Idel group

The ideal group and the ideal class group represent central objects of class field theory in mathematics .

In the local class field theory, the multiplicative group of the local field plays an important role. In the global class field theory this role is taken over by the ideal class group, which is the quotient of the units of the noble ring and the units of the body. The concept of the ideal is a modification of the ideal concept , whereby both concepts are related to each other, see the sentence about the connection between the ideal and the ideal class group . The concept of ideal was introduced in 1936 and 1941 by the French mathematician Claude Chevalley published work under the name "ideal element" (abbreviated: id.el.).

Generalizations of Artin's law of reciprocity lead to the connection of automorphic representations and Galois representations of ( Langlands program ). More precisely, the absolute Galois group operates on the algebraic De Rham cohomology of Shimura varieties with values ​​in the Idel group. These representations are Hodge-Tate with weights (1,2). ${\displaystyle K}$

The ideal group, especially the ideal class group, is used in class field theory, which deals with Abelian field extensions of . The product of the local reciprocity maps in class field theory gives a homeomorphism from the idele group to the Galois group of the maximum Abelian extension over an algebraic number field. Artin's law of reciprocity, which is a generalization of Gauss’s quadratic reciprocity law, says that the product in the multiplicative group of the number field vanishes. Therefore we get the global reciprocity map of the ideal class group from the Abelian part of the absolute Galois group of body extension. ${\displaystyle K}$

Notation: The following is a global solid . This means that either an algebraic number field or an algebraic function field positive characteristic of transcendental degree 1. In the first case this means that it is a finite field extension , in the second case that it is a finite field extension. In the following, one part of The trivial evaluation and the corresponding trivial amount are excluded in the entire article. A distinction is made between finite ( non-Archimedean ) places , which are noted as or , and infinite ( Archimedean ) places, which are noted as. In the following denote the finite set of infinite places of We write for a finite subset of the set of places which contains. Let be the completion of after a position. In a discrete evaluation, denote with the associated discrete evaluation ring of and with the maximum ideal of If this is a main ideal , then write for a uniformizing element. The reader should also be drawn to the unambiguous identification of amounts and evaluations of a body when fixing a suitable constant.The evaluation is assigned to the amount , which is defined as follows: ${\displaystyle K}$ ${\displaystyle K}$ ${\displaystyle K/\mathbb {Q} }$ ${\displaystyle K/\mathbb {F} _{p^{r}}(t)}$ ${\displaystyle v}$ ${\displaystyle K.}$ ${\displaystyle v<\infty }$ ${\displaystyle v\nmid \infty }$ ${\displaystyle v\mid \infty }$ ${\displaystyle P_{\infty }}$ ${\displaystyle K.}$ ${\displaystyle P}$ ${\displaystyle K,}$ ${\displaystyle P_{\infty }}$ ${\displaystyle K_{v}}$ ${\displaystyle K}$ ${\displaystyle v.}$ ${\displaystyle v}$ ${\displaystyle {\mathcal {O}}_{v}}$ ${\displaystyle K_{v}}$ ${\displaystyle {\mathfrak {m}}_{v}}$ ${\displaystyle {\mathcal {O}}_{v}.}$ ${\displaystyle \pi _{v}}$ ${\displaystyle C>1:}$ ${\displaystyle v}$ ${\displaystyle |\cdot |_{v}}$

${\displaystyle |x|_{v}:={\begin{cases}C^{-v(x)}&,{\text{ falls }}x\neq 0\\0&,{\text{ falls }}x=0\end{cases}}\quad \forall x\in K.}$

Conversely, the valuation is assigned to the amount , which is defined as follows: for all This identification is used continuously in the article. ${\displaystyle |\cdot |}$ ${\displaystyle v_{|\cdot |}}$ ${\displaystyle v_{|\cdot |}(x):=-\log _{C}(|x|)}$ ${\displaystyle x\in K^{\times }.}$

Definition of the ideal group of a global body ${\displaystyle K}$

Topology on the unit group of a topological ring

Be a topological ring . Then in general no topological group forms with the subspace topology . We therefore install on the following, coarser topology, which means that fewer sets are open: Look at the inclusion map ${\displaystyle R}$ ${\displaystyle R^{\times }}$ ${\displaystyle R^{\times }}$

${\displaystyle {\begin{aligned}\iota :R^{\times }&\rightarrow R\times R,\\x&\mapsto (x,x^{-1}).\end{aligned}}}$

We install on the topology that is generated from the corresponding subspace topology . That means that we install on the subspace topology of the product topology . By definition, a set is open in the new topology if and only if is open in the subspace topology. This topology becomes a topological group and the inclusion map becomes continuous. It is the coarsest topology that arises from the topology of and that makes it a topological group. ${\displaystyle R^{\times }}$ ${\displaystyle R\times R}$ ${\displaystyle \iota (R^{\times })}$ ${\displaystyle U\subset R^{\times }}$ ${\displaystyle \iota (U)}$ ${\displaystyle (R^{\times },\cdot )}$ ${\displaystyle R^{\times }\hookrightarrow R}$ ${\displaystyle R}$ ${\displaystyle R^{\times }}$

Proof: Take the topological ring. Then the inversion map is not continuous. This can be seen in the following example: Consider the consequence ${\displaystyle \mathbb {A} _{\mathbb {Q} }.}$

${\displaystyle {\begin{aligned}x_{1}&=(2,1,\dotsc )\\x_{2}&=(1,3,1,\dotsc )\\x_{3}&=(1,1,5,1,\dotsc )\\&\vdots \end{aligned}}}$

This sequence converges in the topology against one needle , because for a given area of the we can assume that the following shape: ${\displaystyle \mathbb {A} _{\mathbb {Q} }}$ ${\displaystyle U}$ ${\displaystyle 0}$ ${\displaystyle U}$

${\displaystyle U=\prod _{p{\text{ prim }} \atop p\leq N}U_{p}\times \prod _{p{\text{ prim }} \atop p>N}\mathbb {Z} _{p}.}$

Furthermore, it is true that for everyone and therefore for everyone It follows that for everyone large enough. The image of this sequence under the inversion mapping no longer converges in the subspace topology of (cf. the lemma about the difference between the restricted and unrestricted product topology ). In this new topology, neither the sequence nor its inverse converges . In particular, this example shows that the two topologies are different. So we install the topology described above on the units. This topology becomes a topological group. It remains to show the continuity of the inversion mapping. Let be any open set in the topology defined above; H. is open. It has to be shown that it is open, i. H. is to show that is open. This is the case according to the prerequisite. ${\displaystyle (x_{n})_{p}\in \mathbb {Z} _{p}}$ ${\displaystyle n,p}$ ${\displaystyle (x_{n})_{p}-1\in \mathbb {Z} _{p}}$ ${\displaystyle p.}$ ${\displaystyle x_{n}-1\in U}$ ${\displaystyle n\in \mathbb {N} _{0}}$ ${\displaystyle \mathbb {A} _{\mathbb {Q} }}$ ${\displaystyle R^{\times }}$ ${\displaystyle U\subset R^{\times }}$ ${\displaystyle U\times U^{-1}\subset R\times R}$ ${\displaystyle U^{-1}\subset R^{\times }}$ ${\displaystyle U^{-1}\times (U^{-1})^{-1}=U^{-1}\times U\subset R\times R}$

The ideal group of a global body ${\displaystyle K}$

Be a global body . The unit group of the Adelerring is the so-called Idelegruppe of , which is referred to below with ${\displaystyle K}$ ${\displaystyle K}$

${\displaystyle I_{K}:=\mathbb {A} _{K}^{\times }}$

referred to as. Keep defining

${\displaystyle {\begin{aligned}I_{K,S}&:=\mathbb {A} _{K,S}^{\times },\\I_{K}^{S}&:=(\mathbb {A} _{K}^{S})^{\times }.\end{aligned}}}$

We install the topology that we defined in the previous section on the Idele group. This makes the ideal group a topological group.

