About this blog

 My name is Thuong Dang. This blog was a private corner for me to write various things in mathematics when I was a master's student. CALGANT stands for Cryptography, ALgebra, Geometry And Number Theory. In this blog you will find various topics from pure mathematics that are applied in cryptogrpahy, e.g. elliptic curves, Weil pairings, Fourier analysis on finite groups, etc....

[Functor of Points I] Representable Functors and Zarisky Covering of a Functor

1. Representable Functors.

Let $\mathscr{C}$ be a category, we denote $
\widehat{\mathscr{C}}$ the category of functors from $\mathscr{C}^\text{opp}$ to $Sets$, the Yoneda's lemma tells us there is a faithful functor from $\mathscr{C}$ to $\widehat{\mathscr{C}}$ defined by $X\mapsto h_X$, where $h_X(-):=Hom_{\mathscr{C}}(-,X)$. This implies that we can study $X$ via its associated functor $h_X$. We also recall that a functor $F: \mathscr{C}^\text{opp}\to Sets$ is said to be representable if there exists an object $X$ in $\mathscr{C}$, such that $F\cong h_X$.

If we replace $C$ by the category $Sch$ of schemes, then in general, determining if a functor $F: \widehat{Sch}^\text{opp}\to Sets$ is representable or not is an interesting question. Let us take a look on some examples.

Example 1.1. We define the global section functor $F: Sch^\text{opp}\to Sets$ defined by $X\mapsto \mathscr{O}_X(X)$, then it is representable by $Spec(\mathbb{Z}[t])$. In fact, this follows from the fact that $\mathscr{O}_X(X)\cong Hom_{Rings}(\mathbb{Z}[t], \mathscr{O}_X(X))$, and $Hom_{Rings}(\mathbb{Z}[t], \mathscr{O}_X(X))\cong Hom_{Sch}(X, Spec\mathbb{Z}[t])$ as sets.

Example 1.2. Let $F: Sch^\text{opp}\to Sets$ be a functor defined by $X\mapsto \mathscr{O}_X(X)^\times$, we have $\mathscr{O}_X(X)^\times \cong Hom_{Rings}(\mathbb{Z}[t,t^{-1}], \mathscr{O}_X(X))\cong Hom_{Sch}(X,Spec \mathbb{Z}[t,t^{-1}])$, and we get that $F$ is representable by $Spec \mathbb{Z}[t,t^{-1}]$.

Example 1.3. Let $\mathscr{C}$ be an arbitrary category. We note that in $\widehat{\mathscr{C}}$, the fiber product always exists. Indeed, let $F, G, H$ be functors from $\mathscr C^\text{opp}$ to $Sets$, with $f: F\to H, g:G\to H$ morphisms of functors, we can define

$$(F\times_H G)(T):=F(T)\times_{H(T)}G(T)$$

where the RHS is a fiber product in the category of sets. Then $F\times_H G$ is a well-defined functor from $\mathscr C^\text{opp}$ to $Sets$, and it satisfies the universal properties defining fiber product. Applying this to the situation of schemes, with $H=h_S, F=h_X, G=h_Y$, then we can see that $F
\times_H G$ is representable by $X\times_S Y$. This fact follows easily from the universal property of fiber products.

In the later parts of this series, we will see much more difficult examples, about Grassmanian functors, and Brauer-Severi functors. About $h_X$, we should note that it is deeper than just a functor. We denote $X(T):=h_X(T)$, and we say $X(T)$ is the set of $T$-rational points of $X$. For this notion, I remind to look at my previous notes on $k$-rational points on an affine scheme over fields.

2. Zarisky open covering of a functor.
 
 We are now going to work with schemes basically by their associated functors.

Definition. Let $F,G$ be in $\widehat{Sch}$, a morphism $f: F\to G$ is said to be representable if for all scheme $X$, and all morphism $g: h_X\to G$, the functor $F\times_G h_X$ is representable.

Example 2.1. Let $F:=h_Y, G:=h_Z$, then a morphism from $h_Y\to h_Z$ corresponds to a morphism from $Y\to Z$. In this case, we can form the fiber product $Y\times_Z X$ in the category of schemes, and it represents the functor $h_Y\times_{h_Z}h_X$.

Definition. Let $P$ be a property of morphism of schemes, and $F,G$ in $\widehat{Sch}$. We say that a representable morphism $f: F\to G$ has property $P$ if for all $g: h_X\to G$, and a scheme $Z$ represents $F\times_G h_X$, the corresponding morphism $Z\to X$ has property $P$.

