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youknowwho
youknowwho
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I was somewhat puzzled after I finished learning class field theory for several times. My question is about the relations between "classical simple algebras, Brauer groups" and "modern cohomology theory". (If this question is naive to experts, I apologize sincerely.)

The formalism of class field theory before the introduction of the cohomological formalism is very explicit as follows (My favourite reference is Weil's Basic number theory): The canonical morphism is constructed by cyclic algebras. If $K$ is a local field, then the "pairing" $X_K\times K^*\rightarrow \mathbb{C}^1$ between the character group $X_K$ of $Gal(F^{ab}/F)$$Gal(K^{ab}/K)$ and $K^*$ is given by the following well-known process. First we have to know the Brauer group $B(K)=\mathbb{Q}/\mathbb{Z}$ (identified with the group of all roots of unity in $\mathbb{C}^1$), then given $\chi\in X_K$ and $a\in K^*$, $ker(\chi)$ determines a cyclic extension $K'/K$, we can define the pairing to be the class of cyclic algebra $[K'/K,a]:=[K,\chi,a]\in \mathbb{Q}/\mathbb{Z}\subset \mathbb{C}^1$.

After this explicit construction, we got a map ("canonical morphism") from $K^*$ to the Pontryagin dual of $X_K$, which is exactly $Gal(F^{ab}/F)$$Gal(K^{ab}/K)$ by Pontryagin duality. And in the global context we glue the local data via ideles, and the key point is Hasse's reciprocity, which claims it factors through the ideles class group $\mathbb{I}_K/K^*$.

O.K. That's the "classical" roadmap based on the theory of central simple algebras. In "modern" roadmap of Galois cohomology, one first establish a series of axioms (following Artin-Tate) for a general profinite group $G$ and a continuous $G$-module $A$, coming from this model: namely a series of routine functorical properties (a "formation") and two key points:

(1) Hilbert's satz 90: vanishing of $H^1$ (a "field formation");

(2) A functorial isomorphism of $H^2$ with $\cup_{n\geq 1} \mathbb{Z}/n\mathbb{Z}=\mathbb{Q}/\mathbb{Z}$ (a "class formation").

Then you can start with the huge and powerful machinery of (Tate's) cohomology to obtain Tate's theorem: the cup product map $H^q(G, \mathbb{Z})\rightarrow H^{q+2}(G,A)$ is a functorial isomorphism, which provides the canonical morphism of CFT at $q=-2$: $H^{-2}(G, \mathbb{Z})=G^{ab}$, $H^{0}(G,A)=A^G/N_G A.$ In the local case you take $A=K^*$, and in the global case you first take $A=\mathbb{I}$$A=\mathbb{I}_K$ and then $\mathbb{I}_K/K^*$.

I learnt cohomological aspects from Neukirch's excellent book "Class Field Theory, Bonn Lectures". It's very beautiful and carefully written. I spent much time checking each step and details. I was very striking that this somewhat abstract machinery DOES package the "classical" version.

But so far, I still couldn't get an intuitive explanation: how did this machinery correspond to the classical version? For example, in the "classical" version you get this canonical morphism from the Brauer group very directly, which should correspond to $H^2$ in cohomology. But in the "modern" version you get the cup product morphism from $H^{-2}$ and $H^{0}$. This looks very different. So you must have already done "something" from the class formation axiom for $H^2$ and cohomological machinery to get this. Just as Neukirch wrote on his book (page 78):"Before the introduction of cohomology theory, algebras were used to describe local class field theory; we remark that the use of cohomology has led to considerable simplifications." What makes me puzzled is exactly "how this modern machinery correspond to operations in the classical context".

I was somewhat puzzled after I finished learning class field theory for several times. My question is about the relations between "classical simple algebras, Brauer groups" and "modern cohomology theory". (If this question is naive to experts, I apologize sincerely.)

The formalism of class field theory before the introduction of the cohomological formalism is very explicit as follows (My favourite reference is Weil's Basic number theory): The canonical morphism is constructed by cyclic algebras. If $K$ is a local field, then the "pairing" $X_K\times K^*\rightarrow \mathbb{C}^1$ between the character group $X_K$ of $Gal(F^{ab}/F)$ and $K^*$ is given by the following well-known process. First we have to know the Brauer group $B(K)=\mathbb{Q}/\mathbb{Z}$ (identified with the group of all roots of unity in $\mathbb{C}^1$), then given $\chi\in X_K$ and $a\in K^*$, $ker(\chi)$ determines a cyclic extension $K'/K$, we can define the pairing to be the class of cyclic algebra $[K'/K,a]:=[K,\chi,a]\in \mathbb{Q}/\mathbb{Z}\subset \mathbb{C}^1$.