The Idele group as a restricted product

Be a global body. The following applies: ${\displaystyle K}$

${\displaystyle {\begin{aligned}I_{K,S}&={\widehat {\prod \limits _{v\in S}}}^{{\mathcal {O}}_{v}^{\times }}K_{v}^{\times },\\I_{K}^{S}&={\widehat {\prod \limits _{v\notin S}}}^{{\mathcal {O}}_{v}^{\times }}K_{v}^{\times },\\{\text{und}}\qquad I_{K}&={\widehat {\prod \limits _{v}}}^{{\mathcal {O}}_{v}^{\times }}K_{v}^{\times },\end{aligned}}}$

where equality is to be understood in the sense of topological rings . The restricted product carries the restricted product topology, which is generated by the restricted open rectangles. These have the following form:

${\displaystyle {\begin{aligned}\prod _{v\in E}U_{v}\times \prod _{v\notin E}{\mathcal {O}}_{v}^{\times },\end{aligned}}}$

where is a finite subset of all digits and are arbitrary open sets. ${\displaystyle E}$ ${\displaystyle U_{v}\subset K_{v}^{\times }}$

Proof: We carry out the proof for The other two statements follow analogously. First we consider the set equality. Consider the following chain of equations: ${\displaystyle I_{K}.}$

${\displaystyle {\begin{aligned}I_{K}:=\mathbb {A} _{K}^{\times }:&=\{x=(x_{v})_{v}\in \mathbb {A} _{K}:\exists y=(y_{v})_{v}\in \mathbb {A} _{K}:xy=1\}\\&=\{x=(x_{v})_{v}\in \mathbb {A} _{K}:\exists y=(y_{v})_{v}\in \mathbb {A} _{K}:x_{v}\cdot y_{v}=1\quad \forall v\}\\&=\{x=(x_{v})_{v}:x_{v}\in K_{v}^{\times }\,\,\forall v{\text{ und }}x_{v}\in {\mathcal {O}}_{v}^{\times }\,\,\forall v\,\,{\text{ bis }}{\text{ auf }}{\text{ endlich }}{\text{ viele }}\}\\&={\widehat {\prod \limits _{v}}}^{{\mathcal {O}}_{v}^{\times }}K_{v}^{\times }.\end{aligned}}}$

In the transition from line 2 to 3, note that both as well as in should be so for almost all and almost all for a total of almost all Next we consider that the two topologies coincide us. Obviously, every restricted open rectangle is also open in the topology of the Idele group. On the other hand, be open in the topology of the Idele group, i. H. is open. It follows that for each there is a restricted open rectangle which contains and lies in. It can therefore be represented as a union of restricted open rectangles, i.e. open in the restricted product topology. ${\displaystyle x}$ ${\displaystyle x^{-1}=y}$ ${\displaystyle \mathbb {A} _{K}}$ ${\displaystyle x_{v}\in {\mathcal {O}}_{v}}$ ${\displaystyle v}$ ${\displaystyle x_{v}^{-1}\in {\mathcal {O}}_{v}}$ ${\displaystyle v,}$ ${\displaystyle x_{v}\in {\mathcal {O}}_{v}^{\times }}$ ${\displaystyle v.}$ ${\displaystyle U\subset I_{K}}$ ${\displaystyle U\times U^{-1}\subset \mathbb {A} _{K}\times \mathbb {A} _{K}}$ ${\displaystyle u\in U}$ ${\displaystyle u}$ ${\displaystyle U}$ ${\displaystyle U}$

Further definitions

Using the previous notation, define

${\displaystyle {\widehat {\mathcal {O}}}:=\prod _{v\nmid \infty }{\mathcal {O}}_{v}=\prod _{v<\infty }{\mathcal {O}}_{v}}$

and as the corresponding unit group. It then applies ${\displaystyle {\widehat {\mathcal {O}}}^{\times }}$

${\displaystyle {\widehat {\mathcal {O}}}^{\times }=\prod _{v<\infty }{\mathcal {O}}_{v}^{\times }.}$

The ideal group for a body expansion ${\displaystyle I_{L}}$ ${\displaystyle L/K}$

Alternative description of the id group in the case ${\displaystyle L/K}$

Be a global body and be a finite body extension . Then there is a global body again and the idele group is defined. Define ${\displaystyle K}$ ${\displaystyle L/K}$ ${\displaystyle L}$ ${\displaystyle I_{L}}$

${\displaystyle {\begin{aligned}L_{v}^{\times }&:=\prod _{w\mid v}L_{w}^{\times },\\{\widetilde {{\mathcal {O}}_{v}}}^{\times }&:=\prod _{w\mid v}{\mathcal {O}}_{w}^{\times }.\end{aligned}}}$

Note that both products are finite. The following then applies:

${\displaystyle {\begin{aligned}I_{L}={\widehat {\prod \limits _{v}}}^{{\widetilde {{\mathcal {O}}_{v}}}^{\times }}L_{v}^{\times }.\end{aligned}}}$

Embedding of the id group of K in the id group of L

There is a canonical embedding of Idelegruppe of the Idelegruppe of The Idel the Idel is using for assigned. Therefore can be understood as a subgroup of . An element is in the subgroup if and only if its components satisfy for and if it continues to apply that for and for the same place of ${\displaystyle K}$ ${\displaystyle L.}$ ${\displaystyle a=(a_{v})_{v}\in I_{K}}$ ${\displaystyle a'=(a'_{w})_{w}\in I_{L}}$ ${\displaystyle a'_{w}=a_{v}\in K_{v}^{\times }\subset L_{w}^{\times }}$ ${\displaystyle w\mid v}$ ${\displaystyle I_{K}}$ ${\displaystyle I_{L}}$ ${\displaystyle a=(a_{w})_{w}\in I_{L}}$ ${\displaystyle I_{K},}$ ${\displaystyle a_{w}\in K_{v}^{\times }}$ ${\displaystyle w\mid v}$ ${\displaystyle a_{w}=a_{w'}}$ ${\displaystyle w\mid v}$ ${\displaystyle w'\mid v}$ ${\displaystyle v}$ ${\displaystyle K.}$

The ideal group of an algebra ${\displaystyle K}$ ${\displaystyle A}$

Be a finite - algebra , with a global body. Consider the unit group of The mapping is generally not continuous in the subspace topology. Thus the units do not form a topological group. We therefore provide the topology that we defined in the section on the units on topological rings . Given this topology, we call the unit group of the idele group of The elements of the group are called the idels of . ${\displaystyle A}$ ${\displaystyle K}$ ${\displaystyle K}$ ${\displaystyle \mathbb {A} _{A}.}$ ${\displaystyle x\mapsto x^{-1}}$ ${\displaystyle \mathbb {A} _{A}^{\times }}$ ${\displaystyle A}$ ${\displaystyle \mathbb {A} _{A}^{\times }}$ ${\displaystyle A.}$ ${\displaystyle A}$

