[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)

Local Class Field Theory: Some Words before Beginning

The story begins with Hilbert's twelfth problem, that is: how to construct all abelian extensions of a number fields? It is somewhat equivalent to the question: let $K$ be a number field, construct the maximal abelian extension $K^{\text{ab}}$ of $K$. And this question is a starting point of class field theory. In local fields, one can ask for a similar question, let $K$ be a local field, construct the maximal abelian extension $K^{\text{ab}}$ of $K$. This question can be done completely by Lubin-Tate's theory.

Lubin-Tate's theory is based on formal groups, which is a vast and beautiful topic with many applications. Let $R$ be a (commutative) ring, and $F\in R[[X,Y]]$ be a series, then $F$ is said to be a commutative formal group if

1. $F(X,0)=X, F(0,Y)=Y$.
2. $F(X,F(Y,Z))=F(F(X,Y),Z)$ for all $Z\in R[[X,Y]]$.
3. $F(X,Y)=F(Y,X)$.

If one remember about the group structures of elliptic curves over a base field $k$, then addition law on an elliptic curve around the point $\infty$  is exactly a formal group (Silverman, Chapter IV.1). And one can define homomorphism between two formal group. In elliptic curves case, we obtain an isogeny represented around the two infinity points.

Lubin-Tate's formal group is also a specific example of formal groups. It is defined for Forbenius series over a local field. And by attaching suitable things, we obtain an abelian maximal totally ramified extension $K^{\text{ab.tot}}$ of $K$. We also obtain that $Gal(K^{\text{ab.tot}}/K)\cong \mathscr{O}_K^\times$.

Recall that any extension of local field can be decomposed into two parts: totally ramified part and unramified part. For the unramified part, it is somewhat easy, since to obtain an unramified extension of degree $n$, we just need to add $(q^n-1)$ root of unity, where $q$ is the cardinality of the residue field of $K$. And adjoining them all, we obtain $K^{\text{un}}$-the maximal unramified part of $K$. And from this, one has $Gal(K^{\text{un}}/K)\cong \hat{\mathbb{Z}}:=\varprojlim \mathbb{Z}/n\mathbb{Z}$

And now, $K^{\text{ab}}=K^{\text{ab.tot}}K^{\text{un}}$, with $Gal(K^{\text{ab}}/K)=\mathscr{O}_K^\times\times \hat{\mathbb{Z}}$. More than that, with the Artin map, we can also classify abelian extensions of $K$ with given degree.

[Elliptic Curves II] A Note on Weil Pairing

$E$ in this note is an elliptic curve defined over an algebraically closed field $k$.

1. The definition of Weil's pairing. From the previous note, we know that there exists an isomorphism $\phi$ between $E$ and $Pic^0(E)$, defined by $P \mapsto [P]-[\infty]$. And we know that any principal divisor on $E$ has degree 0. It is natural to ask, when is a divisor of degree 0 is a principal divisor?

Assume that $D:=\sum_{i=1}^n [P_i]-\sum_{i=1}^n [Q_i]\in Div(E)$, where $P_i, Q_i$ are not necessarily distinct. Then $D$ is a principal divisor iff $D\equiv 0$ in $Pic^0(E)$. Via the inverse map $\phi^{-1}$, we get $D\mapsto \sum_{i} P_i - \sum_{i} Q_i$. So $D=0$ in $Pic^0(E)$ iff $\sum_{i} P_i - \sum_{i}Q_i =0$ in $E$.

Now, if we write $D=\sum_{P\in E}n_P[P]\in Div(E)$, then we define $sum(D):=\sum_{P\in E}n_PP$. Note that the later sum is a point in $E$. By our earlier discussion, we get an important

Theorem 1.1. If $D\in Div(E)$ is a divisor, then $D$ is principal iff $\deg(D)=0$ and $sum(D)=\infty$.

Now, let $m$ be a positive integer such that $(m,char(k))=1$, we define $E[m]:=\{P\in E|mP=\infty\}$. It follows from Chapter III of Washington's book that $E[m]\cong \mathbb{Z}_m\oplus \mathbb{Z}_m$. Then for any $T\in E[m]$, we define the divisor $D_T$ as follows

$$D_T:=\sum_{Q\in E, mQ=T}([Q])-\sum_{P\in E[m]}[P]$$

Remark. It can be seen that $[m]: E\to E$ defined by $P\mapsto mP$ is a separable isogeny of $E$, so by our Remark 2, Section 4 of the previous note, $\deg [m]=\#[m]^{-1}(\infty)=E[m]=\#[m]^{-1}(P)=m^2$, for any point $P\in E$. That means the ramification index at all point respected to $[m]$ is just 1. So in this case, the pull-back map of divisors is defined as $[m]^*[P]=\sum_{Q\in E, mQ=P}[Q]$. And it can be seen that in fact, $D_T$ defined above is $[m]^*([T]-[\infty])$ because

$$[m]^*([T]-[\infty])=\sum_{Q\in E, mQ=T}[Q]-\sum_{P\in E, mP=\infty}[P]=\sum_{Q\in E, mQ=T}([Q])-\sum_{P\in E[m]}[P]$$

We will prove that

Lemma 1.1. $D_T$ defined above is a principal divisor.

Proof. Due to the remark, we can see, $\#[m]^{-1}(\infty)=E[m]=\#[m]^{-1}(T)$. This implies there exists a point $Q$ such that $mQ=T$ (because $\#E[m]=m^2$). Then it can be seen for any point $P\in E[m]$, $m(P+Q)=\infty+T=T$. Also, for any $P_1,P_2\in E[m], P_1\ne P_2$, then $P_1+Q\ne P_2+Q$. This yields $[m]^{-1}(T)=\{P+Q|P\in E[m]\}$. And we can now rewrite $D_T$ as

$$D_T=\sum_{P\in E[m]}([P+Q])-[P])$$

It can be seen that $\deg D_T=0$, and $\sum(D_T)=\sum_{P\in E[m]}(P+Q-P)= m^2Q= mT = \infty$. By Theorem 1.1, $D_T$ is a principal divisor. (Q.E.D)

Now, it follows that there exists a rational function $g_T$ in $k(E)$ such that $div(g_T)=D_T$. For $S\in E[m]$, the Weil's pairing is defined as follows

$$e_m(S,T):=\frac{g_T(S+X)}{g_T(X)}$$

for any point $X\in E$ such that $g_T(S+X), g_T(X)\in k^*$. We will prove that

Lemma 1.2. $e_m(S,T)$ is well-defined, i.e. it does not depend on the choice of $X$.

Proof. We actually want to show that $\frac{g_T(S+X)}{g_T(X)}$ is a constant function. That is equivalent to say $div(g(X+S))-div(g(X))=0$. Let $h_T(X):=g_T(S+X)$, it is equivalent to prove that $div(h_T)=div(g_T)$. It can be seen that $g_T$ has zeros at $P+Q$, for some $Q\in E$ such that $mQ=T$, and $P\in E[m]$. This yields the zeros of $h_T$ are $P+Q-S$. Also, poles of $g_T$ are of the form $P$, where $P\in E[m]$. This yields poles of $h_T$ are actually $P-S$. So we get

$$div(h_T)-div(g_T)=\sum_{P\in E[m]}([P-S+Q]-[P-S])-\sum_{P\in E[m]}([P+Q]-[P])$$

But then, we can see that $S\in E[m]$, so $\{P-S|P\in E[m]\}=E[m]$. This yields $div(h_T)-div(g_T)=0$. This yields $h_T/g_T$ is a constant. (Q.E.D)

We will show next that in fact, $e_m(S,T)\in \mu_m$, where $\mu_m$ is the group of $m$-th root of unity in $k^\times$.

