aa r X i v : . [ m a t h . R A ] J un The Jacobian algebras
V. V. Bavula
Abstract
Let P n := K [ x , . . . , x n ] be a polynomial algebra over a field K of charac-teristic zero. The Jacobian algebra A n is the subalgebra of End K ( P n ) generatedby the Weyl algebra A n := D ( P n ) = K h x , . . . , x n , ∂ , . . . , ∂ n i and the elements( ∂ x ) − , . . . , ( ∂ n x n ) − ∈ End K ( P n ). The algebra A n appears naturally in studyof the group of automorphisms of P n . The algebra A n is large since it contains aring M ∞ ( K ) := lim −→ M d ( K ) of infinite dimensional matrices and all (formal) integro-differential operators as all R i = x i ( ∂ i x i ) − ∈ A n . Surprisingly, the algebras A n and A n have little in common: the algebra A n is neither left nor right Noetherian(even contains infinite direct sums of nonzero left and right ideals); not simple; not adomain; contains nilpotent elements; local; prime; central; self-dual; GK ( A n ) = 3 n ;cl . K . dim( A n ) = n ; has only finitely many, say s n , ideals (2 − n + P ni =1 ni ) ≤ s n ≤ n )which are found explicitly (they commute, IJ = J I ; each of them is an idempotentideal, I = I , and GK ( A n /I ) = 3 n if I = A n ). Spec( A n ) is found, it contains exactly2 n elements, each nonzero prime ideal is a unique sum of primes of height 1 (and anysuch a sum is a prime ideal). Each nonzero ideal is a unique product and a unique intersection of incomparable primes; moreover in both presentations the primes arethe same, they are the minimal primes over the ideal. The group of units A ∗ n of A n is huge ( A ∗ n ⊇ K ∗ × (( Z n ) ( Z ) ⋉ GL ∞ ( K ))). A n has only one faithful simple module,namely P n . Key Words: the Jacobian algebras, prime ideal, prime spectrum, unique factor-ization of ideal, minimal primes, group of units, commutant, integro-differential op-erators.Mathematics subject classification 2000: 16D25, 16S99, 16U60, 16U70, 16S60.
Contents
1. Introduction.2. The Jacobian algebras and localizations of the Weyl algebras.3. Unique factorization of ideals of A n and Spec( A n ).4. The group of units A ∗ n of A n . Introduction
Throughout, ring means an associative ring with 1. Let K be a commutative Q -algebra, K ∗ be its group of units, P n := K [ x , . . . , x n ] be a polynomial algebra over K ; ∂ := ∂∂x , . . . , ∂ n := ∂∂x n be the partial derivatives ( K -linear derivations) of P n . General properties of the Jacobian algebras . The
Jacobian algebra A n is thesubalgebra of End K ( P n ) generated by the Weyl algebra A n := K h x , . . . , x n , ∂ , . . . , ∂ n i and the elements H − , . . . , H − n ∈ End K ( P n ) where H := ∂ x , . . . , H n := ∂ n x n . Clearly, A n = A (1) ⊗ · · · ⊗ A ( n ) ≃ A ⊗ n where A ( i ) := K h x i , ∂ i , H − i i ≃ A . The algebra A n contains all the integrations R i : P n → P n , p R p dx i , since R i = x i H − i . In particular, thealgebra A n contains all (formal) integro-differential operators with polynomial coefficients.This fact explains ( i ) significance of the algebras A n for Algebraic Geometry and the theoryof integro-differential operators; ( ii ) why the algebras A n and A n have different properties,and ( iii ) why the group A ∗ n of units of the algebra A n is huge (there are many invertibleintegro-differential operators).Till the end of this section, K is a field of characteristic zero. When n = 1 the group A ∗ is found explicitly, A ∗ ≃ K ∗ × ( Z ( Z ) ⋉ GL ∞ ( K )) (Theorem 4.2) as well as an inversionformula u − for u ∈ A ∗ . This gives explicitly polynomial solutions for all invertible integro-differential operators on an affine line: uy = f ⇒ y = u − f where f ∈ K [ x ] and y is anunknown. For n ≥
2, a description of the group A ∗ n is given (Theorem 4.4), it looks like itis a challenging problem to find an inversion formula for u ∈ A ∗ n (one should go far beyondthe Dieudonn´e determinant). Though, a criterion of invertibility is found (Theorem 4.5).Moreover, the group A ∗ n contains the subgroup K ∗ × (( Z n ) ( Z ) ⋉ GL ∞ ( K )) elements of whichare called minimal integro-differential operators . For each such an operator u one can writedown an inversion formula u − in the same manner as in the case n = 1, and, therefore,one obtains explicitly polynomial solutions for all minimal integro-differential equations uy = f where f ∈ P n .The Weyl algebra A n = A n ( K ) is a simple, Noetherian domain of Gelfand-Kirillovdimension GK ( A n ) = 2 n . The Jacobian algebra A n is neither left nor right Noetherian, itcontains infinite direct sums of nonzero left and right ideals. This means that the concept ofthe left (and right) Krull dimension makes no sense for A n but the classical Krull dimensionof A n is n (Corollary 3.7). The algebra A n is a central, prime algebra of Gelfand-Kirillovdimension 3 n (Corollary 2.7).The canonical involution θ of the Weyl algebra can be extended to the algebra A n (see(15)). This means that the algebra A n is self-dual ( A n ≃ A opn ), and so its left and rightalgebraic properties are the same. Note that the Fourier transform on the Weyl algebra A n can not be lifted to A n . Many properties of the algebra A n = A ⊗ n are determined byproperties of A . When n = 1 we usually drop the subscript ‘1’ in x , ∂ , H , etc. Thealgebra A contains the only proper ideal F = ⊕ i,j ∈ N KE ij where E ij := ( x i − j ( x j ∂ j x j ∂ j − x j +1 1 ∂ j +1 x j +1 ∂ j +1 ) if i ≥ j, ( ∂x ∂ ) j − i ( x j ∂ j x j ∂ j − x j +1 1 ∂ j +1 x j +1 ∂ j +1 ) if i < j. As a ring without 1, the ring F is canonically isomorphic to the ring M ∞ ( K ) := lim −→ M d ( K ) =2 i,j ∈ N KE ij of infinite-dimensional matrices where E ij are the matrix units ( F → M ∞ ( K ), E ij E ij ). This is a very important fact as we can apply concepts of finite-dimensionallinear algebra (like trace, determinant, etc) to integro-differential operators which is notobvious from the outset. This fact is crucial in finding an inversion formula for elementsof A ∗ .The algebra A n = ⊕ α ∈ Z n A n,α is a Z n -graded algebra where A n,α := ⊗ nk =1 A ,α k ( k ) and,for n = 1, (Theorem 2.3) A ,i = x i D if i ≥ , D if i = 0 , D ∂ − i if i ≤ − , where D := L ⊕ ( ⊕ i,j ≥ Kx i H − j ∂ i ) is a commutative, non-Noetherian algebra and L = K [ H ± , ( H + 1) − , ( H + 2) − , . . . ]. This gives a ‘compact’ K -basis for the algebra A (and A n ). This basis ‘behaves badly’ under multiplication. A more conceptual (‘multiplicativelyfriendly’) basis is given in Theorem 2.5. • (Corollary 2.7.(10)) P n is the only faithful, simple A n -module. Spec ( A n ) . 0 is a prime ideal of A n . p := F ⊗ A n − , p := A ⊗ F ⊗ A n − , . . . , p n := A n − ⊗ F, are precisely the prime ideals of height 1 of A n . Let Sub n be the set of all subsets of { , . . . , n } . • (Corollary 3.5) The map