Example 2.2. Let $P=``\text{open immersion}"$, $F:=h_Y, G:=h_Z$, then a morphism $f: h_Y\to h_Z$ corresponds to a morphism $f: Y\to Z$, and it is always representable by the example above. In this case, $h_Y\times_{h_Z}h_X$ is representable by $Y\times_Z X$, and because open immersion is stable under base changes, we can see that the projection map $Y\times_Z X\to X$ is also an open immersion.

We now come to the notion of Zarisky sheaf via functor of points. Let $F: Sch^\text{opp}\to Sets$ be a functor, and $i: U\to X$ is an open immersion of schemes. Let $\psi\in F(X)$, we denote $psi|_{U}:=F(i)(\psi)$.

Definition. Let $F: Sch^\text{opp}\to Sets$ be a functor, we say that $F$ is a Zarisky sheaf if for all scheme $X$, all open covering $\{U_i\}_i$ of $X$, and all $\psi_i\in F(U_i)$ such that for all $i,j$, $\psi_i|_{U_i\cap U_j}=\psi_j|_{U_i\cap U_j}$, there exists a unique $\psi$ in $F(X)$ such that $\psi|_{U_i}=\psi_i$.

Example 2.3. Let $F: Sch^\text{opp}\to Sets$ defined by $F(X):=\mathscr{O}_X(X)$, then by properties of sheaves, $F$ is a Zarisky sheaf.

Example 2.4. Let $F:=h_S$ be a representable functor, then because any morphism from $X$ to $S$ can be glued by morphisms from all $U_i$ to $S$, where $\{U_i\}_i$ is an open covering of $X$, with the condition that these morphism agree on the intersection. We can see that $h_S$ is a Zarisky sheaf.

Definition. Let $F$ be in $\widehat{Sch}$, an open subfunctor of $F$ is an object $F'$ in $\widehat{Sch}$ together with a representable morphism $f: F'\to F$, such that for all $g: h_X\to F$, and $Z$ the scheme representing $F'\times_Fh_X$, the morphism $Z\to X$ is an open immersion.

Example 2.5. Let $F:=h_Z$, and $Y$ is an open subscheme of $Z$, then the corresponding morphism $h_Y\to h_Z$ together with $h_Y$ is an open subfunctor of $h_Z$.

Definition. Let $F$ be a functor, and $\{f_i: F_i\to F\}_i$ be open subfunctors of $F$. We say that $\{f_i:F_i\to F\}$ is an Zarisky open covering of $F$, if for all $g: h_X\to F$, and $Z_i$ is the scheme representing $F_i\times_F h_X$, with the induced map $g_i: Z_i\to X$, then $g_i(Z_i)$ forms an open covering of $X$.

Example 2.6. Let $F:=h_Y$, and $F_i:=h_{U_i}$, where $\{U_i\}_i$ is an open covering of $Y$, then $h_{U_i}$ is an open covering of $h_Y$, since $\{U_i\times_Y X\}_i$ is an open covering of $Y\times_YX=X$.


There is an important theorem, whose proof is more or less the same with the construction of gluing schemes.

Theorem 2.7. Let $F: Sch^\text{opp}\to Sets$ be a functors, then $F$ is representable iff

(i) $F$ is a Zarisky sheaf.

(ii) $F$ has a Zarisky open covering $\{F_i\}_i$, and each $F_i$ is representable.

 
 Application I (Fiber product in the category of schemes). Via the two sections above, we use many examples related to fiber product in the category of schemes. But here is the proof of the existence of fiber product by Grothendieck. First, we can see that there exists fiber product in the category $\widehat{Sch}$ by Example 1.3 and we next use the theorem above to reduce to the affine case. But in the later case, things are done by tensor product.

Later, we will construct Grassmanians over any base scheme, by proving that the Grassmanian functor is representable.

Some words for a new series

One cannot hope to go deep in algebraic geometry without knowledge about commutative algebra and homological algebra. This is a difficult point, but it also makes algebraic geometry more attractive. Recently, I am reading about the theory of quasi-coherent sheaves, and in the case of affine schemes, all things begin with modules. For examples, the category of quasi-coherent sheaf on an affine scheme is equivalent to the category of modules over a ring, and a locally free sheaf on an affine scheme is equivalent to a finitely generated projective module over a ring. A sheaf has its natural geometric intuition, but these kinds of sheaves also reflect a deep connection to algebraic objects.

My first objective is to understand about representable functors, but for the example, to understand about Grassmanian functor, we need much more consideration on quasi-coherent sheaf. Understanding things in details make the subject more interesting, and I will try to type down everything, even about projective/injective modules, which I have never studied seriously before.

[Algebraic Number Theory I] Sum of Two Squares and Gaussian Integers

Reference(s):
[1] Neukirch, J. (1999). Algebraic number theory, volume 322 of Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences].
[2] Shoup, V. (2009). A computational introduction to number theory and algebra. Cambridge university press.



