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In mathematics, the **adele ring** of a global field (also **adelic ring**, **ring of adeles** or **ring of adèles**^{[1]}) is a central object of class field theory, a branch of algebraic number theory. It is the restricted product of all the completions of the global field and is an example of a self-dual topological ring.

An adele derives from a particular kind of **idele**. "Idele" derives from the French "idèle" and was coined by the French mathematician Claude Chevalley. The word stands for 'ideal element' (abbreviated: id.el.). **Adele** (French: "adèle") stands for 'additive idele' (that is, additive ideal element).

The ring of adeles allows one to describe the Artin reciprocity law, which is a generalisation of quadratic reciprocity, and other reciprocity laws over finite fields. In addition, it is a classical theorem from Weil that -bundles on an algebraic curve over a finite field can be described in terms of adeles for a reductive group . Adeles are also connected with the adelic algebraic groups and adelic curves.

The study of geometry of numbers over the ring of adeles of a number field is called **adelic geometry**.

Let be a global field (a finite extension of or the function field of a curve over a finite field). The **adele ring** of is the subring

consisting of the tuples where lies in the subring for all but finitely many places . Here the index ranges over all valuations of the global field , is the completion at that valuation and the corresponding valuation ring.^{[2]}

The ring of adeles solves the technical problem of "doing analysis on the rational numbers ." The classical solution was to pass to the standard metric completion and use analytic techniques there.^{[clarification needed]} But, as was learned later on, there are many more absolute values other than the Euclidean distance, one for each prime number , as was classified by Ostrowski. The Euclidean absolute value, denoted , is only one among many others, , but the ring of adeles makes it possible to comprehend and *use all of the valuations at once*. This has the advantage of enabling analytic techniques while also retaining information about the primes, since their structure is embedded by the restricted infinite product.

The purpose of the adele ring is to look at all completions of at once. The adele ring is defined with the restricted product, rather than the Cartesian product. There are two reasons for this:

- For each element of the valuations are zero for almost all places, i.e., for all places except a finite number. So, the global field can be embedded in the restricted product.
- The restricted product is a locally compact space, while the Cartesian product is not. Therefore, there cannot be any application of harmonic analysis to the Cartesian product. This is because local compactness ensures the existence (and uniqueness) of Haar measure, a crucial tool in analysis on groups in general.

The restricted infinite product is a required technical condition for giving the number field a lattice structure inside of , making it possible to build a theory of Fourier analysis (cf. Harmonic analysis) in the adelic setting. This is analogous to the situation in algebraic number theory where the ring of integers of an algebraic number field embeds

as a lattice. With the power of a new theory of Fourier analysis, Tate was able to prove a special class of L-functions and the Dedekind zeta functions were meromorphic on the complex plane. Another natural reason for why this technical condition holds can be seen by constructing the ring of adeles as a tensor product of rings. If defining the ring of integral adeles as the ring

then the ring of adeles can be equivalently defined as

The restricted product structure becomes transparent after looking at explicit elements in this ring. The image of an element inside of the unrestricted product is the element

The factor lies in whenever is not a prime factor of , which is the case for all but finitely many primes .^{[3]}

The term "idele" (French: *idèle*) is an invention of the French mathematician Claude Chevalley (1909–1984) and stands for "ideal element" (abbreviated: id.el.). The term "adele" (French: *adèle*) stands for additive idele. Thus, an adele is an additive ideal element.

The rationals have a valuation for every prime number , with , and one infinite valuation *∞* with . Thus an element of

is a real number along with a *p*-adic rational for each * of which all but finitely many are **p*-adic integers.

Secondly, take the function field of the projective line over a finite field. Its valuations correspond to points of , i.e. maps over

For instance, there are points of the form . In this case is the completed stalk of the structure sheaf at (i.e. functions on a formal neighbourhood of ) and is its fraction field. Thus

The same holds for any smooth proper curve over a finite field, the restricted product being over all points of .

The group of units in the adele ring is called the **idele group**

- .

The quotient of the ideles by the subgroup is called the **idele class group**

The **integral adeles** are the subring

The Artin reciprocity law says that for a global field ,

where is the maximal abelian algebraic extension of and means the profinite completion of the group.

If * is a smooth proper curve then its Picard group is*^{[4]}

and its divisor group is . Similarly, if is a semisimple algebraic group (e.g. , it also holds for ) then Weil uniformisation says that^{[5]}

Applying this to gives the result on the Picard group.

There is a topology on for which the quotient is compact, allowing one to do harmonic analysis on it. John Tate in his thesis "Fourier analysis in number fields and Hecke Zeta functions"^{[6]} proved results about Dirichlet L-functions using Fourier analysis on the adele ring and the idele group. Therefore, the adele ring and the idele group have been applied to study the Riemann zeta function and more general zeta functions and the L-functions.

If is a smooth proper curve *over the complex numbers*, one can define the adeles of its function field exactly as the finite fields case. John Tate proved^{[7]} that Serre duality on *
*

can be deduced by working with this adele ring . Here *L* is a line bundle on *.
*

Throughout this article, is a global field, meaning it is either a number field (a finite extension of ) or a global function field (a finite extension of for prime and ). By definition a finite extension of a global field is itself a global field.

For a valuation of it can be written for the completion of with respect to If is discrete it can be written for the valuation ring of and for the maximal ideal of If this is a principal ideal denoting the uniformising element by A non-Archimedean valuation is written as or and an Archimedean valuation as Then assume all valuations to be non-trivial.

There is a one-to-one identification of valuations and absolute values. Fix a constant the valuation is assigned the absolute value defined as:

Conversely, the absolute value is assigned the valuation defined as:

A place of is a representative of an equivalence class of valuations (or absolute values) of Places corresponding to non-Archimedean valuations are called *finite*, whereas places corresponding to Archimedean valuations are called *infinite*. Infinite places of a global field form a finite set, which is denoted by

Define and let be its group of units. Then

Let be a finite extension of the global field Let be a place of and a place of If the absolute value restricted to is in the equivalence class of , then lies above which is denoted by and defined as:

(Note that both products are finite.)

If , can be embedded in Therefore, is embedded diagonally in With this embedding is a commutative algebra over with degree

The set of **finite adeles of a global field ** denoted is defined as the restricted product of with respect to the

It is equipped with the restricted product topology, the topology generated by restricted open rectangles, which have the following form:

where is a finite set of (finite) places and are open. With component-wise addition and multiplication is also a ring.

The **adele ring of a global field** is defined as the product of with the product of the completions of at its infinite places. The number of infinite places is finite and the completions are either or In short:

With addition and multiplication defined as component-wise the adele ring is a ring. The elements of the adele ring are called **adeles of ** In the following, it is written as

although this is generally not a restricted product.

**Remark.** Global function fields do not have any infinite places and therefore the finite adele ring equals the adele ring.

**Lemma.**There is a natural embedding of into given by the diagonal map:

**Proof.** If then for almost all This shows the map is well-defined. It is also injective because the embedding of in is injective for all

**Remark.** By identifying with its image under the diagonal map it is regarded as a subring of The elements of are called the **principal adeles** of

**Definition.** Let be a set of places of Define the **set of the -adeles of ** as

Furthermore, if

the result is:

By Ostrowski's theorem the places of are it is possible to identify a prime with the equivalence class of the -adic absolute value and with the equivalence class of the absolute value defined as:

The completion of with respect to the place is with valuation ring For the place the completion is Thus:

Or for short

the difference between restricted and unrestricted product topology can be illustrated using a sequence in :

**Lemma.**Consider the following sequence in :- In the product topology this converges to , but it does not converge at all in the restricted product topology.

**Proof.** In product topology convergence corresponds to the convergence in each coordinate, which is trivial because the sequences become stationary. The sequence doesn't converge in restricted product topology. For each adele and for each restricted open rectangle it has: for and therefore for all As a result for almost all In this consideration, and are finite subsets of the set of all places.

**Definition (profinite integers).** The **profinite integers** are defined as the profinite completion of the rings with the partial order i.e.,

**Lemma.**

**Proof.** This follows from the Chinese Remainder Theorem.

**Lemma.**

**Proof.** Use the universal property of the tensor product. Define a -bilinear function

This is well-defined because for a given with co-prime there are only finitely many primes dividing Let be another -module with a -bilinear map It must be the case that factors through uniquely, i.e., there exists a unique -linear map such that can be defined as follows: for a given there exist and such that for all Define One can show is well-defined, -linear, satisfies and is unique with these properties.