Sub n → Spec( A n ) , I p I := P i ∈ I p i , ∅ 7→ , is a bijection,i.e. any nonzero prime ideal of A n is a unique sum of primes of height 1; | Spec( A n ) | =2 n ; the height of p I is | I | ; and • (Lemma 3.6) p I ⊆ p J iff I ⊆ J . • (Corollary 3.15) a n := p + · · · + p n is the only prime ideal of A n which is completelyprime; a n is the only ideal a of A n such that a = A n and A n / a is a Noetherian (resp.left Noetherian, resp. right Noetherian) ring. Ideals of A n and their unique factorization . The ideal theory of A n is ‘veryarithmetic.’ Let B n be the set of all functions f : { , , . . . , n } → { , } . For each function f ∈ B n , I f := I f (1) ⊗ · · · ⊗ I f ( n ) is the ideal of A n where I := F and I := A . Let C n bethe set of all subsets of B n all distinct elements of which are incomparable (two distinctelements f and g of B n are incomparable if neither f ( i ) ≤ g ( i ) nor f ( i ) ≥ g ( i ) for all i ).For each C ∈ C n , let I C := P f ∈ C I f , the ideal of A n . The next result classifies all the idealsof A n . • (Theorem 3.1) The map C I C := P f ∈ C I f from the set C n to the set of ideals of A n is a bijection where I ∅ := 0 . In particular, there are only finitely many ideals, say s n , of A n . Moreover, − n + P ni =1 ni ) ≤ s n ≤ n (Corollary 3.4).3 Each ideal I of A n is an idempotent ideal, i.e. I = I . • Ideals of A n commute ( IJ = J I ). • (Theorem 3.11) The lattice of ideals of A n is distributive . • (Corollary 2.7.(4,7)) The ideal a n is the largest (hence, the only maximal) ideal of A n distinct from A n , and F ⊗ n is the smallest nonzero ideal of A n . • (Corollary 2.7.(11)) GK ( A n / a ) = 3 n for all ideals a of A n such that a = A n . For each ideal a of A n , Min( a ) denotes the set of minimal primes over a . Two distinctprime ideals p and q are called incomparable if neither p ⊆ q nor p ⊇ q . The algebras A n have beautiful ideal theory as the following unique factorization properties demonstrate. • (Theorem 3.8) 1. Each ideal a of A n such that a = A n is a unique product ofincomparable primes, i.e. if a = q · · · q s = r · · · r t are two such products then s = t and q = r σ (1) , . . . , q s = r σ ( s ) for a permutation σ of { , . . . , n } . Each ideal a of A n such that a = A n is a unique intersection of incomparableprimes, i.e. if a = q ∩ · · · ∩ q s = r ∩ · · · ∩ r t are two such intersections then s = t and q = r σ (1) , . . . , q s = r σ ( s ) for a permutation σ of { , . . . , n } . For each ideal a of A n such that a = A n , the sets of incomparable primes instatements 1 and 2 are the same, and so a = q · · · q s = q ∩ · · · ∩ q s . The ideals q , . . . , q s in statement 3 are the minimal primes of a , and so a = Q p ∈ Min( a ) p = ∩ p ∈ Min( a ) p . • (Corollary 3.10) a ∩ b = ab for all ideals a and b of A n . The next theorem gives all decompositions of an ideal as a product or intersection of ideals. • (Theorem 3.12) Let a be an ideal of A n , and M be the minimal elements with respectto inclusion of a set of ideals a , . . . , a k of A n . Then a = a · · · a k iff Min( a ) = M .2. a = a ∩ · · · ∩ a k iff Min( a ) = M .This is a rare example of a non-commutative algebra of Krull dimension > The group A ∗ of units of A . For each integer i ≥
1, consider the element of A ∗ :( H − i ) − := ( x H ∂ + 1 − x H ∂ if i = 1 ,x H∂ i x i ∂ + P i − j =0 1 j +1 − i π j + π i − if i ≥ , where π j := x j ∂ j x j ∂ j − x j +1 1 ∂ j +1 x j +1 ∂ j +1 . Consider the following subgroup of A ∗ , H := { Y i ≥ ( H + i ) n i · Y i ≥ ( H − i ) n − i | ( n i ) ∈ Z ( Z ) } ≃ Z ( Z ) Z ( Z ) is the direct sum of Z copies of the group Z , see (32) for detail. Let GL ∞ ( K ) := { u ∈ M ∞ ( K ) | det( u ) = 0 } . The group (1 + F ) ∗ of units of the multiplicative monoid1 + F is equal to (1 + F ) ∗ = (1 + M ∞ ( K )) ∗ = GL ∞ ( K ). Note that (1 + F ) ∗ ⊆ A ∗ . • (Theorem 4.2) 1. A ∗ = K ∗ × ( H ⋉ (1 + F ) ∗ ) , each unit a of A is a unique product a = λα (1+ f ) for some elements λ ∈ K ∗ , α ∈ H , and f ∈ F such that det(1+ f ) = 0 . A ∗ = K ∗ × ( H ⋉ GL ∞ ( K )).3. The centre of the group A ∗ is K ∗ . The commutant A ∗ (2)1 := [ A ∗ , A ∗ ] of the group A ∗ is equal to SL ∞ ( K ) := { v ∈ (1 + F ) ∗ = M ∞ ( K ) | det( v ) = 1 } , and A / [ A ∗ , A ∗ ] ≃ K ∗ × H × K ∗ . All the higher commutants A ∗ ( i )1 := [ A ∗ , A ∗ ( i − ] , i ≥ , are equal to A ∗ (2)1 . The group of units A ∗ n of A n . • (Theorem 4.4) 1. A ∗ n = K ∗ × ( H n ⋉ (1 + a n ) ∗ ) where H n := Q ni =1 H ( i ) and a n := p + · · · + p n .2. The centre of the group A ∗ n is K ∗ . Theorem 4.5 is a criterion of when an element of the monoid 1 + a n belongs to its group(1 + a n ) ∗ of units. Question. Is the global dimension of A n equal to n (or ∞ )? In this section, two K -bases for the Jacobian algebras are found (Theorems 2.3 and 2.5),and several properties of the algebras A n are proved: A n is a central, prime, self-dual,non-Noetherian algebra.We start by recalling some properties of generalized Weyl algebras. Some of thesealgebras are factor algebras of the Jacobian algebras. Generalized Weyl Algebras . Let D be a ring, σ = ( σ , ..., σ n ) be an n -tuple ofcommuting automorphisms of D , and a = ( a , ..., a n ) be an n -tuple of (non-zero) elementsof the centre Z ( D ) of D such that σ i ( a j ) = a j for all i = j .The generalized Weyl algebra A = D ( σ, a ) (briefly GWA) of degree n with the base ring D is a ring generated by D and 2 n indeterminates x , ..., x n , y , ..., y n subject to thedefining relations [3], [4]: y i x i = a i , x i y i = σ i ( a i ) ,x i α = σ i ( α ) x i , y i α = σ − i ( α ) y i , α ∈ D, [ x i , x j ] = [ y i , y j ] = [ x i , y j ] = 0 , i = j, x, y ] = xy − yx . We say that a and σ are the sets of defining elements andautomorphisms of A respectively. The GWAs are also known as hyperbolic rings , see thebook of Rosenberg [20]. For a vector k = ( k , ..., k n ) ∈ Z n , let v k = v k (1) · · · v k n ( n ) where,for 1 ≤ i ≤ n and m ≥ v m ( i ) = x mi , v − m ( i ) = y mi , v ( i ) = 1. It follows from thedefinition of the GWA that A = ⊕ k ∈ Z n A k is a Z n -graded algebra ( A k A e ⊆ A k + e , for all k, e ∈ Z n ), where A k = Dv k = v k D .The tensor product (over the base field) A ⊗ A ′ of generalized Weyl algebras of degree n and n ′ respectively is a GWA of degree n + n ′ : A ⊗ A ′ = D ⊗ D ′ (( σ, σ ′ ) , ( a, a ′ )) . Let P n be a polynomial algebra K [ H , . . . , H n ] in n indeterminates and let σ = ( σ , ..., σ n )be an n -tuple of commuting automorphisms of P n such that σ i ( H i ) = H i − σ i ( H j ) = H j , for i = j . Let A n = K h x , . . . , x n , ∂ , . . . , ∂ n i be the Weyl algebra. The algebrahomomorphism A n → P n (( σ , ..., σ n ) , ( H , . . . , H n )) , x i x i , ∂ i y i , i = 1 , . . . , n, (1)is an isomorphism. We identify the Weyl algebra A n with the GWA above via this iso-morphism. Note that H i = ∂ i x i = x i ∂ i + 1. Denote by S n the multiplicative submonoid of P n generated by the elements H i + j , i = 1 , . . . , n , and j ∈ Z . It follows from the abovepresentation of the Weyl algebra A n as a GWA that S n is an Ore set in A n , and, using the Z n -grading, that the (two-sided) localization A n := S − n A n of the Weyl algebra A n at S n is the skew Laurent polynomial ring A n = S − n P n [ x ± , . . . , x ± n ; σ , ..., σ n ] (2)with coefficients from S − n P n = K [ H ± , ( H ± − , ( H ± − , . . . , H ± n , ( H n ± − , ( H n ± − , . . . ] , the localization of P n at S n . We identify the Weyl algebra A n with the subalgebra of A n via the monomorphism, A n → A n , x i x i , ∂ i H i x − i , i = 1 , . . . , n. Let k n be the n ’th Weyl skew field , that is the full ring of quotients of the n ’th Weyl algebra A n (it exists by Goldie’s Theorem since A n is a Noetherian domain). Then the algebra A n is a K -subalgebra of k n generated by the elements x i , x − i , H i and H − i , i = 1 , . . . , n since,for all natural j , ( H i ∓ j ) − = x ± ji H − i x ∓ ji , i = 1 , . . . , n. Clearly, A n ≃ A ⊗ · · · ⊗ A ( n times). Definition . A K -algebra R has the endomorphism property over K if, for each simple R -module M , End R ( M ) is algebraic over K .6 heorem 2.1 [5] Let K be a field of characteristic zero.1. The algebra A n is a simple, affine, Noetherian domain.2. The Gelfand-Kirillov dimension GK ( A n ) = 3 n ( = 2 n = GK ( A n )) .3. The (left and right) global dimension gl . dim( A n ) = n .4. The (left and right) Krull dimension K . dim( A n ) = n .5. Let d = gl . dim or d = K . dim . Let R be a Noetherian K -algebra with d( R ) < ∞ such that R [ t ] , the polynomial ring in a central indeterminate, has the endomorphismproperty over K . Then d( A ⊗ R ) = d( R ) + 1 . If, in addition, the field K isalgebraically closed and uncountable, and the algebra R is affine, then d( A n ⊗ R ) =d( R ) + n . GK ( A ) = 3 is due to A. Joseph [12], p. 336; see also [16], Example 4.11, p. 45.It is an experimental fact that many small quantum groups are GWAs. More aboutGWAs and their generalizations the interested reader can find in [1, 2, 6, 7, 8, 11, 13, 14,15, 17, 18, 19, 21]. Projections . The polynomial algebra P n = ⊕ α ∈ N n Kx α is a left A n -module andEnd K ( P n )-module. For each i = 1 , . . . , n and α ∈ N n , H i ( x α ) = ( α i + 1) x α , and so H i is an invertible map with H − i ( x α ) = ( α i + 1) − x α . Let h i := x i ∂ i . Then, in P n , h i ( x α ) = α i x α , and so ker( h i ) = K [ x , . . . , b x i , . . . , x n ].Note that (in End K ( P n ))( H − ∂ ) i ( xH − ) i = 1 H ( H + 1) · · · ( H + i − , i ≥ . (3)For each α ∈ N n , the following element of End K ( P n ) is invertible,( − α, α ) := ∂ α x α = n Y i =1 H i ( H i + 1) · · · ( H i + α i − . (4) Lemma 2.2
Let K be a commutative Q -algebra and α ∈ N n . Then x α ( − α, α ) − ∂ α is theprojection onto the ideal ( x α ) of P n in the decomposition P n = ( ⊕ β : x β ( x α ) Kx β ) ⊕ ( x α ) .Proof . If x β ( x α ) then ∂ α ( x β ) = 0, and so x α ∂ α x α ∂ α ( x β ) = 0. If x β ∈ ( x α ) then x α ∂ α x α ∂ α ( x β ) = x α ∂ α x α ∂ α x α ( x β − α ) = x α ( x β − α ) = x β . (cid:3) Lemma 2.2 is useful in producing various projections onto homogeneous K -submodulesof P n . Let S be a subset of N n and S ′ be its complement. Then P n = P n,S ⊕ P n,S ′ where P n,S := ⊕ α ∈ S Kx α and P n,S ′ := ⊕ α ∈ S ′ Kx α . Then π S := P α ∈ S π α is the projection onto P n,S . Example . Let a, b ∈ N n with a ≤ b , i.e. a ≤ b , . . . , a n ≤ b n ; and C := { α ∈ N n | a ≤ α ≤ b } be the discrete cube in N n and C ′ be its complement. Then π C :=7 α ∈ C π α is the projection onto P n,C in the decomposition P n = P n,C ⊕ P n,C ′ . Notethat π C = Q ni =1 ( x a i i ∂ aii x aii ∂ a i i − x b i i ∂ bii x bii ∂ b i i ), by Lemma 2.2. In more detail, for each i = 1 , . . . , n , x a i i ∂ aii x aii ∂ a i i is the projection onto the ideal x a i i K [ x i ] in the decomposition K [ x i ] = ( ⊕ a i − j =0 Kx ji ) ⊕ x a i i K [ x i ]. Therefore, p i := x a i i ∂ aii x aii ∂ a i i − x b i i ∂ bii x bii ∂ b i i is the projectiononto Kx a i i ⊕ Kx a i +1 i ⊕ · · · ⊕ Kx b i i . Now, it is obvious that the product p · · · p n is equal to π C . The Jacobian algebra A n . Let K be a commutative Q -algebra. Definition . The
Jacobian algebra A n is the subalgebra of End K ( P n ) generated bythe Weyl algebra A n and the elements H − , . . . , H − n .Surprisingly, the Weyl algebras A n and the Jacobian algebras A n have little in common.For example, the algebra A n contains the infinite direct sum K ( N ) of rings K . In particular, A n is not a domain, and we will see that A n is not left or right Noetherian algebra.By the very definition, A n = A (1) ⊗ A (2) ⊗ · · · ⊗ A ( n ) ≃ A ⊗ n , (5)where A ( i ) := K h x i , ∂ i , H − i i and ⊗ := ⊗ K . The algebra A n contains all the integrations R i = x i H − i , 1 ≤ i ≤ n . In the algebra A n , each element ∂ i has a right inverse, R i : ∂ i R i = id P n ; and each element x i has a left inverse, H − i ∂ i : H − i ∂ i x i = id P n . So, the algebra A n contains all necessary operations of Analysis (like integrations and differentiations) todeal with polynomials. The algebra A n contains all integro-differential operators. By (5),properties of the algebra A n is mainly determined by properties of the algebra A .We pointed out already that the multiplicative submonoid S n of K [ H , . . . , H n ] gener-ated by the elements H i + j , 1 ≤ i ≤ n , j ∈ Z , is a (left and right) Ore set of the Weylalgebra A n and S − n A n = A n . Using the Z n -grading of the Weyl algebra A n coming fromits presentation as a generalized Weyl algebra one can easily verify that the multiplicativesubmonoid S n, + of K [ H , . . . , H n ] generated by the elements H i + j , 1 ≤ i ≤ n , j ∈ N ,is not a (left and right) Ore set of the Weyl algebra A n . This also follows from the factthat the algebra A n is a domain but A n is not (if S n, + were a left or right Ore set then A n ⊆ A n , a contradiction).Consider the case n = 1. Let 2 N be the Boolean algebra of all subsets of N and B bethe Boolean subalgebra generated by all the finite subsets of N . So, a subset S of N is anelement of B iff either S is finite or co-finite (that is, its complement is finite). Note thatthe Jacobian algebra A contains all the projections π S , S ∈ B .In order to make formulae more readable, we drop the subscript 1. So, let, for a moment, x := x , ∂ := ∂ , and H := H . Since ( H − ∂ ) i H − x i = ( H − ∂ ) i x i ( H + i ) − = ( H + i ) − , i ≥