Lemma 1.3. $e_m(S,T)\in \mu_m$.

Proof. It can be seen that $m([T]-[\infty])$ is a principal divisor by Theorem 1.1. This implies there exists a function $f_T\in k(E)$ such that $div(f_T) = m([T]-[\infty])$. We already show that there exists a point $Q\in E$, such that $mQ=P$. And we can see

$$div(f_T\circ [m])=m\sum_{P\in E[m]}([P+Q]-[P])$$

since $f_T$ has zeros at $T$ with order $m$, so $f_T\circ [m]$ has zeros at $[m]^{-1}(T)=\{P+Q|P\in E[m]\}$, and each is of order $m$. Similarly for the poles. But then, the right hand side is just $mD_T=mdiv(g_T)=div(g_T^m)$. This yields $g_T^m=\lambda f_T\circ[m]$, for some $\lambda\in k^\times$. By possibly change $f_T$ by $\lambda f_T$, we can assume $g_T^m=f_T\circ [m]$. That means, for all points $X\in E$, we have $g_T^m(X)=f_T(mX)$, and for $S\in E[m]$, we have $g_T(X+S)^m=f_T(mX+mS)=f_T(mX)$. Hence, $\frac{g_T^m(X+S)}{g_T^m(X)}=1$. This implies $e_m(S,T)$ is a $m$-th root of unity. (Q.E.D)

So, we have proved if $m$ is any positive integer such that $(m,char(k))=1$, then there exists a well-defined map

$$e_m: E[m]\times E[m]\to \mu_m$$

where $\mu_m$ is the multiplicative group of $k^\times$ consisting of all $m$-th root of unity. The next section will be devoted to prove the properties of Weil's pairing.

2. Properties of Weil's pairing.

Proposition 2.1 (Bilinear). The Weil's pairing defined above is bilinear, i.e. for all $S_1, S_2, T_1, T_2\in E[m]$, we have $e_m(S_1+S_2,T)=e_m(S_1,T)e_m(S_2,T)$ and $e_m(S,T_1+T_2)=e_m(S,T_1)e_m(S,T_2)$.

Proof. If we denote $\tau_S(P):=P+S$, for all $P\in E$, i.e. $\tau_S$ is a translation with respect to $S$. Then it can be seen that $e_m(S_1+S_2, T)=\frac{g_T(S_1+S_2+X)}{g_T(X)}$ and $e_m(S_1, T)e_m(S_2,T)=\frac{g_T(S_1+X)}{g_T(X)}\frac{g_T(S_2+X)}{g_T(X)}$. Then $e_m(S_1+S_2,T)=e_m(S_1,T)e_m(S_2,T)$ iff $\frac{g_T(S_1+S_2+X)}{g_T(S_1+X)}=\frac{g_T(S_2+X)}{g_T(X)}=e_m(S_2,T)$. But this follows since $e_m(S_2,T)$ does not depend on the choice of $X$. If we replace $X$ by $S_1+X$, we get $e_m(S_2,T)=\frac{g_T(S_2+S_1+X)}{g_T(S_1+X)}$. And the conclusion follows.

The second property is more difficult. If we let $T_3:=T_1+T_2$, then by Theorem 1.1, we get $[T_1]+[T_2]-[T_3]-[\infty]=div(h)$ for some function $h\in k(E)$. We also have by definition

$$div(g_{T_1})=[m]^*([T_1]-[\infty]), div(g_{T_2})=[m]^*([T_2]-[\infty])$$
$$div(g_{T_3})=[m]^*([T_3]-[\infty])=[m]^*([T_1]+[T_2]-div(h)-2[\infty])=$$
$$=[m]^*([T_1]-[\infty]+[T_2]-[\infty]+div(h))=div(g_1)+div(g_2)+div(h\circ [m])=$$
$$=div(g_1.g_2.(h\circ[m])$$

So, in particular, we get $g_{T_3}=g_1.g_2.(h\circ[m])$. Hence,

$$e_m(S,T_1+T_2)=e_m(S,T_3)=\frac{g_{T_3}(S+X)}{g_{T_3}(X)}=$$
$$=\frac{g_1(S+X)g_2(S+X)h(mS+mX)}{g_1(X)g_2(X)h(mX)}$$

But then, since $mS=\infty$, we have $\frac{h(mS+mX)}{h(mX)}=1$. This yields by our previous calculation $e_m(S, T_1+T_2)=e_m(S,T_1)e_m(S,T_2)$. (Q.E.D)

Proposition 2.2 (Alternating). For any $T\in E[m]$, we have $e_m(T,T)=1$. And one can easily deduce $e_m(S,T)=e_m(T,S)^{-1}$ for all $S,T\in E[m]$.

Proof. Because $E[m]\cong \mathbb{Z}_m\oplus \mathbb{Z}_m$, and the bilinear property, it is sufficient for us to prove $e_m(T,T)=1$, for $T$ is a point with the order exactly $m$. Let $h_i(X):=g_T(X+iT')$, because $div(g_T)=\sum_{P\in E[m]}[P+Q]-[P]$, where $Q\in E, mQ=T$. Let $h_i(X)=g(X+iQ)$, for some $0\le i\le m-1$. Then it can be seen

$$div(h_i)=\sum_{P\in E[m]}([P+Q-iQ]-[P-iQ])=\sum_{P\in E[m]}[P+(1-i)Q]-[P-iQ]$$

Because $mQ=T$, this yields

$$div(\prod_{i=0}^{m-1}h_i)=\sum_{i=0}^{m-1}div(h_i)=\sum_{P\in E[m]}([P-Q]-[P+(m-1)Q])=$$
$$=\sum_{P\in E[m]}[P-Q]-\sum_{P\in E[m]}[P-T-Q]$$

But then, the later divisor is just 0, because $T\in E[m]$. This yields $\prod_{i=0}^{m-1}h_i=1$, i.e, it is a constant. This yields
$$1=g_T(X)\prod_{i=1}^{m-1}g_T(X+iQ)=\prod_{i=0}^{m-1}g_T(X+iQ)=\prod_{i=1}^{m}g_T(X+iT)=\prod_{i=1}^{m-1}g_T(X+iQ)g_T(X+T)$$

Note that the third identity follows when we replace $X$ by $X+Q$, and the last identity follows since $mQ=T$. In particular, we get $g_T(X+T)=g_T(X)$. This implies $e_m(T,T)=1$. The second statement easily follows from Proposition 2.1 and the first statement. (Q.E.D)

Proposition 2.3 (Non-degenerate). The Weil's pairing is non-degenerate, i.e. if $e_m(S,T)=1$ for all points $S\in E[m]$, then $T=\infty$.

Proof.

Remark 2.4. We will contemporarily accept this fact, which will be proved later: if $g\in k(E)$ such that $g\circ \tau_S=g$ for all $S\in E[m]$, i.e. $g$ is invariant under the translation then $g=h\circ [m]$, for some $h\in k(E)$.