**Corollary.**Define This results in an algebraic isomorphism

**Proof.**

**Lemma.**For a number field

**Remark.** Using where there are summands, give the right side receives the product topology and transport this topology via the isomorphism onto

If be a finite extension, then is a global field. Thus is defined, and can be identified with a subgroup of Map to where for Then is in the subgroup if for and for all lying above the same place of

**Lemma.**If is a finite extension, then both algebraically and topologically.

With the help of this isomorphism, the inclusion is given by

Furthermore, the principal adeles in can be identified with a subgroup of principal adeles in via the map

**Proof.**^{[8]} Let be a basis of over Then for almost all

Furthermore, there are the following isomorphisms:

For the second use the map:

in which is the canonical embedding and The restricted product is taken on both sides with respect to

**Corollary.**As additive groups where the right side has summands.

The set of principal adeles in is identified with the set where the left side has summands and is considered as a subset of

**Lemma.**Suppose is a finite set of places of and define- Equip with the product topology and define addition and multiplication component-wise. Then is a locally compact topological ring.

**Remark.** If is another finite set of places of containing then is an open subring of

Now, an alternative characterisation of the adele ring can be presented. The adele ring is the union of all sets :

Equivalently is the set of all so that for almost all The topology of is induced by the requirement that all be open subrings of Thus, is a locally compact topological ring.

Fix a place of Let be a finite set of places of containing and Define

Then:

Furthermore, define

where runs through all finite sets containing Then:

via the map The entire procedure above holds with a finite subset instead of

By construction of there is a natural embedding: Furthermore, there exists a natural projection

Let be a finite dimensional vector-space over and a basis for over For each place of :

The adele ring of is defined as

This definition is based on the alternative description of the adele ring as a tensor product equipped with the same topology that was defined when giving an alternate definition of adele ring for number fields. Next, is equipped with the restricted product topology. Then and is embedded in naturally via the map

An alternative definition of the topology on can be provided. Consider all linear maps: Using the natural embeddings and extend these linear maps to: The topology on is the coarsest topology for which all these extensions are continuous.

The topology can be defined in a different way. Fixing a basis for over results in an isomorphism Therefore fixing a basis induces an isomorphism The left-hand side is supplied with the product topology and transport this topology with the isomorphism onto the right-hand side. The topology doesn't depend on the choice of the basis, because another basis defines a second isomorphism. By composing both isomorphisms, a linear homeomorphism which transfers the two topologies into each other is obtained. More formally

where the sums have summands. In case of the definition above is consistent with the results about the adele ring of a finite extension

^{[9]}

Let be a finite-dimensional algebra over In particular, is a finite-dimensional vector-space over As a consequence, is defined and Since there is multiplication on and a multiplication on can be defined via:

As a consequence, is an algebra with a unit over Let be a finite subset of containing a basis for over For any finite place , is defined as the -module generated by in For each finite set of places, define

One can show there is a finite set so that is an open subring of if Furthermore is the union of all these subrings and for the definition above is consistent with the definition of the adele ring.

Let be a finite extension. Since and from the Lemma above, can be interpreted as a closed subring of For this embedding, write . Explicitly for all places of above and for any

Let be a tower of global fields. Then:

Furthermore, restricted to the principal adeles is the natural injection

Let be a basis of the field extension Then each can be written as where are unique. The map is continuous. Define depending on via the equations:

Now, define the trace and norm of as:

These are the trace and the determinant of the linear map

They are continuous maps on the adele ring, and they fulfil the usual equations:

Furthermore, for and are identical to the trace and norm of the field extension For a tower of fields the result is:

Moreover, it can be proven that:^{[10]}

**Theorem.**^{[11]}For every set of places is a locally compact topological ring.

**Remark.** The result above also holds for the adele ring of vector-spaces and algebras over

**Theorem.**^{[12]}is discrete and cocompact in In particular, is closed in

**Proof.** Prove the case To show is discrete it is sufficient to show the existence of a neighbourhood of which contains no other rational number. The general case follows via translation. Define

is an open neighbourhood of It is claimed that Let then and for all and therefore Additionally, and therefore Next, to show compactness, define:

Each element in has a representative in that is for each there exists such that Let be arbitrary and be a prime for which Then there exists with and Replace with and let be another prime. Then:

Next, it can be claimed that:

The reverse implication is trivially true. The implication is true, because the two terms of the strong triangle inequality are equal if the absolute values of both integers are different. As a consequence, the (finite) set of primes for which the components of are not in is reduced by 1. With iteration, it can be deduced that there exists such that Now select such that Then The continuous projection is surjective, therefore as the continuous image of a compact set, is compact.

**Corollary.**Let be a finite-dimensional vector-space over Then is discrete and cocompact in

**Theorem.**The following are assumed:- is a divisible group.
^{[13]} - is dense.

**Proof.** The first two equations can be proved in an elementary way.

By definition is divisible if for any and the equation has a solution It is sufficient to show is divisible but this is true since is a field with positive characteristic in each coordinate.

For the last statement note that because the finite number of denominators in the coordinates of the elements of can be reached through an element As a consequence, it is sufficient to show is dense, that is each open subset contains an element of Without loss of generality, it can be assumed that

because is a neighbourhood system of in By Chinese Remainder Theorem there exists such that Since powers of distinct primes are coprime, follows.

**Remark.** is not uniquely divisible. Let and be given. Then

both satisfy the equation and clearly ( is well-defined, because only finitely many primes divide ). In this case, being uniquely divisible is equivalent to being torsion-free, which is not true for since but and

**Remark.** The fourth statement is a special case of the strong approximation theorem.

**Definition.** A function is called simple if where are measurable and for almost all

**Theorem.**^{[14]}Since is a locally compact group with addition, there is an additive Haar measure on This measure can be normalised such that every integrable simple function satisfies:- where for is the measure on such that has unit measure and is the Lebesgue measure. The product is finite, i.e., almost all factors are equal to one.

**Definition.** Define the **idele group of ** as the group of units of the adele ring of that is The elements of the idele group are called the **ideles of **

**Remark.** is equipped with a topology so that it becomes a topological group. The subset topology inherited from is not a suitable candidate since the group of units of a topological ring equipped with subset topology may *not* be a topological group. For example, the inverse map in is not continuous. The sequence

converges to To see this let be neighbourhood of without loss of generality it can be assumed:

Since for all for large enough. However, as was seen above the inverse of this sequence does not converge in

**Lemma.**Let be a topological ring. Define:- Equipped with the topology induced from the product on topology on and is a topological group and the inclusion map is continuous. It is the coarsest topology, emerging from the topology on that makes a topological group.

**Proof.** Since is a topological ring, it is sufficient to show that the inverse map is continuous. Let be open, then is open. It is necessary to show is open or equivalently, that is open. But this is the condition above.

The idele group is equipped with the topology defined in the Lemma making it a topological group.

**Definition.** For a subset of places of set:

**Lemma.**The following identities of topological groups hold:- where the restricted product has the restricted product topology, which is generated by restricted open rectangles of the form
- where is a finite subset of the set of all places and are open sets.

**Proof.** Prove the identity for ; the other two follow similarly. First show the two sets are equal:

In going from line 2 to 3, as well as have to be in meaning for almost all and for almost all Therefore, for almost all

Now, it is possible to show the topology on the left-hand side equals the topology on the right-hand side. Obviously, every open restricted rectangle is open in the topology of the idele group. On the other hand, for a given which is open in the topology of the idele group, meaning is open, so for each there exists an open restricted rectangle, which is a subset of and contains Therefore, is the union of all these restricted open rectangles and therefore is open in the restricted product topology.

**Lemma.**For each set of places, is a locally compact topological group.

**Proof.** The local compactness follows from the description of as a restricted product. It being a topological group follows from the above discussion on the group of units of a topological ring.