1, the algebra A contains the subalgebra L := K [ H, H − , ( H +1) − , . . . , ( H + i ) − , . . . ].For each i ≥ ∂ i x i = H ( H +1) ··· ( H + i − ∈ L . Let D := L + P i,j ≥ Kx i H − j ∂ i and V := ⊕ j ≥ KH − j . The next theorem gives a K -basis for the algebra A . Theorem 2.3
Let K be a commutative Q -algebra. Then the Jacobian algebra A = ⊕ i ∈ Z A ,i is a Z -graded algebra ( A ,i A ,j ⊆ A ,i + j for all i, j ∈ Z ) where A , = D , D = L ⊕ ( ⊕ i,j ≥ Kx i H − j ∂ i ) ; and, for each i ≥ , A ,i = x i D and A , − i = D ∂ i . roof . First, let us prove that the sum in the definition of D is the direct one. Supposethat r := l + xv ∂ + x v ∂ + · · · + x s v s ∂ s = 0 is a nontrivial relation for some elements l ∈ L and v i ∈ V := ⊕ j ≥ KH − j . We seek a contradiction. Since L is a subalgebra of End K ( P n ),the relation r is not of the type r = l . So, we can assume that v s = 0 and the naturalnumber s ≥ r . Let r be a nontrivial relation of the leastdegree. The rational function l ∈ K ( H ) can be written as pq where p and q are co-primepolynomials and the polynomial q is a finite product of the type Q i ≥ ( H + i ) n i . Evaluatingthe relation r at 1: 0 = r (1) = l (1) = p (1) q (1) , we see that the polynomial p is equal to ( H − p ′ for some polynomial p ′ ∈ K [ H ]. Suppose that n = 1, then 0 = H − ∂rx = Hσ − ( p ′ q ) + v H where σ : H H − K -automorphism of the polynomial algebra K [ H ] (and of itsfield of fractions K ( H )). It follows that v = − σ − ( p ′ q ) ∈ V ∩ σ − ( L ) = 0, a contradiction.Therefore, s ≥ K [ H ]-module ( V + K [ H ]) /K [ H ] ≃ V has the K -basis { H − i , i ≥ } . In this basis,the matrix of the K -linear map v ( H + λ ) v (where λ ∈ K ) is an upper triangularinfinite matrix with λ on the diagonal. In particular, the matrices of the maps H + 1 , H +2 , . . . , H + s −
1, are invertible, upper triangular. It follows from this fact that the relation H − ∂rx = Hσ − ( p ′ q )+ v H + xv ( H +1) ∂ + · · · + x i − v i ( H + i − ∂ i − + · · · + x s − v s ( H + s − ∂ s − has degree s − s ≥
2. Then, by induction on s , and the fact that each matrix ofthe map H + i −
1, 2 ≤ i ≤ s , is an upper triangular, invertible matrix with i − = 0 onthe diagonal, we have v = · · · = v s = 0 and Hσ − ( p ′ q ) + v H = 0. Then, v = − σ − ( p ′ q ) ∈ V ∩ σ − ( L ) = 0. This means that the relation r is a trivial one, a contradiction. Thisfinishes the proof of the claim.Let ( G, +) be an additive group (not necessarily commutative) and U = ⊕ α ∈ G U α be a G -graded K -module, i.e. a direct sum of K -modules U α . A K -linear map f : U → U hasdegree β ∈ G if f ( U α ) ⊆ U β + α for all α ∈ G . The set E β of all K -linear maps of degree β is a K -submodule of End K ( U ). Clearly, E β = Q β ∈ G Hom K ( U α , U β + α ). In particular, E = Q α ∈ G End K ( U α ). It follows at once that the sum E := P β ∈ G E β ⊆ End K ( U ) is adirect one, E = ⊕ β ∈ G E β and E β E γ ⊆ E β + γ , β, γ ∈ G. (6)So, E := E ( U ) is a G -graded ring.The K -module K [ x ] = ⊕ i ≥ Kx i is naturally Z -graded (even N -graded). By (6), thesum S := P i ≥ D ∂ i + D + P i ≥ x i D is a direct sum since the maps D ∂ i , D , and x i D i have degree − i , 0, and i respectively. In order to prove that A = ⊕ i ∈ Z A ,i it suffices toshow that A ⊆ S . It follows directly from the inclusions: D x i ⊆ x i D , ∂ i D ⊆ D ∂ i , i ≥ ,x i ∂ j ⊆ D ∂ j − i , x j ∂ i ⊆ x j − i D , j ≥ i, that A = P i,j ≥ x i D ∂ j = P s ∈ Z ( P i − j = s x i D ∂ j ). It remains to show that, for s ≥ P i − j = s x i D ∂ j ⊆ x s D ; and, for s < P i − j = s x i D ∂ j ⊆ D ∂ − s .9onsider the case s = 0. We have to show that P i ≥ x i D ∂ i ⊆ D . By the verydefinition of D , this is equivalent to the inclusions x i L∂ i ⊆ D , i ≥
0; and, by the verydefinition of L , this is equivalent to the inclusions x i ( H + j ) − k ∂ i ∈ D , i, j, k ≥ i ≤ j then x i ( H + j ) − k ∂ i = ( H + j − i ) − k x i ∂ i ∈ L ⊆ D .If i > j then x i ( H + j ) − k ∂ i = x i − j H − k x j ∂ i = x i − j H − k ( H − H − · · · ( H − j ) ∂ i − j ∈ x i − j ( V + K [ H ]) ∂ i − j ⊆ D . This proves the case s = 0.If s ≥ P i − j = s x i D ∂ j = x s ( P k ≥ x k D ∂ k ) ⊆ x s D , by the case s = 0.If s ≤ − P i − j = s x i D ∂ j = ( P k ≥ x k D ∂ k ) ∂ − s ⊆ D ∂ − s , by the case s = 0.Thus, the equality A = ⊕ i ∈ Z A ,i is established. By (6), this is a Z -graded algebra sincethe maps A ,i have degree i . (cid:3) Despite the fact that Theorem 2.3 provides a cute K -basis for the algebra D it isunsuitable for computations: to write down the product of the type x i H − j ∂ i · x k H − l ∂ k literally takes half a page. Later, in Corollary 2.4 a more conceptual K -basis is introduced,and which is more important we interpret elements of D as functions from N to K . Wewill see that the ring D is large and has analytic flavour.The polynomial algebra K [ x ] = ⊕ i ≥ Kx i is naturally a Z -graded algebra. Let E = ⊕ i ∈ Z E i be the algebra from (6) for K [ x ]. The E -module K [ x ] is simple. Note that themap E = { f ∈ End K ( K [ x ]) | f ( x i ) = f i x i , i ≥ , f i ∈ K } → K N , f ( f i ) , is an isomorphism of K -algebras. In particular, E is a commutative algebra, and so D is a commutative algebra since D ⊆ E ( K [ H ] is the faithful A -module). It is obviousthat, for i ≥ E i = x i E and E − i = E ∂ i . The algebra K N is the algebra of all functionsfrom N to K . When we identify the set of monomials M := { x i } i ∈ N and N via x i i thealgebras E and K N are identified. So, each element of E can be seen as a function. Thisis a very nice observation indeed as we can use facts and terminology of Analysis. For afunction ϕ : N → K , the set supp( ϕ ) := { i ∈ N | ϕ ( i ) = 0 } is called the support of ϕ . Theset of functions F with finite support is an ideal of the algebra E . Clearly, F = ⊕ i ∈ N Kπ i where π i := x i ∂ i x i ∂ i − x i +1 ∂ i +1 x i +1 ∂ i +1 : K [ x ] → K [ x ]is the projection onto Kx i (Lemma 2.2), i.e. π i ( x j ) = δ ij x j where δ ij is the Kroneckerdelta.Let π − := 0; then xπ i = π i +1 x, ∂π i = π i − ∂, i ≥ . The concept of support can be extended to an arbitrary element f of the algebra E as supp( f ) := { i ∈ N | f ( x i ) = 0 } . The set F of all maps f ∈ E with finite support isa Z -graded algebra F = ⊕ i ∈ Z F i without 1 where F i = F ∩ E i . For i ≥ F i = x i F = ⊕ j ∈ N Kx i π j = ⊕ j ∈ N Kπ j + i x i = F x i and F − i = F ∂ i = ⊕ j ∈ N Kπ j ∂ i = ⊕ j ∈ N K∂ i π j + i = ∂ i F .Note that F = { f ∈ End K ( K [ x ]) | f ( x i K [ x ]) = 0 for some i ∈ N } . It is obvious that F isan ideal of the algebra E , and so F is also an ideal of A since F ⊆ A ⊆ E .Note that h := x∂ ∈ E and h ( x i ) = ix i , i ≥