Let be a finite subset of which contains a - basis of . Be back of the module, which of in is generated. As with the consideration of the ring of adeles, there is a finite subset of the set of places which contains, so that it applies to all that a compact subring of is and contains the units. Furthermore, it holds for each that is an open subset of and that the mapping is continuous on . It follows that the mapping maps the group homeomorphically to its picture below this mapping in . For are those elements of which are shown in the figure above . Thus, an open and compact subgroup of The proof of this statement can be found in Weil (1967) , pp. 71ff. ${\displaystyle \alpha }$ ${\displaystyle A,}$ ${\displaystyle K}$ ${\displaystyle A}$ ${\displaystyle \alpha _{v}}$ ${\displaystyle {\mathcal {O}}_{v}}$ ${\displaystyle \alpha }$ ${\displaystyle A_{v}}$ ${\displaystyle P_{0}}$ ${\displaystyle P_{\infty }}$ ${\displaystyle v\notin P_{0}}$ ${\displaystyle \alpha _{v}}$ ${\displaystyle A_{v}}$ ${\displaystyle v,}$ ${\displaystyle A_{v}^{\times }}$ ${\displaystyle A_{v}}$ ${\displaystyle x\mapsto x^{-1}}$ ${\displaystyle A_{v}^{\times }}$ ${\displaystyle x\mapsto (x,x^{-1})}$ ${\displaystyle A_{v}^{\times }}$ ${\displaystyle A_{v}\times A_{v}}$ ${\displaystyle v\notin P_{0}}$ ${\displaystyle \alpha _{v}^{\times }}$ ${\displaystyle A_{v}^{\times },}$ ${\displaystyle \alpha _{v}\times \alpha _{v}}$ ${\displaystyle \alpha _{v}^{\times }}$ ${\displaystyle A_{v}^{\times }.}$

These considerations can be applied in particular to the endomorphism algebras of vector spaces. Let be a finite-dimensional vector space, where is a global field. Let this be an algebra. The following applies: where a linear mapping is invertible if and only if its determinant is different from . If is a topological field, then is an open subset of because since it is closed and continuous is open. With you can then look at the ideals of as above . ${\displaystyle E}$ ${\displaystyle K}$ ${\displaystyle K}$ ${\displaystyle \operatorname {End} (E):=\{\varphi :E\rightarrow E,\varphi \,\,{\text{ ist }}{\text{ eine }}\,\,K{\text{ -lineare }}{\text{ Abbildung }}\}.}$ ${\displaystyle K}$ ${\displaystyle \operatorname {End} (E)^{\times }=\operatorname {Aut} (E),}$ ${\displaystyle 0}$ ${\displaystyle K}$ ${\displaystyle \operatorname {Aut} (E)}$ ${\displaystyle \operatorname {End} (E),}$ ${\displaystyle \operatorname {End} (E)\setminus \operatorname {Aut} (E)=\operatorname {det} ^{-1}(\{0\}).}$ ${\displaystyle \{0\}}$ ${\displaystyle \operatorname {det} }$ ${\displaystyle \operatorname {Aut} (E)}$ ${\displaystyle A:=End(A)}$ ${\displaystyle A}$

Alternative characterization of the idele group: Let the situation be as before: Let be a finite subset of the set of places which contains. Then ${\displaystyle P}$ ${\displaystyle P_{0}}$

${\displaystyle \mathbb {A} _{A}(P,\alpha )^{\times }:=\prod \limits _{v\in P}A_{v}^{\times }\times \prod \limits _{v\notin P}\alpha _{v}^{\times }}$

an open subgroup of where can be written as the union of , and where all finite subsets of the set of digits pass through. A proof of this statement can be found in Weil (1967) , p. 72. ${\displaystyle \mathbb {A} _{A}^{\times },}$ ${\displaystyle \mathbb {A} _{A}^{\times }}$ ${\displaystyle \mathbb {A} _{A}(P,\alpha )^{\times }}$ ${\displaystyle P\supset P_{0}}$

In the special case you get the following. For every finite subset of the set of places which contains is the group ${\displaystyle A=K}$ ${\displaystyle K,}$ ${\displaystyle P_{\infty }}$

${\displaystyle \mathbb {A} _{K}(P)^{\times }=\prod \limits _{v\in P}K_{v}^{\times }\times \prod \limits _{v\notin P}{\mathcal {O}}_{v}^{\times }}$

An open subgroup of It is still true that the union of all of these subgroups is. ${\displaystyle \mathbb {A} _{K}^{\times }=I_{K}.}$ ${\displaystyle I_{K}}$ ${\displaystyle \mathbb {A} _{K}(P)^{\times }}$

Track and norm

The track cannot simply be transferred to the idle group, but the standard can . Let to this Then we have an injective group homomorphism ${\displaystyle \alpha \in I_{K}.}$ ${\displaystyle \operatorname {con} _{L/K}(\alpha )\in I_{L},}$

${\displaystyle \operatorname {con} _{L/K}:I_{K}\hookrightarrow I_{L}.}$

Since and is therefore invertible, it is also invertible, since it is therefore true that consequently the restriction of the norm mapping provides the following mapping: ${\displaystyle \alpha \in I_{L}}$ ${\displaystyle N_{L/K}(\alpha )}$ ${\displaystyle (N_{L/K}(\alpha ))^{-1}=N_{L/K}(\alpha ^{-1}).}$ ${\displaystyle N_{L/K}(\alpha )\in I_{K}.}$

${\displaystyle N_{L/K}:I_{L}\rightarrow I_{K}.}$

This is continuous and also fulfills the properties of the norm from the lemma about the properties of trace and norm .

properties

${\displaystyle K^{\times }}$ is a discrete subset of ${\displaystyle I_{K}}$

The units of the global body can be embedded diagonally in the idele group: ${\displaystyle K}$

${\displaystyle {\begin{aligned}&K^{\times }\hookrightarrow I_{K,S},\\&a\mapsto (a,a,a,\dotsc ).\end{aligned}}}$

Since this applies to all , the well-definedness and injectivity of this figure follows as in the corresponding sentence about the Adelering. ${\displaystyle K^{\times }\subset K_{v}^{\times }}$ ${\displaystyle v}$

Furthermore, the subgroup is discreet (and therefore in particular closed ) in . This fact follows analogously to the corresponding sentence about the Adelering . ${\displaystyle K^{\times }}$ ${\displaystyle I_{K}}$

In particular is a discrete subset of ${\displaystyle A^{\times }}$ ${\displaystyle \mathbb {A} _{A}^{\times }.}$

The ideal class group

In algebraic number theory, the ideal class group is considered for a given number field . Analogously, the concept of the ideal class group is defined as follows. ${\displaystyle K}$

In analogy to the concept of the main ideal, the elements of in are referred to as the main ideals of . The quotient, i.e. the factor group , is called the ideal class group of . This is related to the ideal class group (cf. the sentence about the connection between the ideal and the ideal class group ) and is the main subject of the considerations in class field theory. ${\displaystyle K^{\times }}$ ${\displaystyle I_{K}}$ ${\displaystyle I_{K}}$ ${\displaystyle C_{K}:=I_{K}/K^{\times },}$ ${\displaystyle K}$

Since in is closed, it follows that it is a locally compact, Hausdorffian, topological group. ${\displaystyle K^{\times }}$ ${\displaystyle I_{K}}$ ${\displaystyle C_{K}}$

For a finite expansion of global bodies, the embedding induces an injective mapping to the ideal class groups: ${\displaystyle L/K}$ ${\displaystyle I_{K}\hookrightarrow I_{L}}$

${\displaystyle {\begin{aligned}C_{K}&\hookrightarrow C_{L},\\\alpha K^{\times }&\mapsto \alpha L^{\times }.\end{aligned}}}$

The well-definition of the mapping follows since the injection obviously maps to a subset of . The injectivity is shown in Neukirch (2007) , p. 388. ${\displaystyle I_{K}\hookrightarrow I_{L}}$ ${\displaystyle K^{\times }}$ ${\displaystyle L^{\times }}$

The ideal group is a locally compact, topological group

For each subset of the set of places of is with the topology of Idelegruppe a locally compact topological group. With the subspace topology, there is generally no topological group because the inversion map is not continuous. ${\displaystyle S}$ ${\displaystyle K}$ ${\displaystyle I_{K,S}}$ ${\displaystyle I_{K,S}}$

This sentence follows from the local compactness of the Adelerring, the construction of the ideal topology and the representation of the ideal group as a restricted product.