Hello everyone, I use this blog, owned by Thuong Dang, for practising my English writing skill in mathematics. This is my fist note and also the beginning of a series about algebraic number theory. The main reference I will use for this note, and this series, is from [1]. 

The main purpose of this note is to prove the following theorem:

Theorem 1 ([1, Theorem 1.1]). Every positive prime number $p$ in $\mathbb{N}$ is a sum of two squares, i.e., $p = a^2 + b^2$ for $a, b \in \mathbb{N}$, if and only if $p \equiv 1 \pmod{4}$.

To prove this theorem, we need to use the ring of Gaussian integers, i.e., the set
$$ \mathbb{Z}[i] = \{a + bi : a, b \in \mathbb{N}\} $$
equipped with the following two operations:
Addition. $(a + bi) + (c + di) = (a + c) + (b + d)i$, and
Multiplication. $(a + bi) * (c + di) = (ac - bd) + (cb + ad) i$.
For simplicity, we write $(a + bi)(c+di)$ in place of $(a + bi) * (c + di)$.

In this note, we only present the proof of Theorem 1. The underlying properties of the ring of Gaussian integers will be discussed in the next note.

We define the norm of a Gaussian integer $\alpha = a + bi$ to be following value:
$$ N(\alpha) = \alpha\overline{\alpha} = (a + bi)\overline{(a + bi)} = (a + bi)(a - bi) = a^2 + b^2.$$
Geometrically, the norm of a Gaussian integer $a + bi$ is the square of the distance from the point $(a,b)$ to the point $(0, 0)$ in a $2$-dimensional space.

Lemma 2 (Wilson's theorem, [2, Theorem 2.22]). Let $p$ be an odd prime. Then, we have
$$\prod_{z \in \mathbb{Z}^{*}_p}z \equiv -1 \pmod{p}.$$
Proof. We make pairs the elements in $\mathbb{Z}^{*}_p - \{-1, 1\}$ such that, for each pair (a, b), its product satisfies
$$a * b \equiv 1 \pmod{p}.$$
For a specific pair (a, b), we clearly see that $a \not\equiv b \pmod{p}$ because the only square roots of $1$ in modulo $p$ are $1$ and $-1$. Since $\mathbb{Z}_p^{*}$ forms a group structure, every element $a$ in $\mathbb{Z}^{*}_p - \{-1, 1\}$ has a unique partner $b \in \mathbb{Z}^{*}_p - \{-1, 1\}$ satisfying $a*b \equiv 1 \pmod{p}$. Therefore,
$$ \prod_{z \in \mathbb{Z} - \{-1, 1\}}z \equiv 1 \pmod{p} $$
and then,
$$ \prod_{z \in \mathbb{Z}}z \equiv -1 \pmod{p}$$
which proves the lemma.

Q.E.D

Lemma 3 ([1]). Let $p = 4n + 1$, $n \in \mathbb{N}$, be a prime. Then, $(2n)!$ is a solution of the modular equation $x^2 \equiv -1 \pmod{p}$.

Proof. Since $p$ is a prime, the elements of the set $\mathbb{Z} - \{0\}$ is exactly the elements of the group $\mathbb{Z}^{*}_p$. By Lemma 2, we have

$$\prod_{z = 1}^{p - 1}\equiv\prod_{z \in \mathbb{Z}^{*}_p}z \equiv -1 \pmod{p}. $$

Therefore, we have

$$ -1 \equiv 1 * 2 * ... * (2n) * [(2n + 1) * ... * 4n] $$ 
$$\equiv 1 * 2 * ... * (2n) * [(p - 1) * (p - 2)* ... * (p - 2n)] $$ 
$$\equiv 1*2*...*(2n)*[(-1) * (-2) *...* (-2n)]$$
$$\equiv 1*2* ...*(2n) * [1 * 2 * ... * (2n)] * (-1)^{2n}$$
$$\equiv [1*2*...*(2n)]^2 \pmod{p}.$$
Thus the lemma is proved.

Q.E.D

Now we prove the main result of this note.

Proof of Theorem 1. For the sufficient condition, since $p \equiv 1 \pmod{4}$, by Lemma 3, we have a solution $x$ of the equation 

$$ x^2 \equiv -1 \pmod{p}.$$

Hence, we have $x^2 + 1 \equiv 0 \pmod{p}$ or $p | (x^2 + 1).$ We see that 

$$x^2 + 1 = (x + i)(x-i). $$

Therefore, $p | (x + i)(x - i)$. Since $p$ is an odd prime, the fraction $\frac{1}{p} \not\in \mathbb{Z}$ and therefore, $p$ is a divisor of neither $x + i$ nor $x - i$. Hence, $x + i = (a + bi)(a'+b'i)$ and $x - i  = (c + di)(c'+d'i)$ such that the product $(a+bi)(c+di) = p$. We also note that the norm of the four Gaussian integers $a + bi, a'+b'i, c+di,$ and $c'+d'i$ are greater than $1$.