0. So, h can be identified with thefunction N → K , i i . Note that H = h + 1. When h runs through 0 , , . . . , H runs10hrough 1 , , . . . . Under the identification E = K N , for i, j ≥ x i H − j ∂ i = ( ( H − H − ··· ( H − i +1)( H − i ) j − if H = i + 1 , i + 2 , . . ., H = 1 , , . . . , i. (7)This means that the element x i H − j ∂ i is a function of the discrete argument H = 1 , , . . . which takes zero value for H = 1 , , . . . , i ; and ( H − H − ··· ( H − i +1)( H − i ) j − for H = i + 1 , i + 2 , . . . .Before the identification this simply means that x i H − j ∂ i ( x k ) = ( ( k +1 − k +1 − ··· ( k +1 − i +1)( k +1 − i ) j − x k if k ≥ i, k = 0 , , . . . , i − . So, the function x i H − j ∂ i is almost a rational function. The case i = j = 1 is rather special,it yields almost a constant function xH − ∂ = ( H = 2 , , . . ., H = 1 . Similarly, for each i = 1 , , . . . , the element ρ i := x i ∂ i x i ∂ i ∈ D is the function ρ i = ( h = i, i + 1 , . . ., h = 0 , , . . . , i − . (8)For each i ≥ π i = ρ i − ρ i +1 ∈ D where ρ := 1, and so F = ⊕ i ≥ Kπ i ⊆ D . Moregenerally, for each i = 1 , , . . . and j ∈ N , let ρ ji := x i H j ∂ i x i ∂ i = 1( H − i ) j ρ i = ( H − i ) j if H = i + 1 , i + 2 , . . ., H = 1 , , . . . , i. (9)Note that all ρ ji ∈ D and ρ i = ρ i . For λ ∈ Z , the element H + λ is invertible in D iff λ = − , − , . . . iff H + λ is invertible in E . For i = 1 , , . . . , ker D ( H − i ) j = ker E ( H − i ) j = Kπ i where ker D ( H − i ) j is the kernel of the map D → D , d ( H − i ) j d . Similarly,ker E ( H − i ) j is defined.For each i ≥
0, let π ′ i := 1 − π i . For natural numbers i, j ≥
1, consider the element H − i ) j π ′ i − of E which is as a linear map defined by the rule1( H − i ) j π ′ i − ( x k ) = ( k +1 − i ) j x k if k = i − , k = i − . As a function, it is almost the rational function H − i ) j but at H = i it takes value 0 ratherthan ∞ as the usual function H − i ) j does. All H − i ) j π ′ i − ∈ D since1( H − i ) j π ′ i − = ( ρ j if i = 1 ,ρ ji + P i − k =0 1( k +1 − i ) j π k if i ≥ . i, j, n, m ≥ H − i ) n π ′ i − · H − i ) m π ′ i − = H − i ) n + m π ′ i − , ( H − i ) m · H − i ) n π ′ i − = H − i ) n − m π ′ i − ;and for j ≥ j = i H − i ) n π ′ i − · H − j ) m π ′ j − ⊆ n X s =1 K H − i ) s π ′ i − + m X t =1 K H − j ) t π ′ j − + Kπ i − + Kπ j − . Clearly, the set L := L ⊕ ⊕ i,j ≥ K H − i ) j π ′ i − (10)is a K -submodule of D . L is not an algebra, though it is an algebra modulo F which isisomorphic to the algebra K [ H ± , ( H ± − , ( H ± − , . . . , ] (see Corollary 2.4 and (11)). Corollary 2.4
Let K be a commutative Q -algebra, ρ ji := x i H j ∂ i x i ∂ i , j ≥ , i ≥ . Then D = L ⊕ F = L ⊕ ( ⊕ i ≥ ,j ≥ Kρ ji ) .Proof . By Theorem 2.3, D = L ⊕ ( ⊕ i,j ≥ Kx i H − j ∂ i ). By (7) and F ⊆ D , we have D ⊆ L + F . By (9) and F ⊆ D , we have the opposite inclusion, and so D = L + F = L ⊕ F since L ∩ F = 0.The K -module M := L + P j ≥ ,i ≥ Kρ ji contains F = ⊕ i ≥ π i since π i = ρ i − ρ i +1 forall i ≥ ρ := 1; and M = L ′ + F where L ′ := L + P j,i ≥ Kρ ji . Consider thefactor module M/F . By (9), ρ ji ≡ H − i ) j π ′ i − mod F , hence L ′ ≡ L mod F . Since D = L ⊕ F and M = L ′ + F , we must have the equality D = M . To finish the proof ofthe second equality of the corollary it suffices to show that L ′ + F = L ⊕ ( ⊕ i,j ≥ Kρ ji ) ⊕ F since then the quality M = L ⊕ ( ⊕ i ≥ ,j ≥ Kρ ji ) follows as F = ⊕ i ≥ Kπ i and π i = ρ i − ρ i +1 .Let l + P i,j ≥ λ ji ρ ji + f = 0 for some l ∈ L , λ ji ∈ K , and f ∈ F . Taking this equalitymodulo F yields l = 0 and λ ji = 0 since ρ ji ≡ H − i ) j π ′ i − mod F . This implies f = 0,and we are done. (cid:3) Note that F is an ideal of D such that F = F and D /F ≃ K [ H ± , ( H ± − , ( H ± − , . . . ] . (11)The equality D = L ⊕ F (Corollary 2.4) means that the set { H i , H + j ) k , H − j ) k π ′ j − , π l | i ∈ Z , l ∈ N , j, k ≥ } is a K -basis for D . Clearly, the set { H i , H + j ) k , H − j ) k π ′ j − | i ∈ Z , j, k ≥ } (12)is a K -basis for L . Similarly, the equality D = L ⊕ ( ⊕ i ≥ ,j ≥ Kρ ji ) means that the set { H j , H + j ) k , ρ ji | j ∈ N , i, k ≥ }
12s a K -basis for D .Clearly, F = ⊕ i,j ≥ KE ij where E ij ( x k ) := δ jk x i , i.e. { E ij } are the ‘elementary matrices’( E ij E kl = δ jk E il ), and E ij = ( x i − j π j if i ≥ j, ( H ∂ ) j − i π j if i < j. (13)For k ∈ N , F ± k = ⊕ i − j = ± k KE ij . Note that E ij = E ik π k E kj for all i, j, k ≥
0. Therefore, F is a simple ring such that F = F , and K [ x ] is a simple faithful F -module. Thering F is neither left nor right Noetherian as the next arguments show: for each natural k ≥
0, let L k := ⊕ i ∈ N , ≤ j ≤ k KE ij and R k := ⊕ ≤ i ≤ k,j ∈ N KE ij then L ⊂ L ⊂ · · · and R ⊂ R ⊂ · · · are strictly ascending sequences of left and right F -modules respectively.This is the main reason why the Jacobian algebra A is also neither left nor right Noetherian(Theorem 2.5.(3)). The ring F is neither left nor right Artinian: for each natural k ≥ L ′ k := ⊕ i,j ∈ N KE i,j k and R k := ⊕ i,j ∈ N KE i k ,j then L ′ ⊃ L ′ ⊃ · · · and R ′ ⊃ R ′ ⊃ · · · are strictly descending sequences of left and right F -modules respectively. Theorem 2.5
Let K be a commutative Q -algebra. Then1. A = ⊕ i ≥ L ∂ i ⊕ L ⊕ ( ⊕ i ≥ x i L ) ⊕ F .2. The set { H i ∂ l , H + j ) k ∂ l , H − j ) k π ′ j − ∂ l , x m H i , x m H + j ) k , x m H − j ) k π ′ j − , E st | i ∈ Z ; j, k, l ≥ m, s, t ∈ N } is a K -basis for A .3. The algebra A is neither a left nor right Noetherian algebra.4. F is an ideal of A , F = F , and the factor algebra A /F is canonically isomorphicto the algebra A (the localization of the Weyl algebra A at S , the multiplicativemonoid generated by H + i , i ∈ Z ).Proof . 1. By Corollary 2.4, D = L ⊕ F . For each natural number i ≥ D ∂ i = L ∂ i ⊕ F − i and x i D = x i L ⊕ F i . Using these equalities together with the equalities A = ⊕ i ≥ D ∂ i ⊕ D ⊕ ( ⊕ i ≥ x i D ) (Theorem 2.3) and F = ⊕ i ≥ F ∂ i ⊕ F ⊕ ( ⊕ i ≥ x i F ),one obtains the equality of statement 1 and (12).2. Since, for all i ≥
1, the maps L → L ∂ i , u u∂ i , and L → x i L , u x i u , areisomorphisms of K -modules, statement 2 follows from statement 1.3. For each i ∈ N , the sum I i := ⊕ ij =0 Kπ j is an ideal of E . Since I i ⊆ D ⊆ E ,one has the strictly ascending chain of ideals of D : I ⊂ I ⊂ · · · . The ascending chain A I ⊂ A I ⊂ · · · of left homogeneous ideals of the algebra A is strictly ascending sincethe zero component of the left ideal A I j = ⊕ i ≥ D ∂ i I j ⊕ I j ⊕ ( ⊕ i ≥ x i I j ) is I j (note that D ∂ i I j ⊆ F − i and x i I j ⊆ F i ). Therefore, A is not a left Noetherian algebra.Similarly, the ascending chain I A ⊂ I A ⊂ · · · of right homogeneous ideals ofthe algebra A is strictly ascending since the zero component of the right ideal I j A = ⊕ i ≥ I j ∂ i ⊕ I j ⊕ ( ⊕ i ≥ I j x i D ) is I j , and so A is not a right Noetherian algebra.4. We proved already that F is an ideal of A such that F = F . The K -module K [ x ] isa topological K -module (even a topological K -algebra) with respect to the m -adic topology13etermined by the m -adic filtration { m i } i ≥ on K [ x ] where m := ( x ). Let End K,c ( K [ x ]) bethe algebra of all continuous K -endomorphisms of K [ x ]. Then E ⊆ End