Due to the assumption, $g_T$ now is invariant under the translation map $\tau_S$, for all $S\in E[m]$. This yields $g_T$ is of the form $h\circ [m]$, for some $h\in k(E)$. Due to the proof of Proposition 2.1, we get $g_T^m=f_T\circ [m]$, i.e. $(h\circ[m])^n=f_T\circ [m]$. This yields for all point $P\in E$, $h(mP)...h(mP)=f_T(mP)$. Since $[m]$ is surjective, we have $h^m=f_T$. But then, the divisor of $f_T$ is $m([T]-[\infty])$. This yields $div(h)=[T]-[\infty]$. Due to Theorem 1.1, we have $T-\infty=\infty$, i.e. $T=\infty$.

So, it remains to prove Remark 2.4. We first begin with an important lemma

Lemma 2.5. If $\alpha$ is a separable isogeny from $E_1$ to $E_2$, then the map $\ker (\alpha) \to Aut(k(E_1)/\phi^*(k(E_2)))$ define by $T\mapsto \tau_T^*$ is a group isomorphism.

Proof. It can be seen first that for any $T\in E_1$, the translation $\tau_T^*: k(E_1)\to k(E_1)$ define by $g\mapsto g\circ \tau_T$ defines an automorphism of field, since its inverse is $\tau_{-T}^*$.

For any $T\in \ker\alpha$, and for all $f\in k(E_2)$, we have $\tau_T^*(\alpha^* f)=f\circ \alpha\circ \tau_T$. So for any point $P\in E_1$, we have $f\circ \alpha\circ \tau_T(P)=f\circ\alpha(P+T)=f\circ\alpha(P)$, since $\alpha(P+T)=\alpha(P)+\alpha(T)=\alpha(P)$. This yields $\tau_T^*(\alpha^* f)=\alpha^* f$, i.e. $\tau_T^*$ fixes the subfield $\phi^*(k(E_2))$ of $k(E_1)$.

Now, it is a group homomorphism since for $T,S\in \ker\alpha$, we have $T+S\mapsto \tau_{T+S}^*=\tau_T^*\circ \tau_S^*$. Now, it follows from Galois theory and our Remark 2 in the previous note that $\#Aut(k(E_1)/\phi^*(k(E_2)))=\deg \alpha =\#\ker \alpha<\infty$. So, we just need to prove the map is injective. But it is obvious, since if $\tau_T^*=\tau_\infty^*$, then for sure $T=\infty$ (Q.E.D)

Now, if we let $\alpha=[m]$ and $E_1=E_2=E$, then $\ker\alpha = E[m]$. Now, $g$ is fixed under the translation $\tau_S$, for all $S\in E[m]$ yields $g\in [m]^*(k(E))$, i.e. there exists $h\in k(E)$ such that $g=[m]^*h=h\circ [m]$. So, the proof of Remark 2.4 follows. (Q.E.D)

Proposition 2.6 (Compatible). Let $\varphi$ be any endomorphism of $E$, then for any $S, T\in E[m]$, we have $e_m(\varphi(S),\varphi(T))=e_m(S,T)^{\deg\varphi}$.

Proof. Let $\deg\varphi =d$. We want to prove for all point $P\in E$,

$$\frac{g_{\varphi(T)}(\varphi(P)+\varphi(S))}{g_{\varphi(T)}(\varphi(S))}=\bigg(\frac{g_T(P+S)}{g_T(P)}\bigg)^d$$

If we denote $\tau_S$ the translation by $S$ (i.e. $P\mapsto P+S$), the statement is equivalent to
\begin{align}
\frac{g_{\varphi(T)}\circ \tau_{\varphi(S)}\circ\varphi}{g_{\varphi(T)}\circ\varphi}=\bigg(\frac{g_T\circ\tau_S}{g_T}\bigg)^d (1)
\end{align}

One can see $\tau_{\varphi(S)}\circ\varphi(P)=\varphi(P+S)=\varphi\circ\tau_S(P)$. Therefore
$$\frac{g_{T\varphi(T)}\circ\tau_{\varphi(S)}\circ\varphi}{g_{\varphi(T)}\circ\varphi}=\frac{g_{\varphi(T)}\circ\varphi\circ\tau_S}{g_{\varphi(T)}\circ\varphi}$$
And for the RHS of (1), because $(g_T\circ\tau_S)^d=g_T^d\circ\tau_S$, we have (1) is equivalent to
$$\frac{g_{\varphi(T)}\circ\varphi\circ\tau_S}{g_{\varphi(T)}\circ\varphi}=\frac{g_T^d\circ\tau_S}{g_T^d}\Leftrightarrow\frac{g_{\varphi(T)}\circ\varphi\circ\tau_S}{g_T^d\circ\tau_S}=\frac{g_{\varphi(T)}\circ\varphi}{g_T^d}\circ\tau_S=\frac{g_{\varphi(T)}\circ\varphi}{g_T^d}$$

That means, we have to prove that $\frac{g_{\varphi(T)}\circ\varphi}{g_T^d}$ is invariant under the translation of any $S\in E[m]$. We need the following

Lemma 2.7. If $r=s\circ[m]$ (i.e. $r=[m]^*s$), where $s\in k(E)$, then $r$ is invariant under the translation of any $S\in E[m]$.

Proof. For any $S\in E[m]$, we have $r(P+S)=s\circ [m](P+S)=s(mP+mS)=s\circ[m](P)=r(P)$, for all $P\in E$. (Q.E.D)

Hence, it is sufficient for us to prove that $F:=\frac{g_{\varphi(T)}\circ\varphi}{g_T^d}=[m]^*[s]$ for some $s\in k(E)$. Let us compute the divisor of $F$, where $div(g_T)=[m]^*([T]-[\infty])$, we have
$$div(F)=(\varphi^*[m]^*)([\varphi(T)]-[\infty])-d[m]^*([T]-[\infty])=[m]^*(\varphi^*([\varphi(T)]-[\infty])-d([T]-[\infty]))$$
And hence, it is sufficient for us to prove that $D:=(\varphi^*([\varphi(T)]-[\infty])-d([T]-[\infty]))$ is a principal divisor. By definition, we have
$$\varphi^*([\varphi(T)]-[\infty])=\sum_{R\in \ker\varphi}e_R([T+R]-[R])$$
where $e_R$ is the ramification index at $R$. Hence, $\deg D=0$, because $\sum_{R\in \ker\varphi}e_R=\sum_{R\in\varphi^{-1}(\infty)}e_R=d$. And by the similar reason, the sum of $D$ is
$$\sum_{R\in \ker\varphi}e_R(T+R-R)-d(T-\infty)=\infty$$
That means, $D$ is a principal divisor, by Theorem 1.1. This implies $F=[m]^*s$, for some $s\in k(E)$. And hence, the statement now follows. (Q.E.D)

Proposition 2.8 (Galois invariant). If $E$ is defined over $k$ (may not be algebraically closed), then for all $\sigma\in Gal(\bar{k}/k)$, we have $e_m(S^\sigma,T^\sigma)=e_m(S,T)^\sigma$.

Proof. It just follows if we consider the action of $\sigma$ on $g$. (Q.E.D)

An interesting article about cryptography (by Neal Koblitz)

Here, one should read to see how mathematical ideas and mathematicians "jump" into cryptography.

(Fourier Analysis on Finite Abelian Groups I) Characters of finite abelian groups

Fourier analysis on finite abelian groups are one of the most important topics in theory and applications: physics, computer science,... We will start the series notes about Fourier transform for the discrete case before going to its applications. The origin content is taken from the book "Fourier Analysis Over Finite Abelian Groups" of Bao Luong.