Since the ideal group with the multiplication form a locally compact group, there is a hair measure on this group. This can be normalized so that this is the normalization at the finite places. Here denotes the set of finite idels, i.e. the unit group of the set of finite adeles. The multiplicative Lebesgue measure is taken from the infinite . ${\displaystyle d^{\times }x}$ ${\displaystyle \textstyle \int _{I_{K,fin}}\mathbf {1} _{\widehat {\mathcal {O}}}\,d^{\times }x=1.}$ ${\displaystyle I_{K,fin}}$ ${\displaystyle \textstyle {\frac {dx}{|x|}}}$

A one- neighborhood basis of the idel group is given by a one-neighborhood basis of . Alternatively, all sets of the following form also form a one-environment basis: ${\displaystyle \mathbb {A} _{K}^{\times }(P_{\infty })}$

${\displaystyle \prod _{v}U_{v},}$

with an environment of in and for almost all ${\displaystyle U_{v}}$ ${\displaystyle 1}$ ${\displaystyle K_{v}^{\times }}$ ${\displaystyle U_{v}={\mathcal {O}}_{v}^{\times }}$ ${\displaystyle v.}$

Amount to and the amount of the ideal from ${\displaystyle I_{K}}$ ${\displaystyle 1}$ ${\displaystyle K}$

Be a global body. We install an amount on the Idel group as follows: For a given Idel define: ${\displaystyle K}$ ${\displaystyle \alpha =(\alpha _{v})_{v}}$

${\displaystyle |\alpha |:=\prod _{v}|\alpha _{v}|_{v}.}$

Since this product is finite and therefore well-defined. The definition of the amount can be extended to the Adelering if we allow infinite products, whereby the convergence in is considered. These products are all making the extended amount on disappears. In the following, the amount mapping to or ${\displaystyle \alpha \in I_{K},}$ ${\displaystyle (\mathbb {R} ,|\cdot |_{\infty })}$ ${\displaystyle 0,}$ ${\displaystyle \mathbb {A} _{K}\setminus I_{K}}$ ${\displaystyle |\cdot |}$ ${\displaystyle \mathbb {A} _{K}}$ ${\displaystyle I_{K}.}$

It is now true that the amount mapping is a continuous group homomorphism, i. H. the mapping is a continuous group homomorphism. This can be seen from the following calculation: Be and Then: ${\displaystyle |\cdot |:I_{K}\rightarrow \mathbb {R} _{>0}}$ ${\displaystyle \alpha }$ ${\displaystyle \beta \in I_{K}.}$

${\displaystyle {\begin{aligned}|\alpha \cdot \beta |&=\prod _{v}|(\alpha \cdot \beta )_{v}|_{v}\\&=\prod _{v}|\alpha _{v}\cdot \beta _{v}|_{v}\\&=\prod _{v}(|\alpha _{v}|_{v}\cdot |\beta _{v}|_{v})\\&=(\prod _{v}|\alpha _{v}|_{v})\cdot (\prod _{v}|\beta _{v}|_{v})\\&=|\alpha |\cdot |\beta |,\end{aligned}}}$

with the transition from line 3 to line 4 it was used that all occurring products are finite. The continuity of the mapping follows by showing sequential continuity and using the fact that the amount mapping is continuous. This can be seen with the reverse triangle inequality. Due to the restricted product topology, only a finite number of places are effectively considered and the claim follows. ${\displaystyle K_{v}}$

We now define the set of -idels as follows: ${\displaystyle 1}$ ${\displaystyle \mathbb {A} _{K}^{1}}$

${\displaystyle \mathbb {A} _{K}^{1}:=\{x\in I_{K}:|x|=1\}=\ker(|\cdot |).}$

The group of -idels are a subgroup of In the literature it is also used for the group of -idels. The notation is used below . ${\displaystyle 1}$ ${\displaystyle I_{K}.}$ ${\displaystyle I_{K}^{1}}$ ${\displaystyle 1}$ ${\displaystyle \mathbb {A} _{K}^{1}}$

It now holds that is a closed subset of , because ${\displaystyle \mathbb {A} _{K}^{1}}$ ${\displaystyle \mathbb {A} _{K}}$ ${\displaystyle \mathbb {A} _{K}^{1}=|\cdot |^{-1}(\{1\}).}$

The topology on corresponds to the subspace topology of on . This statement can be found in Cassels (1967) , pp. 69f. ${\displaystyle \mathbb {A} _{K}}$ ${\displaystyle \mathbb {A} _{K}^{1}}$ ${\displaystyle I_{K}}$ ${\displaystyle \mathbb {A} _{K}^{1}}$

General product formula

Be a global body. The following applies to the homomorphism of to : In other words it means that for all The product formula implies that is. This phrase is known in the literature as "Artin's product formula". ${\displaystyle K}$ ${\displaystyle |\cdot |:z\mapsto |z|}$ ${\displaystyle I_{K}}$ ${\displaystyle \mathbb {R} _{>0}}$ ${\displaystyle K^{\times }\subset \ker(|\cdot |).}$ ${\displaystyle |k|=1}$ ${\displaystyle k\in K^{\times }.}$ ${\displaystyle K^{\times }\subset \mathbb {A} _{K}^{1}}$

There is ample evidence of this statement. This is based on Neukirch (2007) , p. 195. It can also be found in Cassels (1967) , p. 61. The essential idea of ​​the proof is to reduce the general product formula in the algebraic number field case to the special case . The functional body case goes similarly. ${\displaystyle K=\mathbb {Q} }$

Be whatever you want. To show is: ${\displaystyle a\in K^{\times }}$

${\displaystyle \prod _{v}|a|_{v}=1.}$

It is and therefore for everyone for which the associated prime ideal does not appear in the prime ideal decomposition of the main ideal . This is the case for almost everyone . The following applies now: ${\displaystyle v(a)=0}$ ${\displaystyle |a|_{v}=1}$ ${\displaystyle v\nmid \infty ,}$ ${\displaystyle {\mathfrak {p}}_{v}}$ ${\displaystyle (a)}$ ${\displaystyle {\mathfrak {p}}_{v}}$

${\displaystyle {\begin{aligned}\prod _{v}|a|_{v}&=\prod _{p\leq \infty }\prod _{v\mid p}|a|_{v}\\&=\prod _{p\leq \infty }\prod _{v\mid p}|N_{K_{v}/\mathbb {Q} _{p}}(a)|_{p}\\&=\prod _{p\leq \infty }|N_{K/\mathbb {Q} }(a)|_{p},\end{aligned}}}$

with the transition from line 1 to line 2, the generally valid equation was used, with a digit from and a digit from which is above . When moving from line 2 to line 3, a property of the standard was exploited. Note that the norm is in. We can therefore assume without reservation that it is. Then a unique prime decomposition has: ${\displaystyle |a|_{w}=|N_{L_{w}/K_{v}}(a)|_{v}}$ ${\displaystyle v}$ ${\displaystyle K}$ ${\displaystyle w}$ ${\displaystyle L}$ ${\displaystyle v}$ ${\displaystyle \mathbb {Q} }$ ${\displaystyle a\in \mathbb {Q} }$ ${\displaystyle a}$

${\displaystyle a=\pm \prod _{p<\infty }p^{v_{p}},}$

being almost always the Ostrowski's theorem states that the amounts to up to equivalence exactly the -amounts and are. It follows that ${\displaystyle v_{p}\in \mathbb {Z} }$ ${\displaystyle 0.}$ ${\displaystyle \mathbb {Q} }$ ${\displaystyle p}$ ${\displaystyle |\cdot |_{\infty }}$

