.

(v) Let us consider the map $\psi: \mathscr{O}_{K^\flat}\to \mathscr{O}_{K}/\mathfrak{m}_K$ defined by $\psi(\alpha)=\alpha^\sharp\mod \mathfrak{m}_K$. We can see easily that $\psi(\alpha\beta)=\psi(\alpha)\psi(\beta)$. Also,

$$\psi(\alpha+\beta)\equiv(\alpha+\beta)^\sharp\equiv a_0+b_0\mod \omega \mathscr{O}_{K^\flat}\equiv a_0+b_0\mod \mathfrak{m}_K$$

So, $\psi$ is a ring homomorphism. Take any $a_0\in \mathscr{O}_{K}$, we can find $a_1\in \mathscr{O}_{K}$ such that $a_1^q\equiv a_0\mod p\mathscr{O}_{K}$. It follows $a_1^q\equiv a_0\mod \mathfrak{m}_K$ and $a_1^q\equiv a_0\mod \omega \mathscr{O}_{K}$. Continuing this process, we get $\alpha:=(...,a_i\mod \omega \mathscr{O}_{K},...,a_0\mod \omega \mathscr{O}_{K})\in \mathscr{O}_{K^\flat}$, and $\alpha^\sharp\equiv a_0\mod \omega \mathscr{O}_{K}\equiv a_0\mod \mathfrak{m}_K$. And $\psi$ is surjective. From (iii), we have

$$\ker \psi=\{\alpha\in \mathscr{O}_{K^\flat}|\alpha^\sharp\in \mathfrak{m}_K\}=\{\alpha\in \mathscr{O}_{K^\flat}||\alpha^\sharp|<1\}=\{\alpha\in \mathscr{O}_{K^\flat}||\alpha|_\flat<1\}=\mathfrak{m}_{K^\flat}$$

And this yields $\mathscr{O}_{K^\flat}/\mathfrak{m}_{K^\flat}=\mathscr{O}_{K}/\mathfrak{m}_K$.

(vi) It follows from (v) that the map $\theta: \mathscr{O}_{K^\flat}\to \mathscr{O}_{K}/\mathfrak{m}_K$ is surjective, with

$$\ker\theta=\{\alpha\in \mathscr{O}_{K^\flat}|\alpha^\sharp|\le |\omega|\}=\{\alpha\in \mathscr{O}_{K^\flat}||\alpha|_\flat\le |\omega^\flat|_\flat\}=\omega^\flat\mathscr{O}_{K^\flat}$$

So, $\mathscr{O}_{K^\flat}/\omega^\flat \mathscr{O}_{K^\flat}\cong \mathscr{O}_{K}/\omega \mathscr{O}_{K}$. (Q.E.D)

We will conclude this section by the following

Proposition 1.5. $\mathscr{O}_{K^\flat}$ is complete with respect to the norm $|.|_\flat$.

Proof. We have $\mathscr{O}_{K^\flat}=\varprojlim_{(.)^q}\mathscr{O}_{K}/\omega \mathscr{O}_{K}$, and we can equip each $\mathscr{O}_{K}/\omega \mathscr{O}_{K}$ the discrete topology, and $\prod_{\mathbb{N}}\mathscr{O}_{K}/\omega \mathscr{O}_{K}$ the product topology, and $\mathscr{O}_{K^\flat}$ is a topological subgroup of $\prod_{\mathbb{N}}\mathscr{O}_{K}/\omega \mathscr{O}_{K}$, which has a fundamental system of open neighborhoods around $0$ defined as

$$U_m:=\{(...,a_{m+1},0,...,0)\} (m\ge 1)$$

and $U_1\supset U_{2}\supset ...$ forms a filtration. We will prove that with this topology, $\mathscr{O}_{K^\flat}$ is complete, and it coincides with the topology defined by $|.|_\flat$. For the first statement, it is sufficient to prove any Cauchy sequence converges in $\mathscr{O}_{K^\flat}$.

Let $(x_n)_n$ be a Cauchy sequence in $\mathscr{O}_{K^\flat}$. We can represent each $x_n$ as $(...,x_{n,i},...,x_{n,1}, x_{n,0})$, with $x_{n,i+1}^q=x_{n,i}$. And for all $k\ge 0$, there exists some $m_k$ such that $\forall m,n\ge m_k$, $x_m-x_n\in U_{k+1}$, and $m_{k+1}>m_k$. This yields $x_{m,i}-x_{n,i}=0(\forall 0\le i\le k+1; m,n\ge m_k)$.