After this explicit construction, we got a map ("canonical morphism") from $K^*$ to the Pontryagin dual of $X_K$, which is exactly $Gal(F^{ab}/F)$ by Pontryagin duality. And in the global context we glue the local data via ideles, and the key point is Hasse's reciprocity, which claims it factors through the ideles class group $\mathbb{I}_K/K^*$.

O.K. That's the "classical" roadmap based on the theory of central simple algebras. In "modern" roadmap of Galois cohomology, one first establish a series of axioms (following Artin-Tate) for a general profinite group $G$ and a continuous $G$-module $A$, coming from this model: namely a series of routine functorical properties (a "formation") and two key points:

(1) Hilbert's satz 90: vanishing of $H^1$ (a "field formation");

(2) A functorial isomorphism of $H^2$ with $\cup_{n\geq 1} \mathbb{Z}/n\mathbb{Z}=\mathbb{Q}/\mathbb{Z}$ (a "class formation").

Then you can start with the huge and powerful machinery of (Tate's) cohomology to obtain Tate's theorem: the cup product map $H^q(G, \mathbb{Z})\rightarrow H^{q+2}(G,A)$ is a functorial isomorphism, which provides the canonical morphism of CFT at $q=-2$: $H^{-2}(G, \mathbb{Z})=G^{ab}$, $H^{0}(G,A)=A^G/N_G A.$ In the local case you take $A=K^*$, and in the global case you first take $A=\mathbb{I}$ and then $\mathbb{I}_K/K^*$.

I learnt cohomological aspects from Neukirch's excellent book "Class Field Theory, Bonn Lectures". It's very beautiful and carefully written. I spent much time checking each step and details. I was very striking that this somewhat abstract machinery DOES package the "classical" version.

But so far, I still couldn't get an intuitive explanation: how did this machinery correspond to the classical version? For example, in the "classical" version you get this canonical morphism from the Brauer group very directly, which should correspond to $H^2$ in cohomology. But in the "modern" version you get the cup product morphism from $H^{-2}$ and $H^{0}$. This looks very different. So you must have already done "something" from the class formation axiom for $H^2$ and cohomological machinery to get this. Just as Neukirch wrote on his book (page 78):"Before the introduction of cohomology theory, algebras were used to describe local class field theory; we remark that the use of cohomology has led to considerable simplifications." What makes me puzzled is exactly "how this modern machinery correspond to operations in the classical context".

I was somewhat puzzled after I finished learning class field theory for several times. My question is about the relations between "classical simple algebras, Brauer groups" and "modern cohomology theory". (If this question is naive to experts, I apologize sincerely.)

The formalism of class field theory before the introduction of the cohomological formalism is very explicit as follows (My favourite reference is Weil's Basic number theory): The canonical morphism is constructed by cyclic algebras. If $K$ is a local field, then the "pairing" $X_K\times K^*\rightarrow \mathbb{C}^1$ between the character group $X_K$ of $Gal(K^{ab}/K)$ and $K^*$ is given by the following well-known process. First we have to know the Brauer group $B(K)=\mathbb{Q}/\mathbb{Z}$ (identified with the group of all roots of unity in $\mathbb{C}^1$), then given $\chi\in X_K$ and $a\in K^*$, $ker(\chi)$ determines a cyclic extension $K'/K$, we can define the pairing to be the class of cyclic algebra $[K'/K,a]:=[K,\chi,a]\in \mathbb{Q}/\mathbb{Z}\subset \mathbb{C}^1$.

After this explicit construction, we got a map ("canonical morphism") from $K^*$ to the Pontryagin dual of $X_K$, which is exactly $Gal(K^{ab}/K)$ by Pontryagin duality. And in the global context we glue the local data via ideles, and the key point is Hasse's reciprocity, which claims it factors through the ideles class group $\mathbb{I}_K/K^*$.

O.K. That's the "classical" roadmap based on the theory of central simple algebras. In "modern" roadmap of Galois cohomology, one first establish a series of axioms (following Artin-Tate) for a general profinite group $G$ and a continuous $G$-module $A$, coming from this model: namely a series of routine functorical properties (a "formation") and two key points:

(1) Hilbert's satz 90: vanishing of $H^1$ (a "field formation");

(2) A functorial isomorphism of $H^2$ with $\cup_{n\geq 1} \mathbb{Z}/n\mathbb{Z}=\mathbb{Q}/\mathbb{Z}$ (a "class formation").

Then you can start with the huge and powerful machinery of (Tate's) cohomology to obtain Tate's theorem: the cup product map $H^q(G, \mathbb{Z})\rightarrow H^{q+2}(G,A)$ is a functorial isomorphism, which provides the canonical morphism of CFT at $q=-2$: $H^{-2}(G, \mathbb{Z})=G^{ab}$, $H^{0}(G,A)=A^G/N_G A.$ In the local case you take $A=K^*$, and in the global case you first take $A=\mathbb{I}_K$ and then $\mathbb{I}_K/K^*$.