In this note, we present several properties of the character of finite abelian groups $G$.

1. The character of a group

Definition 1. A character of a group is a homomorphism from $G$ to $\mathbb{C}^\times$. The set of all characters of $G$ is denoted by $\hat{G}$, or $\rm{Hom}(G,\mathbb{C}^\times)$. 

Theorem 2.  $\hat{G}$ is an abelian group, with respect to the pointwise product $\chi \chi^{'}(x)= \chi(x)\chi^{'}(x)$ for all $\chi,\chi'\in \hat{G}$.

Proof.

• The pointwise product of two homomorphism $\chi, \psi \in \hat{G}$ is again a homomorphism in $\hat{G}$. Let $a,b$ be elements of $G$, we have $\chi \psi (ab) = \chi (ab) \psi (ab) = \chi (a) \chi (b) \psi (a) \psi (b) = \chi \psi (a) \chi \psi (b)$. 
• The identity of $\hat{G}$ is $1_\hat{G}$.
• An element $\chi (x) \in \hat{G}$ has an inverse element $\chi^{-1}(x) = \chi(x^{-1})$.
• The commutative and associative properties of $\hat{g}$ are inherited by $\mathbb{C}^\times$.

(Q.E.D)

The group $\hat{G}$ is called the dual group of $G$, and in the case of finite abelian group we can prove that $G \cong \hat{G}$. 

Lemma 3. A cyclic group $\mathbb{Z}_n$ is isomorphic with its dual group $\hat{\mathbb{Z}_n}$.

Proof.
Let $\chi: \mathbb{Z}_n \to \mathbb{C^{\times}}$ be a homomorphism. We have $\chi(0)=\chi(na)= \chi(a)^n = 1$. Hence, $\chi(a) \in \mu_n$ with $\mu_n$ is the group of $n^\text{th}$-root of unity.  Since $\mathbb{Z}_n = <1>$ is a cyclic group, we only care about the mapping $\chi(1)$ to $\mu_n$, and others can be constructed as $\chi(a) = a\chi(1)$.
So how is the structure of $\hat{\mathbb{Z}_n}$? We will see it by defined a mapping $\chi_x$ as follows:

$\chi_x: \mathbb{Z}_n \to \mathbb{C}^{\times}$       $a \mapsto \chi_x(a)$ 

with $x \in \mathbb{Z}_n, \omega \in \mu_n$ and $\chi_x(a) = \omega^{xa}$.

1. The mapping $\chi_x \in \hat{\mathbb{Z}_n}$ is a homomorphism: for $\forall a,b \in \mathbb{Z}_n$, $\chi_x(a+b) = \chi_x(a) \chi_x(b) = \omega^{x(a+b)}$.
2.  $\forall a \in \mathbb{Z}_n \text{, } \chi_x(a) = \chi_y(a) \Leftrightarrow x = y \mod n  $.

$\Rightarrow \hat{\mathbb{Z}_n} = \{ \chi_x \text{, with } x \in \mathbb{Z}_n \} $.
Let $\phi: \mathbb{Z}_n \to \hat{\mathbb{Z}_n}$ with $x \mapsto \chi_x$. The mapping $\phi$ is a homomorphism since $\phi (x+y) (a)= \chi_{x+y} (a)= \omega ^{(x+y)a} = \chi_x (a) \chi_y(a)= \phi(x) \phi(y) $. Moreover, $\phi$ is bijective by the construction and (2).

We conclude that $\mathbb{Z}_n \cong \hat{\mathbb{Z}_n}$.

(Q.E.D)

Lemma 4: Let $G_1 \cong G_2$ be finite abelian groups.  Prove that $\hat{G_2} \cong \hat{G_1}$.
Proof.
Let $f$ be an isomorphism from $G_1$ to $G_2$ and $\chi_1 \in \hat{G_1}, \chi_2 \in \hat{G_2} $. We have $\chi_1 = \chi_2 \circ f$.

 

There is a one to one mapping from $\chi_1$ to $\chi_2$. The mapping $h : \hat{G_2} \to \hat{G_1}$ ($ \chi_2 \mapsto \chi_2 \circ f $) is a homomorphism. Hence, $\hat{G_1} \cong \hat {G_2}$.

(Q.E.D)

Let $G_1, G_2$ be finite abelian groups. The mapping $\chi_1 \otimes \chi_2: G_1 \times G_2 \to \mathbb{C}^{\times}$ is called the tensor product of two characters if $\chi_1 \otimes \chi_2(a,b) = \chi_1(a) \chi_2(b)$ with $\chi_1 \in \hat{G_1}, \chi_2 \in \hat{G_2}$. This product is unique, and we denoted $\chi_{xy}=\chi_x \otimes \chi_y$.
The definition of the tensor product of two characters yields to an interesting relation between $\widehat{G_1 \times G_2}$ and $\hat{G_1}, \hat{G_2}$.

Lemma 5: $\hat{G_1} \times \hat{G_2} \cong \widehat{G_1 \times G_2}$, where $G_1$ and $G_2$ are finite abelian group.
Proof.
To make it clear we write the statement above by

$Hom(G_1 , \mathbb{C}^\times) \times Hom(G_2, \mathbb{C}^\times) \cong Hom(G_1 \times G_2, \mathbb{C}^\times) $

Consider the injections $i_1 : G_1 \to G_1 \times G_2$ with $a \mapsto (a,1)$ and $i_2 : G_2 \to G_1 \times G_2$ with $b \mapsto (1, b)$. Let $\chi \in Hom(G_1 \times G_2, \mathbb{C}^{\times})$, the composition $\chi \circ i_i$ is a character of $Hom( G_i, \mathbb{C}^{\times})$ where $i \in \{ 1, 2 \}$. Furthermore, $\chi_{12}(a,b)=\chi(i_1(a)) \chi(i_1(b)) = \chi (a,b)$. It can be seen that an arbitrary character $\chi$ of $\widehat{G_1 \times G_2}$ can be express by a tensor product of two character $\chi_1 \in \hat{G_1}$ and $\chi_2 \in \hat{G_2}$: $\chi = \chi_1 \otimes \chi_2$.

$f: Hom(G_1 , \mathbb{C}^\times) \times Hom(G_2, \mathbb{C}^\times) \cong Hom(G_1 \times G_2, \mathbb{C}^\times) $
$(\chi_1, \chi_2) \mapsto \chi_{12}$

We have $f((\chi_1,\chi_2)(\chi_{1'},\chi_{2'})) (a,b) = f(\chi_{11'}, \chi_{22'}) (a,b) = \chi_1 (a) \chi_{1'}(a)\chi_2 (b) \chi_{2'}(b) $
$= \chi_{12}(a,b) \chi_{1'2'}(a,b) = f(\chi_1,\chi_2)f(\chi_{1'},\chi_{2'})$. The mapping $f$ is homomorphism.

(Q.E.D)

Theorem 6. There is an isomorphism from a finite abelian group $G$ to its dual group $\hat{G}$.

Proof.