Let $x:=(...,x_{m_i,i},...,x_{m_1,1},x_{m_0,0})$. We can see that $x_{m_i,i+1}=x_{m_{i+1},i+1}$, and hence $x_{m_i,i+1}^q=x_{m_i,i}=x_{m_{i+1},i+1}^q$. So, $x\in \mathscr{O}_{K^\flat}$. Now, for any $k\ge 0$, $n\ge m_k$, we have $x-x_n=(x-x_{m_k})-(x_n-x_{m_k})$. It can be seen that for any $0\le i\le k+1$, we have $x_{m_k,i}-x_{m_i,i}=0$, so $x-x_{m_k}\in U_{k+1}$, and $x_n-x_{m_k}\in U_{k+1}$. And this yields $(x_n)_n$ converges to $x$. Hence, with this topology, $\mathscr{O}_{K^\flat}$ is complete. On the other hand, we have

$$U_m=\{\alpha\in \mathscr{O}_{K^\flat}|(\alpha^{1/q^m})^\sharp\in \omega\mathscr{O}_{K}\}=\{\alpha\in \mathscr{O}_{K^\flat}||\alpha^{1/q^m}|_\flat\le |\omega^\flat|_\flat\}=(\omega^\flat)^{q^m}\mathscr{O}_{K^\flat}$$

And hence $\{U_m\}_{m\ge 1}$ also forms a fundamental system of open neighborhoods around $0$ with the topology induced by $|.|_\flat$. It follows that the two topology coincide. And this yields $\mathscr{O}_{K^\flat}$ is complete. (Q.E.D)

For now, it makes sense to talk about $K^\flat$, the fraction field of $\mathscr{O}_{K^\flat}$. It is a field of characteristic $p$. By extending the norm $|.|_\flat$ to $K^\flat$, it is complete and non-archimedean. Also, the inverse map of $\psi$ in Lemma 1.3 can be extended to a multiplicative bijection

$$K^\flat \xrightarrow{\cong} \varprojlim_{(.)^q}K$$

defined by $\alpha \mapsto (...,(\alpha^{1/q^i})^\sharp,...,(\alpha^{1/q})^\sharp,\alpha^\sharp)$.

Definition. $K^\flat$ is called the tilt of $K$.

[Topology on Witt Vectors III] Weak Topology on Witt Vectors

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)

[Topology on Witt Vectors II] Linearly Topologized Rings

For our further purposes, we often use inverse limit to define topology on a commutative ring $A$.

Definition. Let $A$ be a topological group (ring), we say $A$ is linearly topologized group (ring, resp.) if $0$ has a fundamental system of open neighborhoods $\mathscr{N}$ consisting of subgroups (ideals, resp.) of $A$. Such a fundamental system is called linearly topologized fundamental system. And we denote $(A, \mathscr{N})$ for the topological group (ring) $A$ with linearly topologized fundamental system $\mathscr{N}$.

The following lemma is quite useful to detect when are two linearly topoligized fundamental system generate the same topology.

Lemma 2.1. If  $A$ is an abelian group (commutative ring) and $\mathscr{N}, \mathscr{M}$ are two sets containing subgroups of $A$ satisfying conditions of Proposition 1.3 (Proposition 1.6, respectively), then they induces the same topology on $A$ iff for any $I\in \mathscr{N}$, there exists some $J\in \mathscr{M}$ such that $J\subset I$, and vice versa. 

Proof. It follows directly from our construction in Proposition 1.3, since they generate the same base for the topology on $A$. (Q.E.D)

By the following lemma, it makes sense to give the definition of the completion of a linearly topologized ring $A$.

Lemma 2.2. Let $A$ be an abelian group (commutative ring), $\mathscr{M}, \mathscr{N}$ are sets of subgroups of $A$, such that for every $G\in \mathscr{M}$, there exists $H_G\in \mathscr{N}$, such that $G\subset H_G$, and for every $H\in \mathscr{N}$, there exists $G_H\in \mathscr{M}$ such that $G_H\subset H$. Then there is a canonical isomorphism