${\displaystyle {\begin{aligned}|a|&=(\prod _{p<\infty }|a|_{p})\cdot |a|_{\infty }\\&=(\prod _{p<\infty }p^{-v_{p}})\cdot (\prod _{p<\infty }p^{v_{p}})\\&=1.\end{aligned}}}$

There is further evidence of the product formula which can be found in the literature.

characterization of ${\displaystyle \mathbb {A} _{\operatorname {End} (E)}^{\times }}$

Let be a -dimensional vector space. Set Let then the following statements are equivalent ${\displaystyle E}$ ${\displaystyle n}$ ${\displaystyle K}$ ${\displaystyle A:=\operatorname {End} (E).}$ ${\displaystyle a\in \mathbb {A} _{A}.}$

• ${\displaystyle a\in \mathbb {A} _{A}^{\times },}$
• ${\displaystyle \operatorname {det} (a)\in \mathbb {A} _{K}^{\times },}$
• ${\displaystyle x\mapsto ax}$ is an automorphism of ${\displaystyle \mathbb {A} _{E}.}$

If one of the three points is fulfilled, then it is also true that the assignments and homomorphisms are from to or. A proof of this statement can be found in Weil (1967) , pp. 73f. ${\displaystyle |a|=|\operatorname {det} (a)|.}$ ${\displaystyle a\mapsto \operatorname {det} (a)}$ ${\displaystyle a\mapsto |\operatorname {det} (a)|}$ ${\displaystyle \mathbb {A} _{A}^{\times }}$ ${\displaystyle \mathbb {A} _{K}^{\times }}$ ${\displaystyle \mathbb {R} _{>0}.}$

In particular, for a finite-dimensional algebra and the equivalence of the following statements: ${\displaystyle K}$ ${\displaystyle A}$ ${\displaystyle a\in \mathbb {A} _{A}}$

• ${\displaystyle a\in \mathbb {A} _{A}^{\times },}$
• ${\displaystyle N_{A/K}(a)\in \mathbb {A} _{K}^{\times },}$
• ${\displaystyle x\mapsto ax}$ is an automorphism of the additive group ${\displaystyle \mathbb {A} _{A}.}$

If one of the three points is fulfilled, then it is also true that the assignments and homomorphisms are from to or. With this theorem, an alternative proof of the product formula is possible, cf. Weil (1967) , p. 75. ${\displaystyle |a|=|N_{A/K}(a)|.}$ ${\displaystyle a\mapsto N_{A/K}(a)}$ ${\displaystyle a\mapsto |N_{A/K}(a)|}$ ${\displaystyle \mathbb {A} _{A}^{\times }}$ ${\displaystyle \mathbb {A} _{K}^{\times }}$ ${\displaystyle \mathbb {R} _{>0}.}$

${\displaystyle K^{\times }}$is a discrete and co-compact subgroup in the set of -idels ${\displaystyle 1}$

Before we can formulate the sentence, we need the following auxiliary statement:

Lemma: Be a global body. There is a constant which only depends on the global body , so that there exists for all with the property one , so that for all ${\displaystyle K}$ ${\displaystyle C,}$ ${\displaystyle K}$ ${\displaystyle \alpha =(\alpha _{v})_{v}\in \mathbb {A} _{K}}$ ${\displaystyle \textstyle \prod _{v}|\alpha _{v}|_{v}>C}$ ${\displaystyle \beta \in K^{\times }}$ ${\displaystyle |\beta _{v}|_{v}\leq |\alpha _{v}|_{v}}$ ${\displaystyle v.}$

A proof of this statement can be found in Cassels (1967) , p. 66 Lemma.

Corollary: Be a global body, be a place of and be given for all places so that applies to almost all . Then there is a so for everyone ${\displaystyle K}$ ${\displaystyle v_{0}}$ ${\displaystyle K}$ ${\displaystyle \delta _{v}>0}$ ${\displaystyle v\neq v_{0},}$ ${\displaystyle \delta _{v}=1}$ ${\displaystyle v}$ ${\displaystyle \beta \in K^{\times },}$ ${\displaystyle |\beta |\leq \delta _{v}}$ ${\displaystyle v\neq v_{0}.}$

Proof: According to the lemma before, there is a constant that only depends on our (fixed) global body. We denote with uniformizing elements of the corresponding integer rings. Now define the nobility via with minimal so that for all then is almost always. Define with so that this works, because is for almost everyone . According to the above lemma there exists such that holds for all . ${\displaystyle C,}$ ${\displaystyle \pi _{v}}$ ${\displaystyle {\mathcal {O}}_{v}.}$ ${\displaystyle \alpha =(\alpha _{v})_{v}}$ ${\displaystyle \alpha _{v}:=\pi _{v}^{k_{v}}}$ ${\displaystyle k_{v}\in \mathbb {Z} }$ ${\displaystyle |\alpha _{v}|_{v}\leq \delta _{v}}$ ${\displaystyle v\neq v_{0}.}$ ${\displaystyle k_{v}=0}$ ${\displaystyle \alpha _{v_{0}}:=\pi _{v_{0}}^{k_{v_{0}}}}$ ${\displaystyle k_{v_{0}}\in \mathbb {Z} ,}$ ${\displaystyle \textstyle \prod _{v}|\alpha _{v}|_{v}>C.}$ ${\displaystyle k_{v}=0}$ ${\displaystyle v}$ ${\displaystyle \beta \in K^{\times },}$ ${\displaystyle |\beta |_{v}\leq |\alpha _{v}|_{v}\leq \delta _{v}}$ ${\displaystyle v\neq v_{0}}$

Now to the actual sentence:

Sentence: Be a global body. is discrete in and the quotient is compact. ${\displaystyle K}$ ${\displaystyle K^{\times }}$ ${\displaystyle \mathbb {A} _{K}^{1}}$ ${\displaystyle \mathbb {A} _{K}^{1}/K^{\times }}$

Proof: The discrete nature of in implies the discrete nature of in ${\displaystyle K^{\times }}$ ${\displaystyle I_{K}}$ ${\displaystyle K^{\times }}$ ${\displaystyle \mathbb {A} _{K}^{1}.}$

It remains to show that is compact. This proof can be found in Weil (1967) , p. 76 or in Cassels (1967) , p. 70. In the following, Cassel's (1967) idea of ​​proof is reproduced: It is sufficient to show the existence of a compact set so that the natural projection is surjective, since the natural projection is a continuous mapping. Now be given with the property , where is the constant of the lemma formulated at the beginning. Define ${\displaystyle \mathbb {A} _{K}^{1}/K^{\times }}$ ${\displaystyle W\subset \mathbb {A} _{K}}$ ${\displaystyle \pi :W\cap \mathbb {A} _{K}^{1}\rightarrow \mathbb {A} _{K}^{1}/K^{\times }}$ ${\displaystyle \alpha \in \mathbb {A} _{K}}$ ${\displaystyle \prod _{v}|\alpha _{v}|_{v}>C}$ ${\displaystyle C}$

${\displaystyle W:=\{\xi =(\xi _{v})_{v}:|\xi _{v}|_{v}\leq |\alpha _{v}|_{v}\quad \forall v\}.}$

Obviously is compact. Be it in given. We show that there exists such that , by definition of the set of -idels, it holds that ${\displaystyle W}$ ${\displaystyle \beta =(\beta _{v})_{v}}$ ${\displaystyle \mathbb {A} _{K}^{1}}$ ${\displaystyle \eta \in K^{\times }}$ ${\displaystyle \eta \beta \in W.}$ ${\displaystyle 1}$