K,c ( K [ x ]). Let G be the algebra of germs of continuous K -endomorphisms of K [ x ] at 0. An element of G is an equivalence class [ f ] of a continuous K -linear map of the type f : m i → K [ x ], andtwo such maps are equivalent, f ∼ f ′ , if they have the same restriction f | m j = f ′ | m j fora sufficiently large j ( m i is a topological K -module with respect to the induced topologycoming from the inclusion m i ⊆ K [ x ]).The kernel of the K -algebra homomorphism E → G , f [ f ], is F . Thus, the ker-nel of the K -algebra homomorphism g : A → G , f [ f ], is also F since F ⊆ A .The image g ( D ) is naturally isomorphic to the algebra S − K [ H ] since g ( H ) = [ H ] and g ( H − j ) k π ′ j − ) = [ H − j ) k ] for all j, k ≥
1. Now, it follows from statement 1 and the de-composition A = ⊕ i ≥ S − K [ H ] ∂ i ⊕ S − K [ H ] ⊕ ( ⊕ i ≥ x i S − K [ H ]) that the image g ( A )is naturally isomorphic to the algebra A . (cid:3) An ideal I of a ring R such that 0 = I = R is called a proper ideal of R . Corollary 2.6
Let K be a field of characteristic zero. Then1. F is the only proper ideal of the algebra A , hence F is a maximal ideal.2. A /F ≃ A is a simple Noetherian domain.3. GK ( A ) = GK ( A ) = 3 .4. A is a prime ring.5. The algebra A is central, i.e. the centre of A is K .Proof . 2. Statement 2 follows from Theorem 2.5.(4) and Theorem 2.1.(1).1. By statement 2, F is a maximal ideal of A . Let I be a proper ideal of A . Wehave to show that I = F . Let a be a nonzero element of I . Then 0 = F aF ⊆ F ∩ I , andso F aF = F since F is a simple algebra (i.e. a simple F -bimodule). Now, F ⊆ I implies F = I by the maximality of F . So, F is the only proper ideal of the algebra A .3. Since A /F ≃ A , we have GK ( A ) ≥ GK ( A ) = 3 (Theorem 2.1.(2)). Since A = ⊕ i ≥ L ∂ i ⊕ L ⊕ ( ⊕ i ≥ x i L ) ⊕ F and A = ⊕ i ≥ S − K [ H ] ∂ i ⊕ S − K [ H ] ⊕ ( ⊕ i ≥ x i S − K [ H ]),the reverse inequality GK ( A ) ≤ A ) ≤ F do not contribute to the growth of degree 3).4. F is the only proper ideal of A , and so F = F , hence A is a prime ring.5. The field K belongs to the centre of A . Let z be a central element of A . We haveto show that z ∈ K . The algebra A /F ≃ A is central, hence z = λ + f for some λ ∈ K and f ∈ F . Then z − λ = f belongs to the centre of F which is obviously equal to zero.Hence z = λ ∈ K , as required. (cid:3) k ≥ x H − j ) k π ′ j − = 1( H − − j ) k π ′ j x, j ≥ ,∂ H − j ) k π ′ j − = 1( H + 1 − j ) k π ′ j − ∂, j ≥ ,∂ H − k π ′ = 1 H k ∂. By (5), the algebra A n is the tensor product of the Z -graded algebras A ( i ) = ⊕ j ∈ Z A ,j ( i ).Therefore, the algebra A n is a Z n -graded algebra, A n = ⊕ α ∈ Z n A n,α , A n,α := ⊗ ni =1 A ,α i ( i ) . The Z n -grading on A n is the tensor product of the Z -gradings of the tensor multiples, andan element a of A n belongs to A n,α iff a ( x β ) ∈ Kx α + β for all β ∈ N n . Let D n := A n, = D (1) ⊗ D (2) ⊗ · · · ⊗ D ( n ) . The polynomial algebra P n = ⊕ α ∈ N n P n,α is an N n -graded, hence a Z n -graded algebra. Let E = E ( P n ) = ⊕ α ∈ Z n E α be the Z n -graded algebra as in (6). The map E = { f ∈ End K ( P n ) | f ( x α ) = f α x α , f α ∈ K, α ∈ N n } → K N n , f ( f α ) , is a K -algebra isomorphism. Each element α = P ni =1 α i e i ∈ Z n = ⊕ ni =1 Z e i is a uniquedifference α = α + − α − where α + = P α i ≥ α i e i and α − = − P α i ≤ α i e i . For each α ∈ Z n , E α = x α + E ∂ α − , and so E = ⊕ α ∈ Z n x α + E ∂ α − . (14)For each α ∈ Z n , A n,α = A n ∩ E α = x α + D n ∂ α − where D n = A n, = A n ∩ E . The involution θ on A n . Let K be a commutative Q -algebra. The Weyl algebra A n admits the involution θ : A n → A n , x i ∂ i , ∂ i x i , i = 1 , . . . , n, i.e. it is a K -algebra anti-isomorphism ( θ ( ab ) = θ ( b ) θ ( a )) such that θ = id A n . Theinvolution θ can be uniquely extended to the involution of A n by the rule θ : A n → A n , x i ∂ i , ∂ i x i , θ ( H − i ) = H − i i = 1 , . . . , n. (15)Uniqueness is obvious: θ ( H i ) = θ ( ∂ i x i ) = θ ( x i ) θ ( ∂ i ) = ∂ i x i = H i and so θ ( H − i ) = H − i .To prove existence recall that each right module over a ring R is a left module over theopposite ring R op . The involution θ on A n comes from considering the polynomial algebra P n as the right A n -module by the rule pa := θ ( a ) p for all p ∈ P n and a ∈ A n . Since θ ( H i ) = H i , i = 1 , . . . , n , P n is the faithful right A n -module, and this proves the existenceof the involution θ : A n → A n ( θ is injective since P n is a faithful right A n -module, θ is15bviously surjective). So, the algebra A n is self-dual (i.e. it is isomorphic to its oppositealgebra, θ : A n ≃ A opn ). This means that left and right algebraic properties of the algebra A n are the same.For n = 1, F is the only proper ideal of A , hence θ ( F ) = F . Moreover, θ ( E ij ) = i ! j ! E ji (16)where 0! := 1. In more detail, since θ ( H ) = H and E ii = π i = x i − i,i ) ∂ i − x i +1 1( − i − ,i +1) ∂ i +1 ,we have θ ( E ii ) = E ii . For i > j , E ij = x i − j π j , and so θ ( E ij ) = π j ∂ i − j = i ( i − · · · ( j + 1) E ji = i ! j ! E ji . For i < j , E ij = ( H ∂ ) j − i π j , and so θ ( E ij ) = π j ( x H ) j − i = i +1)( i +2) ··· j E ji = i ! j ! E ji .For n = 1, the ring F = ⊕ i,j ∈ N KE ij is equal to the matrix ring M ∞ ( K ) := ∪ d ≥ M d ( K )where M d ( K ) := ⊕ ≤ i,j ≤ d − KE ij . The ring F = M ∞ ( K ) admits the canonical involu-tion which is the transposition ( · ) t : E ij E ji . Let D ! be the infinite diagonal matrixdiag(0! , , , . . . ). Then, for u ∈ F = M ∞ ( K ), θ ( u ) = D − u t D ! . (17)Note that D ! M ∞ ( K ).For an arbitrary n , F ⊗ n = ⊕ α,β ∈ N n KE αβ = M ∞ ( K ) ⊗ n where E αβ := ⊗ ni =1 E α i β i . By(16), θ ( E αβ ) = α ! β ! E βα , (18) θ ( F ⊗ n ) = F ⊗ n . (19)Let D n, ! := D ⊗ n ! . Then, for u ∈ F ⊗ n , θ ( u ) = D − n, ! u t D n, ! (20)where ( · ) t : M ∞ ( K ) ⊗ n → M ∞ ( K ) ⊗ n , E αβ E βα , is the transposition map.Consider the bilinear, symmetric, non-degenerate form ( · , · ) : P n × P n → K given bythe rule ( x α , x β ) := α ! δ α,β for all α, β ∈ N n . Then, for all p, q ∈ P n and a ∈ A n ,( p, aq ) = ( θ ( a ) p, q ) . (21)The Weyl algebra A n admits, so-called, the Fourier transform , it is the K -algebraautomorphism F : A n → A n , x i ∂ i , ∂ i