$$\varprojlim_{G\in \mathscr{M}}A/G\cong \varprojlim_{H\in \mathscr{N}}A/H$$

Proof. For any $G\subset H$: subgroups of $A$, we denote $\psi^G_H$ the canonical map $A/G\to A/H$. Let $(a_G\mod G)_{G\in \mathscr{M}}\in \varprojlim_{G\in \mathscr{M}} A/G$, for $a_G\in A$, due to the assumption, for any $H\in \mathscr{N}$, there is $G_H\in \mathscr{M}$ such that $G_H\subset H$, and we can define a map

$$\psi: \varprojlim_{G\in \mathscr{M}}A/G\to \varprojlim_{H\in \mathscr{N}}A/H$$

by $\psi((a_G\mod G)_G)=(\psi^{G_H}_H(a_{G_H}))_H=(a_{G_H}\mod H)_H$. It is well-defined, since $(\psi^{G_H}_H)$ is compatible with the inverse system $\varprojlim_{H\in \mathscr{M}}A/H$.

Conversely, for any $G\in \mathscr{N}$, there exists $H_G\in \mathscr{M}$, such that $H_G\subset G$, and one can also define a map

$$\theta: \varprojlim_{H\in \mathscr{N}}A/H\to \varprojlim_{G\in \mathscr{M}}A/G$$

as follows $\theta(b_H\mod H)_H=(b_{H_G}\mod G)_{G\in \mathscr{N}}$, and we have the composition $\theta\circ \psi$ would be

$$(a_G\mod G)_G\mapsto (a_{G_H}\mod H)_H\mapsto (a_{G_{H_G}}\mod G)_G$$

We have $G_{H_G}\subset H_G\subset G$, and because we are in inverse systems, we have

$$a_{G_{H_G}}\mod G_{H_G}=a_G\mod G_{H_G}$$

And that means $a_{G_{H_G}}-a_G\in G_{H_G}\subset G$. So, $a_{G_{H_G}}\mod G=a_G\mod G$, and $\theta\circ\psi$ is just the identity map. It is similar to show that $\psi\circ\theta$ is also the identity map. And this yields the desired isomorphism. (Q.E.D)

Via Lemma 2.1 and Lemma 2.2 we have this

Definition. Let $(A, \mathscr{N})$ be a topological group (ring), we say that $A$ is complete if the canonical map $A\to \lim_{I\in \mathscr{N}}A/I$ is surjective.

We can also obtain an algebraic characterization of Hausdorff on a linearly topologized ring $A$.

Proposition 2.3. Let $(A, \mathscr{N})$ be a topological group (ring), then $A$ is Hausdorff iff the canonical map

$$\phi: A\to \varprojlim_{I\in \mathscr{N}} A/I$$
is injective. Also, $A$ is Hausdorff iff $\cap_{I\in \mathscr{N}} I=0$.

Proof. Assume $\phi$ is injective, but $A$ is not Hausdorff, i.e. there exists $0\ne a\in A$, such that all open neighborhoods at $0$ contain $a$. But then, it is a contradiction, since $\phi(a)=0$, and $\phi$ is not injective.

Conversely, assume that $A$ is Hausdorff, and $\phi$ is not injective, that means, there exists $0\ne a\in A$ and $a\in I(\forall I\in \mathscr{N})$. Due to the definition of a fundamental system, if we take any open neighborhood $U$ of 0, then there exists $I\in \mathscr{N}$, and $I\subset U$, and $I$ contains both $0$ and $a$, and hence, so does $U$. That means, $A$ is not Hausdorff, a contradiction.

For the second statement, assume that

$$\phi: A\to \varprojlim_{I\in \mathscr{N}}A/I$$
is injective, and $0\ne a\in \cap_{I\in \mathscr{N}}I$, then $\phi(a)=0$, so $\phi$ is not injective, a contradiction. Conversely, assume that $\cap _{I\in \mathscr{N}}I=0$, and $\phi$ is not injective, i.e. there exists $0\ne a\in A$ such that $\phi(a)=0$. That means $a\in I(\forall I\in \mathscr{N})$, a contradiction. We hence obtain the statements. (Q.E.D)

Remark 2.4. This follows directly from the definition that if $(A, \mathscr{N})$ is Hausdorff then if any Cauchy sequence with respect to $\mathscr{N}$ converges in $A$, then the convergence is unique.