Consider $p^2 = N(p) = N(a + bi)N(c+di).$ Since $N(a+bi) > 1$ and $N(c+di) > 1$, we concludes that $N(a+bi) = p$ and $N(c+di) = p$. Thus, $p = N(a +bi) = a^2 + b^2$ which proves the sufficient condition.

For the necessary condition, consider an integer $z \in \mathbb{Z}$. We always have either $z^2 \equiv 0 \pmod{4}$ or $z^2 \equiv 1 \pmod{4}$. Hence, the value $a^2 + b^2$, for $a, b\in\mathbb{Z}$, cannot be equivalent to $3$ in modulo $4$. Thus the theorem is proved.

Q.E.D

[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)

[Galois Cohomology I] Cohomological Description of Brauer Groups

The main reference for this series will be the Class Field Theory notes of Milne and the book "Algebra II" by Falko Lorenz. Brauer groups over fields are important objects in number theory, and they have applications in class field theory (e.g. the local reciprocity law by Artin), though they were originally defined in the context of semi-simple algebras over fields. The definition of Brauer groups is deserved to be discussed more than just a brief recall, but I don't have time to type down everything. Readers may refer to the second reference for proofs of results in the first section of this note. And in the second section, we will give a cohomological description for Brauer groups. It leads naturally to discuss about group and Galois cohomology later on.

Notation. In this series, we always fix a field $K$, and an algebra $A$ over $K$ is not assumed to be commutative. We also denote $A^o$ the $K$-algebra with the same group structure as $(A,+)$, but the multiplication in $A^o$ is defined to be $a*b:=b.a$, where $.$ is the multiplication in $A$. $D$ is always denoted a division $K$-algebra.

1. Brief recall about Brauer groups.

We recall that $A$ is simple $K$-algebra if $A$ has no proper two-sided ideal. $A$ is said to be a semi-simple $K$-algebra if $A$ is semi-simple left $A$-module. $A$ is said to be an artinian $K$-algebra if $A$ is an artinian left $A$-module. $A$ is said to be central over $K$ if $Z(A)=K$. $A$ is said to be central-simple $K$-algebra if $A$ is central, $A:K<+\infty$ and $A$ is simple as $K$-algebra. By the theorem of Wedderburn, any semi-simple $K$-algebra is a product of simple artinian $K$-algebras, and each simple artinian $K$-algebra is isomorphic to a matrix ring over a division $K$-algebra $M_n(D)$.

On the set of all simple artinian $K$-algebras, we define a relation $\sim$, $A\sim B$ (say: A is similar to $B$) if there exists a division $K$-algebra $D$ such that $A\cong M_r(D), B\cong M_s(D)$ for some non-negative integers $r,s$. If we restrict this definition to the set of all central-simple $K$-algebras, we obtain the definition of the Brauer group over $K$, i.e. $Br K$ is the group of similarity classes of central-simple $K$-algebras, with the group operation is the tensor product over $K$. The most easiest for Brauer group is that when $K$ is an algebraically closed field, we have $Br K$ is trivial, since any finite dimensional division $K$-algebra is $K$ itself.

There are two important theorems that we always have keep in mind about Brauer groups.

Theorem 1.1 (Centralizer Theorem). Let $A$ be a simple, central, artinian $K$-algebra, $B$ a simple subalgebra of finite dimension over $K$, $C:=Z_A(B)$ where $Z_A(B)$ is the centralizer of $A$ in $B$. Then

(i) $C$ is a simple artinian $K$-algebra.

(ii) $C\sim B^0\otimes A$, and $Z(B)=Z(C)$.

(iii) $C:K<+\infty$ iff $A:K<+\infty$, and in this case $A:K=(B:K)(C:K)$.

(iv) If $Z_A(C)=Z_A(Z_A(B))=B$.

(v) If $L$ is a center of $B$, then $A\otimes_K L\sim B\otimes_L C$.

(vi) If $Z(B)=K$, then $A\cong B\otimes_K C$.

Theorem 1.2 (Skolem-Noether). Let $A$ be a simple, central, artinian $K$-algebra, and $B$ a simple algebra of finite dimension over $K$. Let $f, g$ be two $K$-algebra homomorphism from $B\to A$, then there exists a unit $u\in A^\times$, such that $f(b)=u^{-1}g(b)u$, for all $b\in B$.