7→ − x i , i = 1 , . . . , n . Since F ( H i ) = − ( H i − H i is a unit of A n and H i − cannot extend the Fourier transform to A n . The algebra A n is a prime algebra . Consider the ideals of the algebra A n : p := F ⊗ A n − , p := A ⊗ F ⊗ A n − , . . . , p n := A n − ⊗ F. Then A n / p i ≃ ( A /F ) ⊗ A n − ≃ A ⊗ A n − and ∩ ni =1 p i = F ⊗ n . Let a n := p + · · · + p n .Then A n / a n ≃ ( A /F ) ⊗ n ≃ A ⊗ n = A n . (22)16 orollary 2.7 Let K be a field of characteristic zero. Then1. GK ( A n ) = 3 n .2. GK ( M ) ≥ n for all nonzero finitely generated (left or right) A n -modules M .3. The centre of A n is K .4. F ⊗ n is the smallest nonzero ideal of the algebra A n , ( F ⊗ n ) = F ⊗ n .5. The algebra A n is prime.6. If = a ∈ A n and I is a nonzero ideal of A n then aI = 0 , Ia = 0 , and IaI = 0 .7. The ideal a n is the largest ideal of A n distinct from A n ; a n = a n ; hence a n is the onlymaximal ideal of A n .8. A n F ⊗ n ≃ P ( N n ) n is a faithful, semi-simple, left A n -module; F ⊗ n A n ≃ P ( N n ) n A n is a faithful,semi-simple, right A n -module; A n F ⊗ n A n ≃ P ( N n ) n is a faithful, simple A n -bimodule.9. F ⊗ n is the socle of A n considered as a left A n -module, or a right A n -module, or an A n -bimodule.10. A n P n (resp. ( P n ) A n ) is the only faithful, simple, left (resp. right) A n -module.11. GK ( A n / a ) = 3 n for all ideals a of A n such that a = A n .Proof . 1. On the one hand, GK ( A n ) ≥ GK ( A n / a n ) = GK ( A n ) = 3 n (Theorem2.1.(2)); on the other, GK ( A n ) = GK ( A ⊗ n ) ≤ n GK ( A ) = 3 n . Therefore, GK ( A n ) = 3 n .2. Statement 2 is an easy corollary of the inequality of Bernstein : GK ( N ) ≥ n forall nonzero finitely generated (left or right) A n -modules N . Let M be a nonzero finitelygenerated A n -module and 0 = u ∈ M . Then GK A n ( M ) ≥ GK A n ( A n u ) ≥ GK A n ( A n u ) ≥ n . 3. To prove that the centre Z ( A n ) of A n is K we use induction on n . The case n = 1is Corollary 2.6.(5). Suppose that n > A m is central for all m < n .The kernel of the algebra homomorphism A n → Q ni =1 A n / p i is ∩ ni =1 p i = F ⊗ n . Since allthe algebras A n / p i ≃ A ⊗ A n − are central, we have Z ( Q ni =1 A n / p i ) = Q ni =1 Z ( A n / p i ) = Q ni =1 K where Z ( R ) is the centre of R . If z ∈ Z ( A n ) then z + F ⊗ n ∈ Z ( Q ni =1 A n / p i ) = Q ni =1 K , and so z = λ + f for some λ ∈ K and f ∈ F ⊗ n . Now, f = z − λ ∈ Z ( F ⊗ n ) = 0,i.e. z = λ ∈ K , and so the algebra A n is central.4. Clearly, ( F ⊗ n ) = ( F ) ⊗ n = F ⊗ n . It remains to prove minimality of F ⊗ n . Thisis obvious for n = 1 (Corollary 2.6.(1)). To prove the general case we use induction on n . Suppose that n > n ′ < n . Let I be a nonzero idealof A n . We have to show that F ⊗ n ⊆ I . Choose a nonzero element, say a , of I . Since a ∈ A n = A ⊗ A n − , the element a can be written as a sum a = P si =1 a i ⊗ b i for someelements a i ∈ A and b i ∈ A n − such that the elements a , . . . , a s and b , . . . , b s are K -linearly independent elements of the algebras A and A n − respectively. Choose an element,17ay f of F such that f a = 0. Changing a for f a = 0 (and deleting zero terms of the type f a i ⊗ b i ) one may assume that all a i ∈ F . Note that { E kl } is the K -basis of F . So, theelement a can be written as a finite sum a = E kl ⊗ α + E st ⊗ β + · · · + E pq ⊗ γ for some K -linearly independent elements α, β, . . . , γ of A n − and distinct elements E kl , E st , . . . , E pq .Then b := E kk aE ll = E kl ⊗ α ∈ I . By induction, F ⊗ ( n − ⊆ A n − α A n − , and so I ⊇ A n b A n = A E kl A ⊗ A n − α A n − ⊇ F ⊗ F ⊗ ( n − = F ⊗ n .
5. Let I and J be nonzero ideals of A n . By statement 4, they contain the ideal F ⊗ n . Now, IJ ⊇ ( F ⊗ n ) = F ⊗ n = 0. This means that A n is a prime ring.6. Statement 6 follows directly from statement 5. Suppose that aI = 0 for somenonzero element a of A n and a nonzero ideal I . We seek a contradiction. Then 0 = A n aI = A n a A n I = 0 since A n is a prime algebra, a contradiction. Similarly, Ia = 0 (resp. IaI = 0) implies 0 = I A n a A n = 0 (resp. 0 = I A n aI = 0), a contradiction.7. a n ⊇ a n = ( p + · · · + p n ) ⊇ p + · · · + p n = p + · · · + p n = a n since p = p , . . . , p n = p n ,and so a n = a n .It remains to show that a n is the largest ideal distinct from A n , that is a a n implies A n a A n = A n where a ∈ A n . Let B n be the K -basis for A n that is the tensor product ofthe K -bases from Theorem 2.5.(2). B n is the disjoint union of its two subsets M n := { b ∈ B n | b ∈ a n } and N n := { b ∈ B n | b a n } . So, b ∈ M n iff the product b contains a matrixunit E st ( i ) ∈ F ( i ). Clearly, a n = ⊕ µ ∈N n Kν and A n = a ′ n ⊕ a n where a ′ n := ⊕ µ ∈M n Kµ .Elements of the set A n \ a n are called generic . So, an element of A n is generic iff it has atleast one nonzero µ -coordinate for some µ ∈ M n . We have to show that A n a A n = A n forall generic elements a ∈ A n . This is true when n = 1 (Corollary 2.6.(1)). To prove generalcase we use induction on n . So, let n ≥ n ′ < n . Let a be a generic element of A n . Then a = a ⊗ b + · · · + a s ⊗ b s ∈ A ⊗ A n − where a i are nonzero elements of A such that a is generic; b i are distinct elements of thebasis B n − such that b ∈ N n − . By statement 6, F a F = 0, and so E ij a E kl = 0 for some i, j, k, l ∈ N . Then, for each t = 1 , . . . , s , E ij a t E kl = λ t E il for some λ t ∈ K , necessarily λ = 0. Now, E ij aE kl = E il ⊗ u where u := λ b + · · · + λ s b s is a generic element of A n − since λ = 0 and b ∈ N n − . By induction, A n − u A n − = A n − , and so A n a A n ⊇ A n ( E il ⊗ u ) A n = A E il A ⊗ A n − u A n − = F ⊗ A n − = p . By symmetry, all p i ⊆ A n a A n , and so a n = p + · · · + p n ⊆ A n a A n . a n is the maximalideal of A n that is properly contained in the ideal A n a A n (since a is generic), and so A n a A n = A n , as required.8. Let, for a moment, n = 1. One can easily verify that, for all i, j, k ∈ N , x k E ij = E i + k,j , ∂ k E ij = i ( i − · · · ( i − k + 1) E i − k,j , (23) E ij x k = E i,j − k , E ij ∂ k = ( j + k )( j + k − · · · ( j + 1) E i,j + k , (24)where E st := 0 if either s < t <