Proof. Assume that $A$ is Hausdorff, and $(x_n)_n$ is a Cauchy sequence converges in $A$ to $x, x'$, then it is clear that for all $I\in \mathscr{N}$, $x-x'\in I$. But then, since $\cap_{I\in \mathscr{N}}I=0$, we have $x=x'$. (Q.E.D)

For the converse of the remark above, we need to have a more special linearly topologized fundamental system. Let $A$ be an abelian group (commutative ring), and $I_1 \supset I_2\supset ...$ is a chain of ideals in $A$. We can see that $\mathscr{N}:=\{I_1,I_2,...\}$ satisfies conditions Proposition 1.6. Hence, there exists a unique topology on $A$ such that $\mathscr{N}$ is a fundamental system of open neighborhoods around $0$. We say such $\mathscr{N}$ a (countable) filtered fundamental system.

Remark 2.5. The converse of Remark 2.4 holds for a topological group (ring) $(A, \mathscr{N})$, where $\mathscr{N}$ is a filtered fundamental system. That means, in this case, $(A, \mathscr{N})$ is Hausdorff if and only if for any Cauchy sequence with respect to $\mathscr{N}$ converges in $A$, then the convergent value is unique.

Proof. For the converse, assume that $A$ is not Hausdorff, i.e. there exists $0\ne a\in \cap_{I\in \mathscr{N}}I$, then we can choose a sequence $(x_n)_n$ with $x_n\in I_n$, where $I_n\in \mathscr{N}$. This sequence is Cauchy, since we always have $x_{n+1}-x_n\in I_n$, for all $n$. This follows that both $a\ne -a$ are convergent values of $(x_n)_n$, a contradiction. (Q.E.D)

Lemma 2.5. Let $(A, \mathscr{N})$ be a topological group (ring), where $\mathscr{N}:=\{(I_n)_n\}$ is a filtered fundamental system, then the canonical map

$$\phi: A\to \varprojlim_n A/I_n$$

is surjective iff  $A$ is complete with respect to $\mathscr{N}$.

Proof. Assume that $\phi: A\to \varprojlim_n A/I_n$ is surjective, and $(x_n)_n$ is a Cauchy sequence. Then for any $k\ge 1$, there exists $n_k$ such that $n_{k+1}> n_{k}$ and for all $m,n\ge n_k$, we have $x_m-x_n\in I_k$. We can define $y_k:=x_{n_k}$, then $(y_k)_k$ is an element in $\varprojlim_k A/I_k$, because $y_{k+1}-y_k=x_{n_{k+1}}-x_{n_k}\in I_k$. By the surjectivity of $\phi$, there exists $y\in A$ such that $\phi(y)=(y_k)_k$, i.e. $y\mod I_k=y_k, \forall k$ and we can see easily that $y$ is a convergent value of $(y_k)_k$. This also follows that $y$ is a convergent value of $(x_n)_n$ because for any $k$, and for all $n\ge n_k$, $y-x_n=y-x_{n_k}+x_{n_k}-x_n\in I_k$.

Conversely, let $(x_n)_n\in \varprojlim_n A/I_n$ be any element, we can see that $x_{n+1}-x_n\in I_n$, and $(x_n)_n$ is a Cauchy sequence in $A$, and there exists $x\in A$ is a convergent value of $(x_n)_n$. Due to the definition, for every $n$, and there exists some $m_n$ such that for all $m\ge m_n$, $x-x_m\in I_n$. We can choose $m\ge \max\{n,m_n\}$, and $x-x_n=x-x_m+x_m-x_n\in I_n$. And this yields, for all $n$, $x-x_n\in I_n$. Hence, $\phi(x)=(x_n)_n$, and $\phi$ is surjective. (Q.E.D)

Via this lemma, for the case $A$ has a filtered fundamental system, we obtain an equivalence between the notions of sequential complete and complete. We conclude this section by a

Proposition 2.7. Let $A$ be an topological group (ring), and $I_1\supset I_2\supset ...$ a linearly topologized fundamental system around $0$. Then $A$ is Hausdorff and complete iff the canonical map

$$\phi: A\to \varprojlim_n A/I_n$$
is bijective.

[Topology on Witt Vectors I] Topological Groups and Rings

Definition. Let $A$ be an abelian group, and $A$ is also a topological space. Then $A$ is said to be a topological group if the addition map $add: A\times A\to A$, defined by $add(a,b)=a+b$, and the inverse map $inv: A\to A$ defined by $inv(a)=-a$ are continuous, where $A\times A$ is equipped with the product topology.

We begin with

Lemma 1.1. For all $c\in A$, the map $\tau_c: A\to A$ defined by $\tau_c(a)=c+a$ is a homeomorphism.