We also recall that if $L/K$ is a field extension, then there exists a restriction map $res_{L/K}: Br K\to Br L$ which sends $A\mapsto A\otimes_K L$. This map is a homomorphism of groups, and the kernel of this map is denoted $Br(L/K)$, i.e.
$$Br(L/K):=\{[A]\in Br K| A\otimes_K L\cong M_n(L)\text{ for some integer n}\}$$

For $A\in Br(L/K)$, we say that $A$ splits over $L$. The easiest example for Brauer group above (over algebraically closed field) implies that a central-simple $K$-algebra $A$ always splits in an algebraic closure $\overline{K}$ of $K$. And hence $A_{\overline{K}}:=A\otimes_K \overline{K}\cong M_n(\overline{K})$, and hence the dimension $A:K=A_{\overline{K}}:\overline{K}$ is always a square.

And using Theorem 1.1, one can deduce that

Corollary 1.3. Let $A$ be simple, central, artinian $K$-algebra, and $L$ a subalgebra of $A$ of finite dimension over $K$, then the following are equivalent:

(i) $L$ is a maximal commutative subalgebra of $A$, i.e. $Z_A(L)=L$.

(ii) $A:K=(L:K)^2$

If the one (and hence, two) condition(s) are satisfied, then $A$ splits over $L$. When $D:=A$ is a division $K$-algebra, satisfying the both conditions, then maximal commutative subalgebras of $D$ are exactly maximal subfields of $D$ containing $K$.

Another important statement is that for any $A\in Br K$, there always exists $L/K$: finite, Galois such that $A$ splits over $L$. And hence, we obtain
$$Br K=\bigcup_{L/K:\text{fin. Gal.}}Br(L/K)$$

And our goal in the next section is to give a cohomological description for $Br(L/K)$.

2. Cohomological description of Br(L/K).

In this section, we always fix $L/K$ a finite Galois extension, and $A$ a central-simple $K$-algebra, which splits over $L$, $G:=Gal(L/K)$. By the definition, there exists an $L$-algebra isomorphism

$$h: A\otimes_K L\to End_L(V)$$

where $V$ is a $n$-dimensional $L$-vector space. Let $\sigma\in G$, we can extend the action of $\sigma$ to $A\otimes_K L$ as follows

$$(a\otimes \lambda)^\sigma:=a\otimes \lambda^\sigma$$

We also denote $\rho_\sigma: End_L(V)\to End_L(V)$, where $x^{\rho_\sigma}:=x^{h^{-1}\sigma h}$. It can be seen that $\rho_\sigma\rho_\tau=\rho_{\sigma\tau}$, for all $\sigma,\tau\in G$. Since $h$ is an $L$-algebra isomorphism, we have for all $\lambda\in L$, $\lambda^{\rho_\sigma}=\lambda^\sigma$.

One can also consider $End_L(V)$ as a $K$-subalgebra of $End_K(V)$, and since $\sigma$ fixes $K$ for any $\sigma\in G$, we have $\rho_\sigma$ is a $K$-algebra homomorphism. This yields by the Skolem-Noether's theorem that there exists $u_\sigma\in End_K(V)^\times$ such that

$$ x^{\rho_\sigma}=u_\sigma^{-1}xu_\sigma, \forall x\in End_L(V) \text{   (1)}$$

And for all $\lambda\in L$, we have $\lambda^{\rho_\sigma}=\lambda_\sigma=u_\sigma^{-1}\lambda u_\sigma$, which is equivalent to say

$$\lambda u_\sigma=u_\sigma \lambda^\sigma \text{ (2)}$$

We have from (1) that for all $\sigma,\tau\in G$, and $x\in End_L(V)$

$$x^{\rho_{\sigma\tau}}=u_{\sigma\tau}^{-1}xu_{\sigma\tau}$$

And

$$x^{\rho_\sigma\rho_\tau}=(u_\sigma u_\tau)^{-1} x u_\sigma u_\tau$$

If one writes $u_\sigma u_\tau=u_{\sigma\tau}c_{\sigma,\tau}$, for some $c_{\sigma,\tau}\in End_K(V)^\times$, then $c_{\sigma,\tau}\in Z_{End_K(V)}(End_L(V))$. But then, $Z_{End_K(V)}(End_L(V))\supseteq Z(End_L(V))=L$. Assume that $L:K=m$, then $\dim_K End_K(V)=m^2n^2$, $\dim_K (End_L(V))=mn^2$, and hence, the by the centralizer theorem above, $\dim_K Z_{End_K(V)}(End_L(V))=\dim_K L= m$. And hence, $c_{\sigma,\tau}\in L^\times$.