${\displaystyle \prod _{v}|\beta _{v}|_{v}=1,}$

and therefore

${\displaystyle \prod _{v}|\beta _{v}^{-1}|_{v}=1.}$

It follows that

${\displaystyle \prod _{v}|\beta _{v}^{-1}\alpha _{v}|_{v}=\prod _{v}|\alpha _{v}|_{v}>C.}$

Because of the previous lemma, there exists such that for all It follows that this implies the assertion. ${\displaystyle \eta \in K^{\times },}$ ${\displaystyle |\eta |_{v}\leq |\beta _{v}^{-1}\alpha _{v}|_{v}}$ ${\displaystyle v.}$ ${\displaystyle \eta \beta \in W.}$

Some isomorphisms in the case ${\displaystyle K=\mathbb {Q} }$

In the case there is a canonical isomorphism. Furthermore, it holds that is is a system of representatives . This means that the amount also induces the following isomorphisms of topological groups: ${\displaystyle K=\mathbb {Q} }$ ${\displaystyle \mathbb {A} _{\mathbb {Q} }^{1}/\mathbb {Q} ^{\times }\cong {\widehat {\mathbb {Z} }}^{\times }.}$ ${\displaystyle {\widehat {\mathbb {Z} }}^{\times }\times \{1\}}$ ${\displaystyle \mathbb {A} _{\mathbb {Q} }^{1}/\mathbb {Q} ^{\times }}$ ${\displaystyle ({\widehat {\mathbb {Z} }}^{\times }\times \{1\})\mathbb {Q} ^{\times }=\mathbb {A} _{\mathbb {Q} }^{1}.}$

${\displaystyle {\begin{aligned}I_{\mathbb {Q} }&\cong \mathbb {A} _{\mathbb {Q} }^{1}\times (0,\infty ){\text{ und }}\\\mathbb {A} _{\mathbb {Q} }^{1}&\cong I_{\mathbb {Q} ,fin}\times \{\pm 1\}.\end{aligned}}}$

It follows that there is a representative system of . This sentence is part of sentence 5.3.3 on page 128 in Deitmar (2010) . ${\displaystyle {\widehat {\mathbb {Z} }}^{\times }\times (0,\infty )}$ ${\displaystyle I_{\mathbb {Q} }/\mathbb {Q} ^{\times }}$

Proof: Define the mapping via This mapping is obviously well-defined since it applies to all and thus . The mapping is a continuous group homomorphism. For injectivity, therefore, there exists such that through a comparison at the infinite point, it follows and therefore for surjectivity is given. Since the absolute value of this element is, It follows that So is and thus the mapping is surjective, because for all cf. The representation of The other isomorphisms are given by: via and via The proof that these are isomorphisms is left to the reader for exercise. ${\displaystyle \varphi :{\widehat {\mathbb {Z} }}^{\times }\rightarrow \mathbb {A} _{\mathbb {Q} }^{1}/\mathbb {Q} ^{\times },}$ ${\displaystyle (a_{p})_{p}\mapsto ((a_{p})_{p},1)\mathbb {Q} ^{\times }.}$ ${\displaystyle |a_{p}|_{p}=1}$ ${\displaystyle p}$ ${\displaystyle (\prod _{p<\infty }|a_{p}|_{p})\cdot 1=1}$ ${\displaystyle ((a_{p})_{p},1)\mathbb {Q} ^{\times }=((b_{p})_{p},1)\mathbb {Q} ^{\times }.}$ ${\displaystyle q\in \mathbb {Q} ^{\times },}$ ${\displaystyle ((a_{p})_{p},1)q=((b_{p})_{p},1).}$ ${\displaystyle q=1}$ ${\displaystyle (a_{p})_{p}=(b_{p})_{p}.}$ ${\displaystyle ((\beta _{p})_{p},\beta _{\infty })\mathbb {Q} ^{\times }\in \mathbb {A} _{\mathbb {Q} }^{1}/\mathbb {Q} ^{\times }}$ ${\displaystyle 1}$ ${\displaystyle \textstyle |\beta _{\infty }|_{\infty }={\frac {1}{\prod _{p}|\beta _{p}|_{p}}}\in \mathbb {Q} .}$ ${\displaystyle \beta _{\infty }\in \mathbb {Q} .}$ ${\displaystyle \textstyle ((\beta _{p})_{p},\beta _{\infty })\mathbb {Q} ^{\times }=(({\frac {\beta _{p}}{\beta _{\infty }}})_{p},1)\mathbb {Q} ^{\times }}$ ${\displaystyle \textstyle {\frac {|\beta _{p}|_{p}}{|\beta _{\infty }|_{p}}}=1}$ ${\displaystyle p,}$ ${\displaystyle |\beta _{\infty }|_{\infty }.}$ ${\displaystyle \psi :I_{\mathbb {Q} }\rightarrow \mathbb {A} _{\mathbb {Q} }^{1}\times (0,\infty ),}$ ${\displaystyle a=(a_{p})_{p\leq \infty }\mapsto ((a_{p})_{p<\infty },a_{\infty }/|a|,|a|)}$ ${\displaystyle {\widetilde {\psi }}:I_{\mathbb {Q} ,fin}\times \{\pm 1\}\rightarrow \mathbb {A} _{\mathbb {Q} }^{1},}$ ${\displaystyle (a_{fin},\epsilon )\mapsto (a_{fin},\epsilon |a_{fin}|^{-1}).}$

Relationship between ideal class group and ideal class group

For an algebraic number field we define that : ${\displaystyle K}$ ${\displaystyle C_{K,fin}:=I_{K,fin}/K^{\times }.}$

${\displaystyle {\begin{aligned}J_{K}&\cong I_{K,fin}/{\widehat {\mathcal {O}}}^{\times }\qquad \quad {\text{und}}\\Cl_{K}&\cong C_{K,fin}/{\widehat {\mathcal {O}}}^{\times }K^{\times }\qquad {\text{und}}\\Cl_{K}&\cong C_{K}/({\widehat {\mathcal {O}}}^{\times }\times \prod _{v\mid \infty }K_{v}^{\times })K^{\times }.\end{aligned}}}$

Here refers to the group of broken ideals in two with the product ideal as a group link. This is by a group called the Ideal Group write We for the ideal class group of Dedekind ring so is the integer ring of the algebraic number field By definition applies now ${\displaystyle J_{K}}$ ${\displaystyle K}$ ${\displaystyle J_{K}}$ ${\displaystyle K.}$ ${\displaystyle Cl_{K}}$ ${\displaystyle {\mathcal {O}}:=\{\alpha \in K:\exists f\in \mathbb {Z} [x]{\text{ normiert: }}f(\alpha )=0\},}$ ${\displaystyle {\mathcal {O}}}$ ${\displaystyle K.}$ ${\displaystyle Cl_{K}=J_{K}/K^{\times }.}$

Proof: In the following we use the facts that for an algebraic number field there is a one-to-one relationship between the finite digits of and the prime ideal not equal to zero of : ${\displaystyle K}$ ${\displaystyle K}$ ${\displaystyle {\mathcal {O}}}$

Let be a finite place of and be a representative of the equivalence class Define ${\displaystyle v}$ ${\displaystyle K}$ ${\displaystyle |\cdot |_{v}}$ ${\displaystyle v.}$

${\displaystyle {\mathfrak {p}}_{v}:=\{x\in {\mathcal {O}}:|x|_{v}<1\}.}$

Then there is a prime ideal in The map is a bijection between the set of all finite places of and the set of prime ideals of The inverse map is given by: ${\displaystyle {\mathfrak {p}}_{v}}$ ${\displaystyle {\mathcal {O}}.}$ ${\displaystyle v\mapsto {\mathfrak {p}}_{v}}$ ${\displaystyle K}$ ${\displaystyle {\mathfrak {p}}\neq 0}$ ${\displaystyle {\mathcal {O}}.}$