Proof. One can see that the restriction of the topology on $A\times A$ to $\{c\}\times A$ makes $\{c\}\times A$ homeomorphic to $A$ via the map $pr(c,b) = b$. Also, the map $add|_{\{c\}\times A}$ is continuous and bijective. This yields $\tau_c:=add|_{\{c\}\times A}\circ pr^{-1}: A\to A$ is also continuous and bijective. Its inverse map $\tau_{-c}$, by similar arguments, is also continuous. Hence $\tau_c$ is a homeomorphism (Q.E.D)

For most of cases, we want to equip topology for a given abelian group $A$. Here is a way to do this.

Definition. Let $A$ be a topological group, $x\in A$ be a point, then a fundamental system of open neighborhoods around $x$ is a set $\mathscr{N}_x$ containing open neighborhoods of $x$, such that for all $x\in U$, where $U\subset A$ open, there exists $V\in \mathscr{N}_x$ such that $V\subset U$. We denote a fundamental system of open neighborhoods around 0 as $\mathscr{N}$, instead of $\mathscr{N}_0$.

Lemma 1.2. The following holds for $\mathscr{N}$:

(T.G.1) $\forall U\in \mathscr{N}, \forall c\in U$, there exists $V\in \mathscr{N}$, such that $c+V\subset U$.

(T.G.2) $\forall U\subset \mathscr{N}$, there exists $V\in \mathscr{N}$, such that $V+V\subset U$.

Let $\mathscr{B}:=\{a+U|a\in A, U\in \mathscr{N}\}$, then $\mathscr{B}$ forms a base for the topology on $A$.

Proof. For (T.G.1), by Lemma 1.1, we have $U-c$ is also an open neighborhood of $0$, and hence, due to the definition of $\mathscr{N}$, there exists $V\in \mathscr{N}$, such that $V\subset U-c$, i.e. $V+c\subset U$.

For (T.G.2), let us look at $add^{-1}(U)$, it is an open neighborhood of $(0,0)$, and hence, there exists $V_1,V_2$: open neighborhoods of $0$ such that $add(V_1\times V_2)\subset U$. Take $V\in \mathscr{N}$, and $V\subset V_1\cap V_2$, we get $add(V+V)\subset U$.

Now, it follows by Lemma 1.1. that if $\mathscr{N}$ is a fundamental system around $0$, then $\{x+U|U\in \mathscr{N}\}$ is a fundamental system around $x$. Let $x\in A$ be any element, and $U_x$ is an open neighborhood of $x$, then $U_x-x$ is an open neighborhood of $0$, and by (T.G.1), there exists $V\in \mathscr{N}$, such that $V\subset U_x-x$, i.e. $x+V\subset U_x$, and $x+V\in \mathscr{B}$, and $x+V$ is an open neighborhood of $x$. This yields, any open subset of $A$ is a union of elements in $\mathscr{B}$. Hence, $B$ is a base for the topology on $A$. (Q.E.D)

We are now ready to see how to define a suitable topology for an abelian group so that it becomes a topological group.

Proposition 1.3. Let $\mathscr{N}$ be a set containing subsets of $A$, which contain $0$, such that $\mathscr{N}$ satisfies (T.G.1), (T.G.2) and

(T.G.3) $\forall U\in \mathscr{N}$, $-U\in \mathscr{N}$.

(T.G.4) $\forall U,V\in \mathscr{N}$, there exists $W\in \mathscr{N}$, such that $W\subset U\cap V$

then there exists a unique topology on $A$ that makes $A$ become a topological group, and $\mathscr{N}$ a fundamental system around 0. 

Proof. We first define a topology on $A$ as follows. $U\subset A$ is open if either $U$ is empty, or for all $x\in U$, there exists $V\in \mathscr{N}$, such that $x+V\subset U$. With this kind of open sets, we now prove $A$ is a topological space.

i. $\emptyset, A$ are open.

ii. Let $U_i(i\in I)$ be open subsets of $A$, then it is obvious to see that $\cup_{i\in I} U_i$ is also open.

iii. Let $U_1,...,U_n$ be open subsets of $A$, we will prove that $U_1\cap ...\cap U_n$ is also open. By induction, it is sufficient for us to prove the case $n=2$. If $U_1\cap U_2=\emptyset$, then there is nothing to prove. Otherwise, let $x\in U_1\cap U_2$, then there exists $V_i(i=1,2)$, such that $V_i\in \mathscr{N}$, and $x+V_i\subset U_i(i=1,2)$. By (T.G.4), we can choose $V\subset V_1\cap V_2$ and $V\in \mathscr{N}$, and this yields $x+V\subset U_1\cap U_2$. Hence, $U_1\cap U_2$ is also open.