• We know that a finite abelian group is isomorphism with the direct product of cyclic groups (The Fundamental theorem of finite Abelian group). Hence, we can express $G \cong \mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times \ldots \times \mathbb{Z}_{nm}$. 
• Apply Lemma 5 and Lemma 4, we have $\hat{G} \cong \widehat{\mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times \ldots \times \mathbb{Z}_{nm}} \cong \hat{\mathbb{Z}_{n_1}} \times \hat{\mathbb{Z}_{n_2}} \times \ldots \times \hat{\mathbb{Z}_{nm}} \cong G$. 

(Q.E.D)

2. The orthogonal relation

Let $f: G \to C$, we see that $V^G = \{ f \}$ is a vector space over the $\mathbb{C}$.

• The addition of two functions in $V^G$ is defined as: $(fg)(x) = f(x) g(x) \Rightarrow fg \in V^G$ .
• The scalar product of $f \in V^G$ and $a \in \mathbb{C}$ is $(af)(x) = a(f(x)) \Rightarrow af \in V^G$.
• $(a+b)f = af +bf$.
• $a(f+g) =af + bg $
• The identity mapping in $V^G$ is $1_{V^G}(x) = 1$.
• The inverse element of $f \in V^G$ is $f^{-1}(x) = \frac{1}{f(x)}$. 

It is natural to ask about dimensions of the vector space $V^G$. We need to construct a set of linearly independent vector $\{\delta_i\}$ that spans $V^G$. The cardinality of $\{\delta_s\}$ is $dim(V^G)$. Let's try to build the mapping $\delta_i$ (it is also called the Kronecker delta) as follows:

$\delta_i: G \to C$ 

with $G = \{ a_1, \ldots, a_n\}$ and $\delta_i(a_j) = 1$ if $i=j$  and $\delta_i(a_j) = 0$ if $i \ne j$.

• $\{\delta_i\}^n_{i=1}$ is linearly independent since if there is existed a collection of coefficients $c_i \in \mathbb{C}$ satisfied: $\forall a \in G, \sum^{n}_{i=1}c_i \delta_i (a) = 0$. Then, $a = a_i$ leads to $c_i = 0$. Hence, we only obtain a trivial solution $\{c_i=0 \}^{n}_{i=1}$.
• Every function $f \in V^G$ can be expressed as $f(a) = \sum^{n}_{i=1}{f(a) \delta_i(a)}$.

The set $\{ \delta_i \}^n_{i=1}$ forms the standard basis of $V^G$. Moreover,  $dim(V^G) = |\{\delta_i\}^n_{i=1}| = |G|$.

Lemma 7: The dimensions of $V^G$ is |G|.

(Q.E.D)

 Endowing the vector space $V^G$ with a suitable inner product $<,>$ , we can see a beautiful result about the orthogonal basis of $V^G$. More precisely, this basis is the dual goup of $G$.

Now, we define the inner product of two functions $f,g \in V^G$

$<f,g> = \sum_{a \in G}{f(a) \overline{g(a)}} $

with $\bar{g(a)}$ is the complex conjugate.

The inner product $<f,g>$ is satisfied the following axioms:

• Conjugate symmetry: $<f,g> = \overline{<g,f>}$:  $\overline{<g,f>}=\overline{\sum_{a \in G}{g(a) \overline{f(a)}}} = \sum_{a \in G}{f(a) \overline{g(a)}} = <f,g>$. 
• Linearity: 
• $\forall \alpha \in \mathbb{C}, <\alpha f, g> = \alpha<f, g> = \alpha\sum_{a \in G}{f(a) \overline{g(a)}}$. 
• $<f_1 + f_2,g> = <f_1,g> +  <f_2,f> $ .

Lemma 8. $\forall \chi \in \hat{G},\sum_{a}{\chi(a)} = 0$. Note that, $\chi$ is not the trivial character $\chi_0$ with $\forall a \in G, \chi_0(a) = 1$.
Proof.
We can always choose $b \ne 1$ such that $\chi(b) \ne 0$  (the proof is found in the previous post, Exercise 3).
$\Rightarrow \sum_{a}{\chi(b)\chi(a)} = \sum_{a}{\chi(ab)} = 0 \Rightarrow \sum_{a}{\chi(a)} = 0$.

(Q.E.D)

So, what is the inner product of two characters $\chi_1, \chi_2 \in \hat{G}$? Let's try to compute it step by step to see 

$<\chi, \psi> = \begin{cases}
0   &  \chi \ne \psi \\
|G| &  \chi = \psi
\end{cases}
$.

More precisely, $\forall \chi, \psi \in \hat{G}, <\chi, \psi> = \sum_{a \in G}{\chi(a) \overline{\psi (a)}} = \sum_{a \in G}{\chi(a) \psi^-1 (a)}$. If $\chi \ne \psi $, then by Lemma 7, $<\chi, \psi> = 0$, otherwise $<\chi, \psi> = |G|$. 

Theorem 9: The group $\hat{G}$ forms the orthogonal basis of the vector space $V^G$.

Proof.

The set $\hat{G}$ is orthogonal, then $\hat{G}$ is linearly independent. In Lemma 7, we proved that $dim(V^G) = |G| = |\hat{G}|$ $\Rightarrow $ $\hat{G}$ is an orthogonal basis of $V^G$.

(Q.E.D)

Remark 10: In a inner product vector space $V$, an arbitrary vector $v \in V$ is represented by the following equation

$v = \sum^n_{i=1}{c_i u_i}$ (*) 

, where $U = \{ v_1, \ldots, v_n\}$ is a orthogonal basis of $V$. By the bilinear properties of inner product, we obtain $<v,u_j> = <\sum^n_{i=1}{c_i u_i}, u_j > =\sum^n_{i=1}{c_i<u_i,u_j>} = c_j <u_j, u_j> $.

Rewrite (*):

$v = \sum^n_{i=1}{\frac{<v,u_i>}{<u_i,u_i>} u_i}$

Back to this section, $\hat{G}$ is an orthogonal basis of $V^G$, then 

                                  $ f = \frac{1}{|G|} \sum_{\chi \in \hat{G}}{<f,\chi>{\chi}}$.

Simplify a character $\chi$ by $B_{\chi} = \frac{1}{|G|} \chi $, the orthonormal basis of $V^G$ is $B = \{ B_{\chi} \text{ with } \chi \in \hat {G} \}$. Hence, every function $f \in V^G$ can be represented as:

$f = \sum_{B_{\chi} \in B}{<f,B_{\chi}>B_{\chi}}$.

The result obtained by the Fourier transform of $f$ is $\hat{f} = <f,B_{\chi}>$. In the next post, several ideas about Fourier Transform will be yields. And we hope it can make you happy.

Equations on finite fields

I copy my note from our old blog on trello.com.

PART 1

First, we begin with the circle $(C): x^2 + y^2 = 1$ over a finite field $F_q$, whose characteristic is not $2$. And the question we are interested in is that how many points lie on this circle? 

If $-1$ is a square in $F_q$, we assume that $-1=a^2$, then $x^2+y^2=x^2-a^2y^2=(x-ay)(x+ay)=1$. Hence, $(x,y)\in (C)$ is equivalent $x-ay = b, x+ay = b^{-1}$. And from this $x = (b+b^-1)/2$, and $y = (b-b^{-1})/2a$. An EXERCISE is that counting the number of points in this case.

If $-1$ is not square in $F_q$, we consider the quadratic extensions $F_q(i)/F_q$, where $i$ is the square root of $-1$ in $\overline{F_q}$. The field extension is Galois, and there exists two elements of the Galois group $\{id, x\mapsto x^q\}$, or equivalent $\{id, x+iy \mapsto x-iy\}$.