A given prime ideal is assigned the evaluation given by ${\displaystyle {\mathfrak {p}}}$ ${\displaystyle v_{\mathfrak {p}}}$

${\displaystyle {\begin{aligned}v_{\mathfrak {p}}(x)&:=\operatorname {max} \{k\in \mathbb {N} _{0}:x\in {\mathfrak {p}}^{k}\}\quad \forall x\in {\mathcal {O}}^{\times },\\v_{\mathfrak {p}}({\frac {x}{y}})&:=v_{\mathfrak {p}}(x)-v_{\mathfrak {p}}(y)\quad \forall x,y\in {\mathcal {O}}^{\times }.\end{aligned}}}$

Now for the real evidence. The following figure is well defined:

${\displaystyle {\begin{aligned}(\cdot ):\,&I_{K,fin}\rightarrow J_{K},\\&\alpha =(\alpha _{v})_{v}\mapsto \prod _{v\nmid \infty }{\mathfrak {p}}_{v}^{v(\alpha _{v})},\end{aligned}}}$

where is the prime ideal associated with the position . The mapping is obviously a surjective group homomorphism. It is true that the first isomorphism from the proposition now follows with the homomorphism proposition . ${\displaystyle {\mathfrak {p}}_{v}}$ ${\displaystyle v}$ ${\displaystyle (\cdot )}$ ${\displaystyle \ker((\cdot ))={\widehat {\mathcal {O}}}^{\times }.}$

Now we divide out on both sides . This is possible because ${\displaystyle K^{\times }}$

${\displaystyle {\begin{aligned}(\alpha )&=((\alpha ,\alpha ,\dotsc ))\\&=\prod _{v\nmid \infty }{\mathfrak {p}}_{v}^{v(\alpha )}\\&=(\alpha )\end{aligned}}}$

for all Note the misuse of the notation: On the left-hand side in line 1, the bracket stands for the previously defined figure. Then the embedding of in is used. In line 2 the definition of the mapping rule is applied and finally in line 3 we use the fact that the integer ring is a Dedekind ring and therefore every ideal, especially the main ideal , can be prime factorized. The mapping is therefore an -equivariant group homomorphism. Consequently, the above mapping induces a surjective homomorphism ${\displaystyle \alpha \in K^{\times }.}$ ${\displaystyle K^{\times }}$ ${\displaystyle I_{K,fin}}$ ${\displaystyle {\mathcal {O}}}$ ${\displaystyle (\alpha ),}$ ${\displaystyle (\cdot )}$ ${\displaystyle K^{\times }}$

${\displaystyle {\begin{aligned}\varphi :\,&C_{K,fin}\rightarrow Cl_{K},\\&\alpha K^{\times }\mapsto (\alpha )K^{\times }.\end{aligned}}}$

We now show that . Let Then is there for all Let now be reversed with Then it follows that there is a representative for which holds: Consequently, and therefore, the second isomorphism from the proposition is proven. ${\displaystyle \ker(\varphi )={\widehat {\mathcal {O}}}^{\times }K^{\times }}$ ${\displaystyle \xi =(\xi _{v})_{v}\in {\widehat {\mathcal {O}}}^{\times }.}$ ${\displaystyle \textstyle \varphi (\xi K^{\times })=\prod _{v}{\mathfrak {p}}_{v}^{v(\xi _{v})}K^{\times }=K^{\times },}$ ${\displaystyle v(\xi _{v})=0}$ ${\displaystyle v.}$ ${\displaystyle \xi K^{\times }\in C_{K,fin}}$ ${\displaystyle \varphi (\xi K^{\times })={\mathcal {O}}K^{\times }.}$ ${\displaystyle \textstyle K^{\times }=\prod _{v}{\mathfrak {p}}_{v}^{v(\xi _{v})}K^{\times }.}$ ${\displaystyle \textstyle \prod _{v}{\mathfrak {p}}_{v}^{v(\xi '_{v})}={\mathcal {O}}.}$ ${\displaystyle \xi '\in {\widehat {\mathcal {O}}}^{\times }}$ ${\displaystyle \xi K^{\times }=\xi 'K^{\times }\in {\widehat {\mathcal {O}}}^{\times }K^{\times }.}$

To show the last isomorphism from the theorem, we note that the mapping is a surjective group homomorphism ${\displaystyle \varphi }$

${\displaystyle {\widetilde {\varphi }}:C_{K}\rightarrow Cl_{K}}$

induced. It holds that this shows the theorem. ${\displaystyle \textstyle \ker({\widetilde {\varphi }})=({\widehat {\mathcal {O}}}^{\times }\times \prod _{v\mid \infty }K_{v}^{\times })K^{\times }.}$

Note: The mapping is continuous in the following sense: On we have the usual ideal topology. On we install the discrete topology. The continuity follows when we can show that there is open to every Now is open, so that ${\displaystyle (\cdot )}$ ${\displaystyle I_{K,fin}}$ ${\displaystyle J_{K}}$ ${\displaystyle (\{{\mathfrak {a}}\})^{-1}}$ ${\displaystyle {\mathfrak {a}}\in J_{K}.}$ ${\displaystyle (\{{\mathfrak {a}}\})^{-1}=\alpha {\widehat {\mathcal {O}}}^{\times }}$ ${\displaystyle \alpha =(\alpha _{v})_{v}\in \mathbb {A} _{K,fin},}$ ${\displaystyle \textstyle {\mathfrak {a}}=\prod _{v}{\mathfrak {p}}_{v}^{v(\alpha _{v})}.}$

Decomposition of and ${\displaystyle I_{K}}$ ${\displaystyle C_{K}}$

Be a global body. If has characteristic , then If has characteristic , then where is a closed subset of which is isomorphic to . The following also applies: ${\displaystyle K}$ ${\displaystyle K}$ ${\displaystyle p>1}$ ${\displaystyle I_{K}\cong \mathbb {A} _{K}^{1}\times \mathbb {Z} .}$ ${\displaystyle K}$ ${\displaystyle 0}$ ${\displaystyle I_{K}\cong M\times \mathbb {A} _{K}^{1},}$ ${\displaystyle M}$ ${\displaystyle I_{K}}$ ${\displaystyle \mathbb {R} _{>0}}$

${\displaystyle C_{K}=I_{K}/K^{\times }\cong \mathbb {A} _{K}^{1}/K^{\times }\times N,}$

where if or if is. ${\displaystyle N\cong \mathbb {Z} ,}$ ${\displaystyle \operatorname {char} (K)>0}$ ${\displaystyle N\cong \mathbb {R} _{>0},}$ ${\displaystyle \operatorname {char} (K)=0}$

Proof: Let the characteristic of be equal For every place of, it holds that the characteristic of is the same , so that for each in the subgroup of is which is generated by. Consequently, this also applies to any where This is equivalent to the fact that the image of homomorphism is a discrete subgroup of which lies in. Since this is not trivial, i.e. H. is, it is generated by one , for a choice such that Then the direct product of and the subgroup generated by is discrete and is therefore isomorphic . ${\displaystyle K}$ ${\displaystyle p>1.}$ ${\displaystyle v}$ ${\displaystyle K}$ ${\displaystyle K_{v}}$ ${\displaystyle p}$ ${\displaystyle |x|_{v}}$ ${\displaystyle x\in K_{v}^{\times }}$ ${\displaystyle \mathbb {R} _{>0}}$ ${\displaystyle p}$ ${\displaystyle |z|,}$ ${\displaystyle z\in I_{K}.}$ ${\displaystyle z\mapsto |z|}$ ${\displaystyle \mathbb {R} _{>0}}$ ${\displaystyle p^{\mathbb {Z} }}$ ${\displaystyle {\text{Bild}}(|\cdot |)\neq \{1\}}$ ${\displaystyle Q=p^{N}}$ ${\displaystyle N\in \mathbb {N} .}$ ${\displaystyle z_{1}\in I_{K},}$ ${\displaystyle |z_{1}|=Q.}$ ${\displaystyle I_{K}}$ ${\displaystyle \mathbb {A} _{K}^{1}}$ ${\displaystyle z_{1}}$ ${\displaystyle \mathbb {Z} }$