We now prove that with this kind of topology, $A$ becomes a topological group, that means, we have to prove $add$ and $inv$ are continuous. First, let $U\subset A, U\ne \emptyset$ be an open subset, then $inv^{-1}(U)=-U$. Take any $x\in -U$, we have $-x\in U$, and there exists $V\in \mathscr{N}$, such that $-x+V\subset U$, i.e. $x-V\subset U$. By (T.G.3), $-V$ is also in $\mathscr{N}$. This yields $-U$ is open, and $inv$ is continuous.

Next, we will prove that if $U\subset A, U\ne \emptyset$ is an open subset, then for any $c\in A$, $c+U$ is also open. Taking any $b\in c+U$, i.e. $b-c\in U$, and there exists $V\in \mathscr{N}$, such that $b-c+V\subset U$, i.e. $b+V\subset c+U$. And this yields $c+U$ is open.

Now, let $U\ne \emptyset, U\subset A$ be an open subset, consider $(c,d)\in add^{-1}(U)$, i.e. $c+d\in U$, then there exists $V\in \mathscr{N}$, such that $c+d+V\subset U$. By (T.G.2), there exists $W\in \mathscr{N}$ such that $W+W\subset V$, and this yields $(c+W)+(d+W)\subset U$. And by our previous argument, both $c+W, d+W$ are open and $(c+W)\times (d+W)\subset add^{-1}(U)$. Hence, $add$ is also continuous. Hence, $A$ is a topological group with this topology.

By (T.G.1), all elements in $\mathscr{N}$ are open subsets of $A$, and they form a fundamental neighborhood of $0$, since for any $U\subset A$: open, containing 0; there exists $V\in \mathscr{N}$, such that $0+V=V\subset U$. By Lemma 1.2, $\mathscr{B}:=\{a+U|a\in A,U\in \mathscr{N}\}$ forms a base for the topology on $A$, and this yields the uniqueness part. (Q.E.D)

We now come to the definition of topological rings.

Definition. Let $(A,+,.)$ be a commutative ring, then $A$ is said to be a topological ring if $A$ is also a topological space, and 

i. (A,+) is a topological group.

ii. The map $mult: A\times A\to A$ defined by $mult(a,b)=ab$ is continuous.

Lemma 1.4. Let $A$ be a topological ring, and $a\in A$, then the map $m_c:A\to A$ defined by $m_c(a)=ac$ is continuous.

Proof. We have $\{c\}\times A$ is homeomorphic to $A$ via the map $pr(c,b)=b$, and the restriction of $mult|_{\{c\}\times A}$ is continuous, and hence, the composition $m_c=mult|_{\{c\}\times A}\circ pr^{-1}$ is continuous. (Q.E.D)

Similarly to earlier argument, we obtain

Lemma 1.5. Let $\mathscr{N}$ be a fundamental system around 0 of a topological ring $A$, then

(T.R.1) $\forall U\in \mathscr{N}, \forall c\in A$, there exists $V\in \mathscr{N}$, such that $cV\subset U$.

(T.R.2) $\forall U\in \mathscr{N}$, there exists $V\in \mathscr{N}$, such that $V.V:={uv|u,v\in V}\subset U$.

Proof. For (T.R.1), using Lemma 1.4, the map $m_c:A\to A$ is continuous, and this yields $m_c^{-1}(U)\subset A$ is also open, and contains $0$. So, there exists $V\in \mathscr{N}$, such that $V\subset m_c^{-1}(U)$, i.e. $cV\subset U$.

For (T.R.2), the map $mult: A\times A\to A$ is continuous, and hence, $mult^{-1}(A)$ is open in $A\times A$. Let $V_1\times V_2\subset mult^{-1}(U)$, and $V_1, V_2$ are open neighborhoods of $0$, we can now choose $v\in \mathscr{N}$ and $V\subset V_1\cap V_2$, and $V.V=mult(V\times V)\subset U$. (Q.E.D)

And similar to commutative group, here is a way to put a topology on a commutative ring $A$ so that it becomes a topological ring.