Because of the definition of norm, let $u=x+iy$, we can compute its norm by two ways: $Norm(u) = x^2 + y^2 = u^{q+1}$. And hence, points on $(C)$ are exactly the kernel of the norm map.

Because $F_q(i)^\times$ is cyclic, whose order $q^2-1$ and generator $g$, the kernel of norm is the cyclic group generated by $g^{q-1}$, and has order $q+1$. That means, $(C)$ will have $(q+1)$ points.

(NOTE: at this stage, we will also conclude that Norm is surjective, because $\#Norm(F_q(i)^\times)=|F_q(i)^\times|/|Ker(Norm)|=q-1=|F_q^\times|$)

Note that there is a 1-1 corresponding between points on $(C)$ and $Ker(Norm)$. Via this, we can define the group law for points in $(C)$ (See my earlier comment on the card named "Pairing on Linear Varieties and Curves of Genus 0").

An interesting corollary of NOTE is that: the norm map is surjective, and hence, any element in $F_q$ can be represented under the form $x^2+y^2$. And we obtain the following

Theorem 1. Let $F_q$ be a finite field with characteristic not 2, and $\alpha,\beta,\gamma$ are not zero in $F_q$. Then the equation $\alpha x^2+\beta y^2+ \gamma z^2$ always has non-trivial solutions over $F_q$.

(Or equivalent statement by another language: Any quadratic form of rank $\ge 3$ is isotropic over $F_q$)

We NOTE that the equation $\alpha x^2 + \beta y^2 = 0$ does not always has non-trivial solution over $F_q$, since in this case $\frac{-\alpha}{\beta}$ must be a square in $F_q^\times$.

The interesting story now begins. One may ask how is the case of cubic forms? That means, for what $n$, we can ensure that the equation $a_1x_1^3+a_2x_2^3+...+a_nx_n^3$ always has non-trivial solution? And more generally, $d$-forms?

PART 2

The aim of this part is to prove the theorem of Chevalley, conjectured by E. Artin.

Theorem 1. Let $f(x_1,...,x_n) \in F_q[x_1,...,x_n]$ be a multi-variables of degree $d$, where $n>d$, and $f(0,0,..,0)=0$, then $f$ has at least two solutions.

Two solutions are mentioned in the book of K. Ireland and M. Rosen "A Classical Introduction to Modern Number Theory", Chapter X, and we just explain one of them, with some further comments. 

(NOTE: that the remaining solution is very interesting too, and close to the method for supersingular elliptic curves discussed in the book of L. C. Washington about EC, Chapter IV.)

Assume that $f$ has no other solutions. That means, $f(a_1,...,a_n)\ne 0$, for all $(a_1,...,a_n)\ne (0,...,0)$, that means $1-f^{q-1}$ takes value 1 at $(0,..,0)$, and 0 otherwise. It can be seen that, in fact, it is "equivalent" (this will be defined later) to the polynomial $g(x_1,...,x_n)=(1-x_1^{q-1})...(1-x_n^{q-1})$. because $g$ takes value 1 at $(0,...,0)$, and 0 otherwise (*).

The term "equivalent" can be explained as follows.

Definition 2. The two polynomial $f,g$ over $F_q[x_1,...,x_n]$ is called equivalent if $f(a_1,...,a_n)=g(a_1,...,a_n)$ for all $a_i\in F_q$.

The properties Frobenius automorphism allows us to prove that $x^q$ and $x$ are equivalent. And hence, $x^{aq+r}$ is equivalent to $x^r$. And we can extend this observation to the following

Proposition 3. Let $f\in F_q[x_1,...,x_n]$, then $f$ is equivalent to $g$, where any monomial term of $g$ has the form $x_1^{a_1}...x_n^{a_n}$, and $a_i<q$. 

Such a polynomial $g$ is called reduced polynomial of $f$, and it is easy to see that the reduced polynomial of a given polynomial is unique.

Proposition 4. The reduced form of a polynomial is unique.

Proof. It is equivalent to prove that if $f$ and $g$ are reduced forms, i.e. their monomial term has the forms $x_1^{a_1}...x_n^{a_n}$, where $a_i<q$, and identical at all values in $F_q^n$, then they are identical. Let $h=f-g$, and fix the value $a_2,...,a_n\in F_q$, we can see $k(x_1)=h(x_1,a_2,...,a_n)$ is the polynomial of variable $x_1$, with the degree does not exceed $q-1$, and it takes $q$ zeros. Hence $k(x_1)$ is identical zero. Extend this method, it is obvious to see that $h$ is zero, and hence $f\equiv g$ (Q.E.D).

Back to (*), we can see that actually, $g$ is a reduced polynomial of $f$, and hence, $\deg(1-f^{q-1})=(q-1)\deg(f) \ge n(q-1)$. This implies $d\ge n$, a contradiction with the hypothesis of our theorem. Hence, $f$ must have another solution other than zero. And the theorem of Chevalley is now completely proved (Q.E.D).

As a consequence, we can see in any positive characteristic, the equation $\alpha x_1^2 + \beta x_2^2 + \gamma x_3^2$, where $\alpha, \beta, \gamma \ne 0$ always has non-trivial solution (NOTE that in the first part, we implied this in the case characteristic is not 2). And for $d$-form, the number of variables must be greater than $d$ to ensure that our equation always has non trivial solution.

How about in the case $n=d$? A very interesting counter example is built in the EXERCISE section of this book (ex. 17). We will conclude this part by proving the following

FACT 5. There is a $d$-form in $d$ variables such that it has no solution other than $(0,...,0)$.

Proof. We now need some Galois theory to point out this. Consider the extension $F_{q^d}/F_q$. The Galois group in this case is cyclic of order $d$, and generated by the Frobenius automorphism $x \rightarrow x^q$, and all of its elements is $\{id, x\rightarrow x^{q}, ..., x\rightarrow x^{q^{d-1}}\}$. Let $a_1,...,a_d$ is the basis for the vector space $F_{q^d}$ over $F_q$. Consider the polynomial

$$f(x_1,...,x_m)=\prod_{i=0}^{d-1}(a_1^{q^i}x_1+...+a_d^{q^i}x_d)$$

Now, $f$ is invariant under the action from the Galois group (CHECK THIS). Hence, it is defined in $F_q[x_1,...,x_d]$. And the basic facts for the basis of a vector space implies that $f=0$ iff $x_1=...=x_d=0$. And this is our counter example. (Q.E.D)

The book gave a hint for the polynomial and the solution is mine. I don't know how the idea for this counter example comes from. And I didn't think that the Galois theory helps in this case. But actually, it does!

PART 3

IMPORTANT: Fields in this note have characteristic different from 2.

For a long time (maybe 2 weeks), I did not take note on anything, there is many things here (and, of course, quite far from Math), and it makes me feel a little bit of tired. I want to write something about the orthogonal groups over finite fields today.

It should be first noted that we need to know something about quadratic forms first, Witt's theorems (Cancelation Theorem and Extension Theorem). It is not difficult, but really interesting, because it reflects the geometric intuition, and one may be curious how to define "geometry" over arbitrarily fields. 

I took a note in the past, but today I look it back, and find some stupid mistakes. I did not bring the other notes with me. But I will edit the mistakes, and then make an outline to compute the order of the orthogonal group over finite fields (associated with a given quadratic form).