From (1), we can see that $u_1\in L^\times$, by the similar argument as above. Our first goal is to prove that the similarity class of $A$ depends only on $c_{\sigma,\tau}$, but $c_{\sigma,\tau}$ is not uniquely determined by $A$.

Theorem 2.1. Let $L, K, G$ be as above, and $\Gamma$ a $K$-algebra, containing $L$ as a subalgebra. Assume further that $\Gamma$ is generated by $\lambda\in L$, and $u_\sigma$, for $\sigma\in G$ such that $u_1\in L^\times$, and

(i) $\lambda u_\sigma=u_\sigma\lambda^\sigma$.
(ii) $u_\sigma u_\tau=u_{\sigma\tau}c_{\sigma,\tau}$

for $c_{\sigma,\tau}\in L^\times$, then $(u_\sigma)_{\sigma\in G}$ forms a basis of $\Gamma$ as $L$-vector space, and $\Gamma:K=(L:K)^2$. Furthermore, $\Gamma$ is a central-simple $K$-algebra, with $L$ is its maximal commutative subalgebra, and hence, $\Gamma\in Br(L/K)$. And the $c_{\sigma,\tau}$ satisfy

$$c_{\sigma,\tau}^\rho c_{\sigma\tau,\rho}=c_{\sigma,\tau\rho}c_{\tau,\rho} \text{ (3)}$$

Proof. First, we have $u_\sigma u_{\sigma^{-1}}=u_1c_{\sigma,\sigma^{-1}}\in L^\times$, and this yields $u_\sigma$ is invertible, for all $\sigma\in G$. For any $\lambda\in L$, consider the left multiplication map by $\lambda$, $\lambda_l: \Gamma\to \Gamma$ defined by $\lambda_l(x)=\lambda x$. Then for any $\gamma\in L$, we have

$$\lambda_l(x\lambda)=\lambda x \gamma=\lambda_l(x)\gamma$$

And so, we can consider $\lambda_l$ as an endomorphism of $L$-right vector space. And the fact that $\lambda u_\sigma=u_\sigma\lambda^{\sigma}$ implies that $u_\sigma$ is an eigenvector and $\lambda^\sigma$ is an eigenvalue of $\lambda_l$. By the primitive element theorem, there exists $\lambda\in L$, such that $(\lambda^\sigma)_{\sigma\in L}$ forms a basis of $L/K$, and this yields $u_\sigma$ are linearly independent over $L$, since they are eigenvectors of distinct eigenvalues. On the other hand, since $u_\sigma u_\tau=u_{\sigma\tau}c_{\sigma,\tau}$, which means that $\Gamma:L\le \# G=L:K$. And this implies $\Gamma:K=(L:K)^2$.

To prove that $\Gamma$ is simple, it is sufficient to prove that any surjective ring homomorphism $\phi$ from $\Gamma$ to $\Gamma'$ where $\Gamma'\ne 0$ has the zero kernel. First, we can equip $\Gamma'$ the structure of $L$-vector space via the map $\phi$ as follows $\phi(x)\lambda=\phi(x)\phi(\lambda)$, for all $x\in \Gamma,\lambda\in L$. And the map $\phi$ can be now considered as a map between $L$-algebras. And due to the map $\phi$, we also have $(\phi(u_\sigma))_\sigma$ satisfying the same conditions (i) and (ii). And this yields by our previous argument that $\Gamma':K=\Gamma:K=(L:K)^2$. Hence $\phi$ is an isomorphism, and $\ker \phi=0$.

We will now prove that $Z(\Gamma)=K$. In the case $L\equiv K$ then there is nothing to prove. Otherwise, for any $z\in Z(\Gamma)$, we can represent $z=\sum_{\sigma\in G}u_\sigma\lambda_\sigma$ with $\lambda_\sigma\in L$. Then for any $\lambda\in L$, we have $z\lambda = \lambda z$, where

$$z\lambda = \sum_{\sigma\in G}u_\sigma\lambda_\sigma \lambda \text{ and } \lambda z=\sum_{\sigma\in G}\lambda u_\sigma \lambda_\sigma=\sum_{\sigma\in G}u_\sigma \lambda^{\sigma} \lambda_\sigma$$

And by comparing coefficients, we have $\lambda_\sigma\lambda=\lambda^{\sigma}\lambda_\sigma$ for all $\sigma\in G, \lambda\in L$. Taking $\sigma\ne 1$, if $\lambda_\sigma\ne 0$, we obtain $\lambda=\lambda^\sigma$, for all $\lambda\in L$, which yields $L=K$, a contradiction. Hence for all $\sigma\ne 1$, $\lambda_\sigma= 0$, and this yields $z=u_1\lambda_1\in L$. But then,

$$u_\sigma z=zu_\sigma = u_\sigma z_\sigma$$

Since $u_\sigma$ is invertible, we obtain $z=z^{\sigma}$, and this yields $z\in K$. Hence, $Z(\Gamma)=K$. And this yields $\Gamma$ is central-simple over $K$, and $\Gamma:K=(L:K)^2$, hence, $L$ is a maximal commutative subalgebra of $\Gamma$. Due to Corollary 1.3, we have $\Gamma$ splits over $L$, and hence $\Gamma\in Br(L/K)$.