Proposition 1.6. Let $\mathscr{N}$ be a set containing subsets of $A$, which contain $0$, such that $\mathscr{N}$ satisfies (T.G.1)-(T.G.4) and (T.R.1), (T.R.2), then there exists a unique way to equip $A$ a topology such that $\mathscr{N}$ becomes a fundamental system around $0$. 

Proof. By Proposition 1.3, there is a unique way to equip $A$ a topology such that $A$ becomes a topological group. So, it is sufficient to prove that, with this kind of topology, $A$ now becomes a topological ring. That means we have to check that the map $mult$ is continuous.

First, let $m_c: A\to A$ be defined as above, then for any $U\subset A$: open, non-empty, and $x\in m_c^{-1}(U)$, i.e. $cx\in U$. By (T.G.1), there exists $V\in \mathscr{N}$, such that $cx+V\subset U$, and by (T.R.1), there exists $W\in \mathscr{N}$, such that $cW\subset V$, and hence, $cx+cW\subset U$, that means,  $x+W\subset m_c^{-1}(U)$. And this yields $m_c$ is continuous.

We will now prove that $mult$ is continuous. Let $U$ be as above, take $(c,d)\in mult^{-1}(U)$, i.e. $cd\in U$. By (T.G.1), there exists $V\in \mathscr{N}$ such that $cd+V\subset U$. And by (T.R.2), there exists $W\in \mathscr{N}$ such that $W.W\subset V$. Now, by (T.R.1), we can choose $W',W''\in \mathscr{N}$ such that $dW'\subset V, cW''\subset V$. By (T.G.4), there exists $W_c, W_d\in \mathscr{N}$ such that $W_c\subset W'\cap W, W_d\subset W''\cap W$, then

$$(c+W_c)(d+W_d)=cd+dW_c+cW_d+W_cW_d\subset U$$

Hence, $mult$ is a continuous map. This yields $A$ is a topological ring. (Q.E.D)

We now discuss about Cauchy sequence on a topological ring $A$.

 Definition. Let $A$ be a topological ring with $\mathscr{N}$ is a fundamental system of open neighborhoods around $0$, and $(x_n)_n$ a sequence of elements in $A$, then $(x_n)_n$ is said to be a Cauchy sequence with respect to $\mathscr{N}$ if for every $U\in \mathscr{N}$, there exists an integer $n_U$ such that for all $m,n\ge n_U$, we have $x_m-x_n\in U$. We say that $x$ is a convergent value of $(x_n)_n$ if for every $U\in \mathscr{N}$, there exists an integer $n_U$, such that for all $m\ge n_U$, we have $x-x_m\in U$. We say $(A, \mathscr{N})$ is sequential complete with respect to $\mathscr{N}$ if any Cauchy sequence with respect to $\mathscr{N}$ converges in $A$.

Proposition 1.7. Let $A$ be a topological ring, and $\mathscr{N}$ is a fundamental system of open neighborhoods around $0$. If $\mathscr{M}$ is another fundamental system of open neighborhoods around $0$, then $(x_n)_n$ is a Cauchy sequence with respect to $\mathscr{N}$ iff $(x_n)_n$ is a Cauchy sequence with respect to $\mathscr{M}$. Hence, the definition of Cauchy sequences, and sequential complete do not depend on the choice of fundamental systems.

Proof. Assume that $(x_n)_n$ is a Cauchy sequence with respect to $\mathscr{N}$, then by definition, for any $V\in \mathscr{M}$, there exists $U\in \mathscr{N}$, such that $U\subset V$, then if we choose $m,n\ge n_U$, we get $x_m-x_n\in U\subset V$. Hence, $(x_n)_n$ is also a Cauchy sequence with respect to $V$. The converse is the same.

Assume that a Cauchy sequence $(x_n)_n$ with respect to $\mathscr{N}$ converges to $x\in A$, then it is clear from our earlier argument that $(x_n)_n$ also converges to $x$ with respect to $\mathscr{M}$. Hence, the definition of completion also does not depend on the choice of fundamental systems.

 (Q.E.D)

Witt Vectors: A Complete Introduction

Witt vectors are my old friends. I had a very hard time to digest it in the book of Jacobson at the first time we met. However, things are easier now when I read the book of Schneider about Galois representation. His treatment about Witt vectors is general for local fields. The main thing is that for Witt's vectors, Frobenius acts in a nice way. We don't have it in the ring of integer of a local field of characteristic $0$, though they are isomorphic, if we construct Witt vectors from the residue field with coefficients in our given local field.