But does it have anything to do with the equations over finite fields? Well, as we will see later, one of the most important steps is: compute the number of solutions for the equation $a_1x_1^2+a_2x_2^2+...+a_nx_n^2 = 1$ over $F_q$. And this can be done with the help of Jacobi and Gauss's sum over finite fields. 

A simple note on Dirichlet's series (Not finished)

The note consists of four sections. The first sections introduces the Euler's proof for infiniteness of primes. The second section mentions some intuition about Dirichlet's series and Dirichlet's product, that leads to a loose proof of Moebius's inversion theorem. A more compact analytic theory of Section 2 will be mentioned in Section 3. And the fourth section includes a proof of Euler's product for Dirichlet's series, and some interesting factorizations.

As an application of Moebius' inversion theorem, one may refer to my thesis about counting irreducible polynomials over finite fields of a certain degree.

1. Euler's factorization.

We begin this section by recall some facts about series.

Theorem 1.1. Let $f$ be a one-to-one map from $\mathbb{Z}^+$ to $\mathbb{Z}^+$, and $\sum a_n$ is the absolutely convergent series. Then $\sum a_{f(n)}$ is also absolutely convergent. And $\sum a_n=\sum a_{f(n)}$.

Proof. Left as an exercise.

We can see that the series $\sum a_{f(n)}$ can be considered as a rearrangement of the series $\sum a_n$. That means, if $\sum a_n$ is absolutely convergent, then we can permute its elements, and the obtained series is also absolutely convergent, and they have the same sum.

We can use this to prove the following

Theorem 1.2. There exists infinitely many primes.

Proof. If there exists finitely many primes. Consider the following product $\prod_p (\frac{1}{1-p^-1})$. Using the identity $\frac{1}{1-x}=1+x+x^2+...$ for all $|x|<1$, our product is exactly $\prod_p (1+\frac{1}{p}+\frac{1}{p^2}+...)$. Because there exists finite primes, our product is in fact convergent. And hence, we can rearrange all terms in the product, and using the unique factorization of the integer, we obtain the sum $\sum_{n\ge 1}\frac{1}{n}$. But this sum diverges, and this leads the contradiction. Hence, there exists infinitely many primes. (Q.E.D)

For stronger result, we will prove

Theorem 1.3. The sum $\sum_p \frac{1}{p}$ that runs over all primes diverges.

Proof. If $\sum \frac{1}{p}$ converges, then $ \sum_p \frac{1}{p}+\sum_p (\frac{1}{p^2}+...+\frac{1}{p^n}+...)$ converges, since $\frac{1}{p^2}+\frac{1}{p^3}+...=\frac{1}{p^2}(1+\frac{1}{p}+...)=\frac{1}{p^2}\frac{1}{1-p^{-1}}<\frac{2}{p^2}$, and hence $\sum_p (\frac{1}{p^2}+...+\frac{1}{p^n}+...)<2(\sum_{n\ge 1}\frac{1}{n^2})$, which converges.

Because $\frac{1}{p^m}\ge \frac{1}{mp^m}$, we can see that the sum $\sum_p \frac{1}{p}+\sum_p (\frac{1}{2p^2}+...+\frac{1}{np^n}+...)$ converges. And we can rearrange the series so that it has the form $\sum_p (\frac{1}{p}+\frac{1}{2p^2}+\frac+...+\frac{1}{np^n}+...)$. Using the identity $\log(\frac{1}{1-x})=\sum_{n\ge 1}\frac{x^n}{n}$, we have $\sum_p (\frac{1}{p}+\frac{1}{2p^2}+\frac+...+\frac{1}{np^n}+...)=\sum_p \log(\frac{1}{1-p^-1})$ converges. That means, the sum $\sum_p \log(\frac{1}{1-p^{-1}})$ is bounded by a positive real number $M$. This implies $e^M > \prod_p (\frac{1}{1-p^{-1}})$. And hence, our product is convergent. This leads the contradiction to Theorem 1.2. And hence $\sum_p \frac{1}{p}$ is divergent. (Q.E.D)

From Theorem 1.1, one may think that Euler comes very closely to the identity $\sum_n \frac{1}{n}=\prod_p (\frac{1}{1-p^{-1}})$. But actually, this does not make sense, because the two sides of the identity go to infinity.  And it is natural to think about the convergent sum $\sum_n \frac{1}{n^s}$, where $Re(s)>1$. Can one obtain the Euler's factorization? That means, $\sum_n \frac{1}{n^s}=\prod_p (\frac{1}{1-p^{-s}})$? Amazingly, it is true! In the second section, we will give a more general context for the Euler's product.

2. Dirichlet's series, Dirichlet's product and Moebius inversion theorem.

To begin with this section, we recall about the product of two series.

Theorem 2.1. Let $\sum_m a_m$ and $\sum_n b_n$ be two absolutely convergent series, and their values are $A,B$ respectively. Then their Cauchy's product, defined by $\sum_k c_k$, where $c_k=\sum_{m+n=k}a_mb_n$ also converges absolutely, and $\sum_k c_k = AB$. Also, their Dirichlet's product, defined by $\sum_k d_k$, where $d_k = \sum_{d|k}a_db_{\frac{k}{d}}$, also converges absolutely, and $\sum_k d_k = AB$.

Proof. See T. M. Apostol "Mathematical Analysis", Chapter 8.

The Cauchy's product is very similar to the product of polynomials, and the Dirichlet's product appear naturally when we consider about product of two Dirichlet series, defined as follows.

Definition 2.2. Let $f:\mathbb{Z^+}\rightarrow \mathbb{C}$ be a map, then the Dirichlet series corresponds to $f$ is $\sum_n \frac{f(n)}{n^s}$.

Formally, one can take the product of two Dirichlet's series without focusing on their convergent information, and represent the result as another Dirichlet's series.

$$(\sum \frac{a_n}{n^s})(\sum \frac{b_n}{n^s})=\sum \frac{c_n}{n^s}$$

where $c_n=\sum_{d|n}a_db_{n/d}$. If $(\sum \frac{a_n}{n^s})$ and $(\sum \frac{b_n}{n^s})$ are absolutely convergent, then their Dirichlet's product will converge to their product, as Theorem 2.1 pointed out. The very important example will be constructed below.

Definition 2.3. The Moebius $\mu$ functions is defined as $\mu(1)=1, \mu(n)=(-1)^k$ where $n=\prod_{i=1}^k p_i$, where $p_i$'s are distinct primes. If $n$ is not square free, then $\mu(n)=0$.

We have an interesting property for Moebius's function.

Lemma 2.4. For all $n>1$, $\sum_{d|n}\mu(d)=0$.

Proof. Left as an exercise.

We recall that the series $\zeta(s)=\sum_{n\ge 1}\frac{1}{n^s}$ absolutely converges for all complex number $s$, such that $Re(s)>1$ (For $s=a+bi$, then $|n^s|=|e^{s\log n}|=|e^{(a+bi)\log n}|=e^{a\log n}=n^a$, and $\sum \frac{1}{n^a}$ converges for $a>1$). Let us consider next the series $M(s)=\sum_{n\ge1}\frac{\mu(n)}{n^s}$. Because the value of Moebius function does not exceed $1$, $M(s)$ is absolutely converge. And, their Dirichlet's product will be

$$\zeta(s) M(s) = \sum_n \sum_{d|n}(\frac{\mu(d)}{n^s})$$

By Lemma 2.4, we have $\zeta(s) M(s)$ is actually 1. And hence $\zeta(s)$ is the inverse of $M(s)$. And we have constructed the main ingredient for the Moebius' inversion theorem, which stated

Theorem 2.5. Let $F, f$ be two arithmetical functions (that means they go from $\mathbb{Z}^+$ to $\mathbb{C}$, and $F(n)=\sum_{d|n}f(d)$ for all $n\in \mathbb{Z}^+$, then $f(n)=\sum_{d|n}\mu(d)F(n/d)$ for all $n$.