I learnt cohomological aspects from Neukirch's excellent book "Class Field Theory, Bonn Lectures". It's very beautiful and carefully written. I spent much time checking each step and details. I was very striking that this somewhat abstract machinery DOES package the "classical" version.

But so far, I still couldn't get an intuitive explanation: how did this machinery correspond to the classical version? For example, in the "classical" version you get this canonical morphism from the Brauer group very directly, which should correspond to $H^2$ in cohomology. But in the "modern" version you get the cup product morphism from $H^{-2}$ and $H^{0}$. This looks very different. So you must have already done "something" from the class formation axiom for $H^2$ and cohomological machinery to get this. Just as Neukirch wrote on his book (page 78):"Before the introduction of cohomology theory, algebras were used to describe local class field theory; we remark that the use of cohomology has led to considerable simplifications." What makes me puzzled is exactly "how this modern machinery correspond to operations in the classical context".

Source Link
youknowwho
youknowwho
  • 709
  • 3
  • 8

Essence of relations between central simple algebras and Galois cohomology in canonical morphism of class field theory

I was somewhat puzzled after I finished learning class field theory for several times. My question is about the relations between "classical simple algebras, Brauer groups" and "modern cohomology theory". (If this question is naive to experts, I apologize sincerely.)

The formalism of class field theory before the introduction of the cohomological formalism is very explicit as follows (My favourite reference is Weil's Basic number theory): The canonical morphism is constructed by cyclic algebras. If $K$ is a local field, then the "pairing" $X_K\times K^*\rightarrow \mathbb{C}^1$ between the character group $X_K$ of $Gal(F^{ab}/F)$ and $K^*$ is given by the following well-known process. First we have to know the Brauer group $B(K)=\mathbb{Q}/\mathbb{Z}$ (identified with the group of all roots of unity in $\mathbb{C}^1$), then given $\chi\in X_K$ and $a\in K^*$, $ker(\chi)$ determines a cyclic extension $K'/K$, we can define the pairing to be the class of cyclic algebra $[K'/K,a]:=[K,\chi,a]\in \mathbb{Q}/\mathbb{Z}\subset \mathbb{C}^1$.

After this explicit construction, we got a map ("canonical morphism") from $K^*$ to the Pontryagin dual of $X_K$, which is exactly $Gal(F^{ab}/F)$ by Pontryagin duality. And in the global context we glue the local data via ideles, and the key point is Hasse's reciprocity, which claims it factors through the ideles class group $\mathbb{I}_K/K^*$.

O.K. That's the "classical" roadmap based on the theory of central simple algebras. In "modern" roadmap of Galois cohomology, one first establish a series of axioms (following Artin-Tate) for a general profinite group $G$ and a continuous $G$-module $A$, coming from this model: namely a series of routine functorical properties (a "formation") and two key points:

(1) Hilbert's satz 90: vanishing of $H^1$ (a "field formation");

(2) A functorial isomorphism of $H^2$ with $\cup_{n\geq 1} \mathbb{Z}/n\mathbb{Z}=\mathbb{Q}/\mathbb{Z}$ (a "class formation").

Then you can start with the huge and powerful machinery of (Tate's) cohomology to obtain Tate's theorem: the cup product map $H^q(G, \mathbb{Z})\rightarrow H^{q+2}(G,A)$ is a functorial isomorphism, which provides the canonical morphism of CFT at $q=-2$: $H^{-2}(G, \mathbb{Z})=G^{ab}$, $H^{0}(G,A)=A^G/N_G A.$ In the local case you take $A=K^*$, and in the global case you first take $A=\mathbb{I}$ and then $\mathbb{I}_K/K^*$.

I learnt cohomological aspects from Neukirch's excellent book "Class Field Theory, Bonn Lectures". It's very beautiful and carefully written. I spent much time checking each step and details. I was very striking that this somewhat abstract machinery DOES package the "classical" version.

But so far, I still couldn't get an intuitive explanation: how did this machinery correspond to the classical version? For example, in the "classical" version you get this canonical morphism from the Brauer group very directly, which should correspond to $H^2$ in cohomology. But in the "modern" version you get the cup product morphism from $H^{-2}$ and $H^{0}$. This looks very different. So you must have already done "something" from the class formation axiom for $H^2$ and cohomological machinery to get this. Just as Neukirch wrote on his book (page 78):"Before the introduction of cohomology theory, algebras were used to describe local class field theory; we remark that the use of cohomology has led to considerable simplifications." What makes me puzzled is exactly "how this modern machinery correspond to operations in the classical context".