The final statement follows easily from the comparison between $(u_\sigma u_\tau)u_{\rho}$ and $u_\sigma (u_\tau u_\rho)$. (Q.E.D)

Example. Let $L,K,G$ be as above. Consider the $K$-algebra $End_K(L)$, we note that $G\subset End_K(L)$, and for all $\lambda,\gamma\in L$, we have

$$\sigma(\lambda_l(\gamma))=(\lambda\gamma)^\sigma=\lambda^\sigma \gamma^\sigma=(\lambda^\sigma)_l(\sigma(\gamma))$$

So, in the algebra $\Gamma:=End_K(L)^o$, we have $\lambda*\sigma =\sigma*\lambda^\sigma$. If we denote $u_\sigma:=\sigma$ in $\Gamma$, then $u_1=1$, and $u_\sigma u_\tau=u_{\sigma\tau}$, and this yields $c_{\sigma,\tau}=1$, for all $\sigma,\tau\in G$.

We can now come back to our computation before the statement of Theorem 1. Let $\Gamma$ be a $K$-subalgebra of $End_L(V)$ generated by $L$ and $u_\sigma (\sigma\in G)$, then by Theorem 1, $\Gamma$ is central-simple. We will next prove that $\Gamma^o\sim A$.

Lemma 2.2. $\Gamma\subseteq Z_{End_K(V)}(A^h)$.

Proof. Taking any $x\in A$, then $x^\sigma=x$ for all $\sigma\in G$, hence $h(x)^{\rho_\sigma}=h(x)$. And this implies that $h(x)$ is commutative with $u_\sigma$ for all $\sigma\in G$ due to (1). The commutativity between $h(x)$ and $\lambda\in L$ is clear. So, $\Gamma\subseteq Z_{End_K(V)}(A^h)$. (Q.E.D)

To prove that the equality happen in Lemma 2, we need the following theorem

Theorem 2.3. Let $A$ be a central-simple $K$-algebra, and $L/K$ is a finite extension of fields such that $A$ splits over $L$, then there exists a central-simple $K$-algebra $A'$, such that $A'\sim A$, and $L$ is a maximal commutative subalgebra of $A'$.

Proof. Let $n^2:=A:K$, then there exists an isomorphism of $L$-algebras

$$h: A\otimes_K L\to End_L(V)$$

where $V\cong L^n$. We denote $B:=A^h$ the image of $A$, and $C:=Z_{End_K(V)}(B)$. By the centralizer theorem, we have $Z(C)=Z(B)=K$, and hence, $C$ is a central-simple $K$-algebra, and $C$ contains $L$ since it is obvious that $L\subseteq Z_{End_K(V)}(B)$. Using the theorem again yields

$$End_K(V):K=(C:K)(B:K)$$

And this implies that $C:K=(L:K)^2$. Hence, $L$ is a maximal commutative subalgebra of $C$. And due to the centralizer theorem again, we have $C\sim A^o\otimes_K End_K(V)\sim A^o$. Let $A'=C^o$, then $A'\sim A$. (Q.E.D)

From the theorem above, it follows that $\Gamma=Z_{End_K(V)}(A^h)$, and hence $\Gamma^0\sim A$, as in the proof of Theorem 2.3. This follows that the similarity class of $A$ in $Br(L/K)$ is completely determined by $c_{\sigma,\tau}$.

We will now reason why $[A]$ does not determine $c_{\sigma,\tau}$ exactly. If we change $u_\sigma$ by $u'_\sigma:=u_\sigma a_\sigma$, for some $a_\sigma\in L^\times$, then the shape of $\Gamma$ does not change, since we will obtain another basis $(u'_\sigma)$ for $\Gamma$. But now, by simple computation, we have

$$c'_{\sigma,\tau}=c_{\sigma,\tau}a^\tau_\sigma a_\tau a_{\sigma\tau}^{-1}$$

And this leads us to the following

Definition. Let $L, K, G$ be as above, a tuple $(c_{\sigma,\tau})_{\sigma,\tau\in G}$ where $c_{\sigma,\tau}\in L^\times$ is said to be a cocycle if $c_{\sigma,\tau}$ satisfy the condition of (3) in Theorem 2.1. $(c_{\sigma,\tau})$ is said to be a coboundary if $d_{\sigma,\tau}=a^\tau_\sigma a_\tau a_{\sigma\tau}^{-1}$ for some $a_\sigma\in L^\times,\sigma\in G$.