Remark. Note that we will give here a loose proof, that may not make sense if we do not consider carefully about convergence of Dirichlet's series. Amazingly, though we will build a more compact theory latter, we also cannot prove the full theorem. Actually, the proof of the theorem is purely algebraic, when one considers the set of all arithmetical functions $f$, such that $f(1)\ne 0$, under the Dirichlet's product forms an abelian group.

Proof.  Let us consider $\sum \frac{F(n)}{n^s}, \sum \frac{f(n)}{n^s}$, the Dirichlet's series corresponding to $F$ and $f$, respectively. Then $F(n)=\sum_{d|n}f(d)$ implies that $\sum \frac{F(n)}{n^s}$ is the Dirichlet's product of two series $\sum \frac{f(n)}{n^s}$ and $\zeta(s)$. Hence, $M(s)(\sum \frac{F(n)}{n^s})=\sum \frac{f(n)}{n^s}$. And hence, when we consider all coefficients at each $n^s$, we obtain $f(n)=\sum_{d|n}\mu(d)F(n/d)$. (Q.E.D)

Look at the proof above, one may ask, if we omit all conditions about convergence, how can we make sure that if $\sum \frac{a(n)}{n^s}=\sum \frac{b^n}{n^s}$, then $a(n)$ is equal $b(n)$ for all $n$? Actually, it is the very important properties of Dirichlet's series, known under the name Uniqueness Theorem, as we will see later.

3. Some analytic properties of Dirichlet's series.



4. Euler's product.


5. An application of Moebius inversion theorem.

In this section, we will give an interesting application of Moebius inversion theorem to compute the $n^{th}$ cyclotomic polynomial. Let $\xi$ be the $n^{th}$ primitive root of unity in $\mathbb{C}$. The $n^{th}$ cyclotomic polynomial $\Phi_n(x)$ is defined

$$\Phi_n(x)=\prod_{k\le n,(k,n)=1}(x-\xi^k)$$

It can be seen that $\Phi_1(x)=(x-1), \Phi_2(x) = (x+1), \Phi_3(x) = (x^2+x+1)$. The natural question is that is $\Phi_n(x) \in \mathbb{Q}(x)$? And how to compute them? We first note that

Lemma 5.1. $(x^n-1)=\prod_{d|n}\Phi_d(x)$.

Proof. First, we can see that $x^n-1 = \prod_{k=1}^{n}(x-\xi^k)$, and The set $N = \{1,2,...,n\}$ can be partitioned into disjoint subsets $A(d) = \{k\in N|gcd(k,n)=d\}$, where $d|n$. Hence,

$$(x^n-1)=\prod_{d|n}\prod_{k\in A(d)}(x-\xi^k)$$

Furthermore, the product $\prod_{k\in A(d)}(x-\xi^k\big)$ is actually $\Phi_{n/d}(x)$, because $\xi^d$ is the $\frac{n}{d}^{th}$ primitive root of unity. This shows $(x^n-1)=\prod_{d|n}\Phi_{n/d}(x)=\prod_{d|n}\Phi_d(x)$ (Q.E.D).

From this, we can deduce $\log(\prod_{d|n}\Phi_d(x))=\log(x^n-1)$. Hence, $\sum_{d|n}\log(\Phi_d(x))=\log(x^n-1)$. Let us denote $F(n)=\log(x^n-1)$, and $f(d)=\log(\Phi_d(x))$. Then $F(n) = \sum_{d|n}f(d)$. Applying the Moebius inversion theorem, $f(d) = \sum_{d|n}\mu(d)F(n/d)$. That means,

$$\log(\Phi_n(x))=\sum_{d|n}\mu(d)\log(x^{n/d}-1)=\log(\prod_{d|n}(x^{n/d}-1)^{\mu(d)})$$

And hence,

$$\Phi_n(x)=\prod_{d|n}(x^{n/d}-1)^{\mu(d)}$$

Because the Moebius function $\mu(n)$ can receive only two values $1$ or $-1$. We can see that $\Phi_n(x)$ is a rational function in $\mathbb{Q}$. But it is actually a polynomial in $\mathbb{C}[x]$. We now prove the rationality of $\Phi_n(x)$.

Theorem 5.2. $\Phi_n(x)\in \mathbb{Q}[x]$.

Proof. Let $\Phi_n(x) = \frac{p(x)}{q(x)}$ where $p(x),q(x)\in\mathbb{Q}[x]$. And over $\mathbb{C}[x]$, we have $p(x) = q(x)\Phi_n(x)$. Assume that $\Phi_n(x)\notin \mathbb{Q}[x]$, and $\phi_l$ is the least coefficient that does not lie in $\mathbb{Q}$ of $\Phi_n(x)$.  Then the $l^{th}$ coefficient of $p(x)$ (denoted by $p_l$) will be $q_0\phi_l+q_1\phi_{l-1}+...+q_l\phi_0$, where $q_i$'s are coefficients of $q(x)$. From this, we can see that $p_l\notin \mathbb{Q}$, that leads to the contradiction. Hence $\Phi_n(x)\in\mathbb{Q}[x]$ (Q.E.D).

Via the Moebius inversion theorem, one can compute the $n^{th}$ cyclotomic polynomial. The rationality of it can also be obtained by Galois theory. We now discuss about it. First, it can be seen that $\mathbb{Q}(\xi)/\mathbb{Q}$ is Galois extension, because $\mathbb{Q}(\xi)$ is the splitting field of $(x^n-1)$, which is separable over $\mathbb{Q}$. And the Galois group can be computed by just looking at the action on $\xi$.

Let $\tau\in Gal(\mathbb{Q}(\xi)/\mathbb{Q})$, then $\tau(\xi)$ is another root of $(x^n-1)$, and because $\tau$ is a field automorphism, we must have $\tau(\xi)=\xi^k$, where $(k,n)=1$. And hence, $Gal(\mathbb{Q}(\xi)/\mathbb{Q})\cong (\mathbb{Z}/n\mathbb{Z})^\times$. And the minimum polynomial of $\xi$ over $\mathbb{Q}$ is

$$\prod_{\tau\in Gal(\mathbb{Q}(\xi)/\mathbb{Q})}(x-\xi^\tau)=\prod_{k\le n,(k,n)=1}(x-\xi^k)=\Phi_n(x)$$

And the rationality of $\Phi_n(x)$ now follows. We can deduce that $\Phi_n(x)\in\mathbb{Z}[x]$, by noting that $\xi$ is an algebraic integer, and then, so are all coefficients of $\Phi_n(x)$. This implies $\Phi_n(x)\in\mathbb{Z}[x]$.

Let us try with two interesting two exercises.

Problem 5.3. Prove that $p$ is a prime number iff $\Phi_p(x)=1+x+...+x^{p-1}$.

Problem 5.4. Prove that there exists infinitely many primes of the form $kn+1$.