A neighbourhood system of is a neighbourhood system of Alternatively, take all sets of the form:

where is a neighbourhood of and for almost all

Since the idele group is a locally compact, there exists a Haar measure on it. This can be normalised, so that

This is the normalisation used for the finite places. In this equation, is the finite idele group, meaning the group of units of the finite adele ring. For the infinite places, use the multiplicative lebesgue measure

**Lemma.**Let be a finite extension. Then:- where the restricted product is with respect to

**Lemma.**There is a canonical embedding of in

**Proof.** Map to with the property for Therefore, can be seen as a subgroup of An element is in this subgroup if and only if his components satisfy the following properties: for and for and for the same place of

^{[15]}

Let be a finite-dimensional algebra over Since is not a topological group with the subset-topology in general, equip with the topology similar to above and call the idele group. The elements of the idele group are called idele of

**Proposition.**Let be a finite subset of containing a basis of over For each finite place of let be the -module generated by in There exists a finite set of places containing such that for all is a compact subring of Furthermore, contains For each is an open subset of and the map is continuous on As a consequence maps homeomorphically on its image in For each the are the elements of mapping in with the function above. Therefore, is an open and compact subgroup of^{[16]}

**Proposition.**Let be a finite set of places. Then- is an open subgroup of where is the union of all
^{[17]}

**Corollary.**In the special case of for each finite set of places- is an open subgroup of Furthermore, is the union of all

The trace and the norm should be transfer from the adele ring to the idele group. It turns out the trace can't be transferred so easily. However, it is possible to transfer the norm from the adele ring to the idele group. Let Then and therefore, it can be said that in injective group homomorphism

Since it is invertible, is invertible too, because Therefore As a consequence, the restriction of the norm-function introduces a continuous function:

**Lemma.**There is natural embedding of into given by the diagonal map:

**Proof.** Since is a subset of for all the embedding is well-defined and injective.

**Corollary.**is a discrete subgroup of

**Defenition.** In analogy to the ideal class group, the elements of in are called **principal ideles of ** The quotient group is called idele class group of This group is related to the ideal class group and is a central object in class field theory.

**Remark.** is closed in therefore is a locally compact topological group and a Hausdorff space.

**Lemma.**^{[18]}Let be a finite extension. The embedding induces an injective map:

**Definition.** For define: Since is an idele this product is finite and therefore well-defined.

**Remark.** The definition can be extended to by allowing infinite products. However, these infinite products vanish and so vanishes on will be used to denote both the function on and

**Theorem.**is a continuous group homomorphism.

**Proof.** Let

where it is used that all products are finite. The map is continuous which can be seen using an argument dealing with sequences. This reduces the problem to whether is continuous on However, this is clear, because of the reverse triangle inequality.

**Definition.** The set of -idele can be defined as:

is a subgroup of Since it is a closed subset of Finally the -topology on equals the subset-topology of on ^{[19]}^{[20]}

**Artin's Product Formula.**for all

**Proof.**^{[21]} Proof of the formula for number fields, the case of global function fields can be proved similarly. Let be a number field and It has to be shown that:

For finite place for which the corresponding prime ideal does not divide , and therefore This is valid for almost all There is:

In going from line 1 to line 2, the identity was used where is a place of and is a place of lying above Going from line 2 to line 3, a property of the norm is used. The norm is in so without loss of generality it can be assumed that Then possesses a unique integer factorisation:

where is for almost all By Ostrowski's theorem all absolute values on are equivalent to the real absolute value or a -adic absolute value. Therefore:

**Lemma.**^{[22]}There exists a constant depending only on such that for every satisfying there exists such that for all

**Corollary.**Let be a place of and let be given for all with the property for almost all Then there exists so that for all

**Proof.** Let be the constant from the lemma. Let be a uniformising element of Define the adele via with minimal, so that for all Then for almost all Define with so that This works, because for almost all By the Lemma there exists so that for all

**Theorem.**is discrete and cocompact in

**Proof.**^{[23]} Since is discrete in it is also discrete in To prove the compactness of let is the constant of the Lemma and suppose satisfying is given. Define:

Clearly is compact. It can be claimed that the natural projection is surjective. Let be arbitrary, then:

and therefore

It follows that

By the Lemma there exists such that for all and therefore proving the surjecti