It can be seen that the set of cocycles forms an abelian group, and the set of coboundaries forms a subgroup of the former group. We denote $H^2(G, L^\times)$ the quotient of the two groups, and it is called the second cohomology group of $G$ with coefficients in $L^\times$.

And the remaining of this note is devoted for the proof of the fact that there exists an isomorphism $Br(L/K) \cong H^2(G,L^\times)$. We first begin with the map

$$f: \{\text {central-simple K algebras splitting over L}\}\to H^2(G,L^\times)$$

as in the beginning of the second section. We note that it is well-defined, and does not depend on the choice of an isomorphism $h$, due to the Stokem-Noether's theorem. And at the first step, we will prove that this depend only on the choice of the similarity class of $A$.

Lemma 2.4. The map $f$ is multiplicative, i.e. $f(A\otimes_K A')=f(A)f(A')$.

Proof. Let us denote $h: A\otimes_K L\to End_L(V)$, and $h': A\otimes_K L\to End_L(V')$ two $L$-algebra isomorphisms. We can introduce the map

$$(h,h'): (A\otimes_K L)\otimes (A'\otimes_K L)\to End_L(V)\otimes_K End_L(V')$$

Let us denote $(u_\sigma, c_{\sigma,\tau}),(u'_\sigma, c'_{\sigma,\tau})(U_\sigma, C_{\sigma,\tau})$ the pairs for $(A\otimes_K A')\otimes_K L$ as in the beginning of the section, then it can be seen that $U_\sigma=u_\sigma\otimes u'_\sigma$. And this follows easily that $C_{\sigma,\tau}=c_{\sigma,\tau}c'_{\sigma,\tau}$. (Q.E.D)


Lemma 2.5. 

(i) If $A\sim K$, then $f(A)=1$.

(ii) $f(A^o)=f(A)^{-1}$.

(iii) If $[A]=[B]$ in $Br(L/K)$, then $f(A)=f(B)$.

Proof.

(i) When $A\sim K$ we have $A\cong M_n(K)$, and $A\otimes_K L\cong M_n(K)\otimes_K L\cong M_n(L)$, and in this case, the map $h$ can be seen as the identification maps $1\otimes e_i\mapsto e_i$, where $(e_i)$ is a basis of $L$ over $K$. And one obtains $h\sigma=\sigma h$, for all $\sigma\in G$. And this yields $x^{\rho_\sigma}=x^\sigma$, for all $\sigma\in G$.

And for $x=(x_{ij})\in M_n(L)$, the action of $\sigma\in G$ is defined to be $x^\sigma:=(x^\sigma_{ij})$. And due to the construction of $u_\sigma$, we need to find $u_\sigma\in End_K(L^n)^\times$, such that $xu_\sigma=u_\sigma x^\sigma$. Define

$$u_\sigma((y_i)):=(y_i^{\sigma^{-1}})_i$$

Then it can be seen that $u_\sigma\in End_K(L^n)^\times$, and it is easy to check that $xu_\sigma=u_\sigma x^\sigma$, for all $x\in M_n(L)$. Also, it can be checked easily that $u_\sigma u_\tau=u_{\sigma\tau}$. Hence $c_{\sigma,\tau}=1$, for all $\sigma,\tau \in G$.

(ii) This follows easily from Lemma 2.4 and (i).

(iii) If $A\sim B$, then $A\otimes_K A^o \sim B\otimes_K A^o\sim K$, and $f(B\otimes_K A^o)=1=f(B)f(A)^{-1}$. Hence $f(A)=f(B)$. (Q.E.D)

Hence, one can see that the map $f$ depends only on the similarity class of $A$, and one obtains, by abusing notations, a map

$$f: Br(L/K)\to H^2(G,L^\times)$$

By Lemma 2.4, it is a homomorphism of groups. And it is surjective, due to Theorem 2.1. To prove that $f$ is injective, we can assume that $[A]$ is mapped to $[c]$, where $c_{\sigma,\tau}=1$, for all $\sigma,\tau\in G$. This yields, by our example below Theorem 2.1 that in fact, $A\sim End_K(L)^0\sim K$. And hence, the map $f$ is an isomorphism of groups. In the next part, we will discuss on some number theoretic applications of this observation, for the simplest, but fundamental importance case, which is cyclic extension.