If the characteristic of is the same then write for the ideal for which is valid at the finite places of and is valid at all infinite places of . Here then the mapping is an isomorphism of into a closed subgroup of and the following applies: The isomorphism is given by multiplication: ${\displaystyle K}$ ${\displaystyle 0,}$ ${\displaystyle z(\lambda )}$ ${\displaystyle (z_{v})_{v},}$ ${\displaystyle z_{v}=1}$ ${\displaystyle K}$ ${\displaystyle z_{w}=\lambda }$ ${\displaystyle K}$ ${\displaystyle \lambda \in \mathbb {R} _{>0}.}$ ${\displaystyle \lambda \mapsto z(\lambda )}$ ${\displaystyle \mathbb {R} _{>0}}$ ${\displaystyle M}$ ${\displaystyle I_{K}}$ ${\displaystyle I_{K}\cong M\times \mathbb {A} _{K}^{1}.}$

${\displaystyle {\begin{aligned}\phi :M\times \mathbb {A} _{K}^{1}&\rightarrow I_{K},\\((\alpha _{v})_{v},(\beta _{v})_{v})&\mapsto (\alpha _{v}\beta _{v})_{v}\end{aligned}}}$

Obvious is a homomorphism. On injectivity: Let Da for follow for Furthermore there exists such that for From this it follows that for Da additionally is, it follows that is, where the number of infinite places of is. It follows and with it injectivity. Let the surjectivity be given. We define and further we define for and for Define. It is now true that Surjectivity follows. ${\displaystyle \phi }$ ${\displaystyle (\alpha _{v}\beta _{v})_{v}=1.}$ ${\displaystyle \alpha _{v}=1}$ ${\displaystyle v<\infty ,}$ ${\displaystyle \beta _{v}=1}$ ${\displaystyle v<\infty .}$ ${\displaystyle \lambda \in \mathbb {R} _{>0},}$ ${\displaystyle \alpha _{v}=\lambda }$ ${\displaystyle v\mid \infty .}$ ${\displaystyle \beta _{v}=\lambda ^{-1}}$ ${\displaystyle v\mid \infty .}$ ${\displaystyle \textstyle \prod _{v}|\beta _{v}|_{v}=1}$ ${\displaystyle \lambda ^{n}=1}$ ${\displaystyle n}$ ${\displaystyle K}$ ${\displaystyle \lambda =1}$ ${\displaystyle \gamma =(\gamma _{v})_{v}\in I_{K}}$ ${\displaystyle \lambda :=|\gamma |^{1/n}}$ ${\displaystyle \alpha _{v}=1}$ ${\displaystyle v<\infty }$ ${\displaystyle \alpha _{v}=\lambda }$ ${\displaystyle v\mid \infty .}$ ${\displaystyle \textstyle \beta ={\frac {\gamma }{\alpha }}.}$ ${\displaystyle \textstyle |\beta |={\frac {|\gamma |}{|\alpha |}}={\frac {\lambda ^{n}}{\lambda ^{n}}}=1.}$

The second statement follows with a similar consideration.

Characterization of the Idele Group

Let be an algebraic number field. There is a finite set of so that: ${\displaystyle K}$ ${\displaystyle S}$ ${\displaystyle K,}$

${\displaystyle I_{K}=(I_{K,S}\times \prod _{v\notin S}{\mathcal {O}}_{v}^{\times })K^{\times }=(\prod _{v\in S}K_{v}^{\times }\times \prod _{v\notin S}{\mathcal {O}}_{v}^{\times })K^{\times }.}$

Proof: We use as a prerequisite that the class number is finite. Be ideals that represent the classes in . These are composed of a finite number of prime ideals . Let now be a finite set of primes that contains the digits belonging to this prime ideal and the infinite digits. It should be noted that we take advantage of the one-to-one identification between prime positions and positions in the body. Then the assertion from the sentence is fulfilled . To see this, we use the following isomorphism ${\displaystyle {\mathfrak {a}}_{1},\dotsc ,{\mathfrak {a}}_{h}}$ ${\displaystyle h}$ ${\displaystyle Cl_{K}}$ ${\displaystyle {\mathfrak {p}}_{1},\dotsc ,{\mathfrak {p}}_{n}}$ ${\displaystyle S}$ ${\displaystyle S}$

${\displaystyle I_{K}/(\prod _{v\nmid \infty }{\mathcal {O}}_{v}^{\times }\times \prod _{v\mid \infty }K_{v}^{\times })\cong J_{K},}$

which is induced by the image . ${\displaystyle \textstyle (\alpha _{v})_{v}\mapsto \prod _{v<\infty }{\mathfrak {p}}_{v}^{v(\alpha _{v})}}$

In the following we show the assertion of the theorem only in the finite places, since it is clear in the infinite places.

The inclusion " " is clear. ${\displaystyle \supset }$

Let the assigned ideal belong to a class , i. H. with a main ideal The ideal is mapped onto the ideal under our illustration . This means that since the prime ideals occurring in are in, is for all (here again prime ideals and places are identified with one another), i. H. for all Therefore thus ${\displaystyle \alpha \in I_{K,fin},}$ ${\displaystyle \textstyle (\alpha )=\prod _{v\nmid \infty }{\mathfrak {p}}_{v}^{v(\alpha _{v})}}$ ${\displaystyle {\mathfrak {a}}_{i}K^{\times }}$ ${\displaystyle (\alpha )={\mathfrak {a}}_{i}(a)}$ ${\displaystyle (a).}$ ${\displaystyle \alpha '=\alpha a^{-1}}$ ${\displaystyle I_{K,fin}\rightarrow J_{K}}$ ${\displaystyle {\mathfrak {a}}_{i}}$ ${\displaystyle \textstyle {\mathfrak {a}}_{i}=\prod _{v\nmid \infty }{\mathfrak {p}}_{v}^{v(\alpha '_{v})}.}$ ${\displaystyle {\mathfrak {a}}_{i}}$ ${\displaystyle S}$ ${\displaystyle v(\alpha '_{v})=0}$ ${\displaystyle v\notin S}$ ${\displaystyle \alpha '_{v}\in {\mathcal {O}}_{v}^{\times }}$ ${\displaystyle v\notin S.}$ ${\displaystyle \alpha '=\alpha a^{-1}\in I_{K,S},}$ ${\displaystyle \alpha \in I_{K,S}K^{\times }.}$

In Weil (1967) , p. 77 the above theorem is shown for an arbitrary global body . ${\displaystyle K}$

literature

• John Cassels , Albrecht Fröhlich : Algebraic number theory: proceedings of an instructional conference, organized by the London Mathematical Society, (a NATO Advanced Study Institute) . Academic Press, London 1987, ISBN 0-12-163251-2 .
• Jürgen Neukirch : Algebraic number theory. unchangeable Reprint of the 1st edition. Springer, Berlin 2007, ISBN 978-3-540-37547-0 .
• André Weil : Basic number theory . Springer, Berlin / Heidelberg / New York 1967, ISBN 978-3-662-00048-9 .
• Anton Deitmar: Automorphic forms . Springer, Berlin / Heidelberg et al. 2010, ISBN 978-3-642-12389-4 .
• Serge Lang : Algebraic number theory, Graduate Texts in Mathematics 110. 2nd edition. Springer-Verlag, New York 1994, ISBN 0-387-94225-4 .