0. By (23), for each j ∈ N , the left A -module C j := ⊕ i ∈ N KE ij is isomorphic to the left A -module K [ x ], and so C j is a faithful, simple,18eft A -module. The left A -module F = ⊕ i ∈ N C i ≃ K [ x ] ( N ) (25)is a direct sum of N copies of K [ x ], and so A F is the faithful, semi-simple, left A -module.For an arbitrary n , the left A n -module C i ⊗ · · · ⊗ C i n is isomorphic to P n . Therefore, A n F ⊗ n ≃ ⊕ i ,...,i n ∈ N C i ⊗ · · · ⊗ C i n ≃ P ( N n ) n (26)is a semi-simple, left A n -module which is faithful, by statement 6.Recall that the structure of the right A n -module P n is given by the rule: p ∗ a := θ ( a ) p where p ∈ P n and a ∈ A n . By (19), θ ( F ⊗ n ) = F ⊗ n . Then, by (26), the K -module F ⊗ n ≃ P ( N n ) n has the natural structure of right A n -module, namely, f ∗ a := θ ( a ) f where f ∈ F ⊗ n and a ∈ A n . In order to distinguish this structure of right A n -module from theobvious structure (as a right ideal of A n ) we write F ⊗ nθ ≃ P ( N n ) n . The map θ : F ⊗ n → F ⊗ nθ ≃ P ( N n ) n , f θ ( f ) , (27)is an isomorphism of right A n -modules (since θ ( f a ) = θ ( a ) θ ( f ) = θ ( f ) ∗ a ). Therefore, F ⊗ n A n ≃ F ⊗ nθ ≃ P ( N n ) n is the faithful, semi-simple, right A n -module by the proved left versionof this fact.The A n -bimodule F ⊗ n is simple since the ring F ⊗ n ≃ M ∞ ( K ) ⊗ n ≃ M ∞ ( K ) is simple.The map 1 ⊗ θ : A n ⊗ A opn → A n is an isomorphism of K -algebras such that 1 ⊗ θ ( F ⊗ n ⊗ F ⊗ n ) = F ⊗ n . Using this equality and A n F ⊗ n A n ≃ A n ⊗ A opn F ⊗ n ≃ A n F ⊗ n , we see that the A n -bimodule F ⊗ n is faithful: if a F ⊗ n = 0 for some nonzero ideal a of A n then F ⊗ n ⊆ a ,and so 0 = a F ⊗ n ⊇ F ⊗ n · F ⊗ n = F ⊗ n F ⊗ n F ⊗ n = F ⊗ n = 0, a contradiction.9. Let soc( A n ) be the socle of the module A n A n (resp. A n A n A n ). By statement 8, F ⊗ n ⊆ soc( A n ). Suppose that F ⊗ n = soc( A n ). Then soc( A n ) = F ⊗ n ⊕ M for a nonzeromodule A n M (resp. A n M A n ). On the one hand, F ⊗ n soc( A n ) ⊆ F ⊗ n and so F ⊗ n M = 0. Onthe other hand, F ⊗ n M = 0, by statement 6, a contradiction. Therefore, soc( A n ) = F ⊗ n .The algebra A n admits the involution θ such that θ ( F ⊗ n ) = F ⊗ n . Therefore, soc( A n A n ) = F ⊗ n since soc( A n A n ) = F ⊗ n .10. Let M be a faithful, simple, left (resp. right) A n -module. Then F ⊗ n M = 0(resp. M F ⊗ n = 0). Choose a nonzero element, say m , of M such that F ⊗ n m = 0 (resp. mF ⊗ n = 0). Then M = F ⊗ n m (resp. M = mF ⊗ n ), by simplicity of M . There is theepimorphism F ⊗ n → F ⊗ n m = M , f f m (resp. F ⊗ n → mF ⊗ n = M , f mf ) of left(resp. right) A n -modules. Now, the result follows from statement 8.11. 3 n = GK ( A n ) ≥ GK ( A n / a ) ≥ GK ( A n / a n ) = GK ( A n ) = 3 n , and so GK ( A n / a ) =3 n . (cid:3) A n and Spec( A n ) In this section, all the results on ideals that are mentioned in the Introduction are proved.19et B n be the set of all functions f : { , , . . . , n } → Z := { , } where Z := Z / Z isa field. B n is a commutative ring with respect to addition and multiplication of functions.For f, g ∈ B n , we write f ≥ g iff f ( i ) ≥ g ( i ) for all i = 1 , . . . , n where 1 >
0. Then ( B n , ≥ )is a partially ordered set. For each function f ∈ B n , I f denotes the ideal I f (1) ⊗ · · · ⊗ I f ( n ) of A n which is the tensor product of the ideals I f ( i ) of the tensor components A ( i ) in A n = A (1) ⊗ · · · ⊗ A ( n ) where I := F and I := A . f ≥ g iff I f ⊇ I g . For f, g ∈ B n , I f I g = I f ∩ I g = I fg . By induction on the number of functions one immediately provesthat, for f , . . . , f s ∈ B n , s Y i =1 I f i = ∩ si =1 I f i = I f ··· f s . Let C n be the set of all subsets of B n all distinct elements of which are incomparable (twodistinct elements f and g of B n are incomparable iff f g and g f ). For each C ∈ C n ,let I C := P f ∈ C I f , the ideal of A n . The next result classifies ideals of the algebra A n . Theorem 3.1
Let K be a field of characteristic zero. Then1. The map C I C := P f ∈ C I f from the set C n to the set of ideals of A n is a bijectionwhere I ∅ := 0 . In particular, there are only finitely many ideals of A n .2. Each ideal I of A n is an idempotent ideal, i.e. I = I .3. Ideals of A n commute ( IJ = J I ).Proof . 1. Statement 1 follows from Lemma 3.2.2. The result is obvious for I = 0. So, let I = 0. By statement 1, I = P f ∈ C I f forsome C ∈ C n . Then I = P f ∈ C I f + P f = g I f I g = P f ∈ C I f + P f = g I f I g = P f ∈ C I f = I .3. I f I g = I fg = I gf = I g I f for all f, g ∈ B n . The result is obvious if either I = 0 or J = 0. So, let I = 0 and J = 0. By statement 1, I = I C and J = I D for some C, D ∈ C n .Then IJ = ( P f ∈ C I f )( P g ∈ D I g ) = ( P g ∈ D I g )( P f ∈ C I f ) = J I . (cid:3) Let B n be the K -basis for the algebra A n that is the tensor product of the K -basesfrom Theorem 2.5.(2) (see the proof of Corollary 2.7.(7) for details). For each element b = b ⊗ · · · ⊗ b n of B n , one can attach an element f b of B n by the rule f b ( i ) := ( b i F , b i ∈ F .
Let a = P b ∈ B n λ b b be a nonzero element of A n where { λ b } are the coordinates of a withrespect to the basis B n . One has the well-defined map A n → C n , a Max( a ), whereMax( a ) are the maximal elements (in B n ) of the subset { f b | λ b = 0 } of B n where Max(0) := ∅ . Lemma 3.2
Let K be a field of characteristic zero and = a = P b ∈ B n λ b b ∈ A n . Then A n a A n = P b ∈ Max( a ) I f b = P { b | λ b =0 } I f b . roof . It suffices to prove only the first equality since the second follows from the first:for any b ∈ B n such that λ b = 0 there exists c ∈ Max( a ) such that λ c = 0 and f c ≥ f b , andso I f c ⊇ I f b . Hence, P b ∈ Max( a ) I f b = P { b | λ b =0 } I f b .Fix b ∈ Max( a ). Up to order of the tensor multiples in the tensor product A n = ⊗ ni =1 A ( i ), one may assume that f b ( i ) := ( ≤ i ≤ s, s + 1 ≤ i ≤ n. We have to show that A n a A n ⊇ A s ⊗ F ⊗ ( n − s ) . a = λ b + · · · + λ t b t + λ t +1 b t +1 + · · · + λ r b r where b := b , all λ i = 0, f b = · · · = f b t and f b j = f b for j = t + 1 , . . . , r . By the choiceof b , for each j such that t + 1 ≤ j ≤ r , either f b j < f b or, otherwise, the functions f b j and f b are incomparable. b = c ⊗ f for unique elements c ∈ M s and f = E p ,q ( s +1) · · · E p n − s ,q n − s ( n ) ∈ N n − s where M s and N n − s were defined in the proof of Corollary2.7.(7). Let E pp := E p ,p ( s + 1) · · · E p n − s ,p n − s ( n ) and E qq := E q ,q ( s + 1) · · · E q n − s ,q n − s ( n ).Then E pp f E qq = f . Note that E pp A n − s E qq = KE pq = Kf . Changing the element a forthe element E pp aE qq and deleting zero terms of the type E pp λ ν b ν E qq , one may assume that a = ( P li =1 µ i c i ) ⊗ f where all µ i = 0 and µ i ∈ K ; all c i ∈ B s ; f c = · · · = f c k = 1 forsome k such that 1 ≤ k ≤ l , i.e. c , . . . , c k ∈ M s ; and f c j < f c where k + 1 ≤ j ≤ l .The element c := P li =1 µ i c i a s , hence A s c A s = A s . Now, A n a A n = A n ( c ⊗ f ) A n = A s c A s ⊗ A n − s f A n − s = A s ⊗ F ⊗ ( n − s ) , as required. (cid:3) The next result is a useful criterion of when one ideal contains another.
Corollary 3.3
Let K be a field of characteristic zero and C, C ′ ∈ C n . Then I C ⊆ I C ′ iff C ≤ C ′ (this means that, for each f ∈ C , there exists f ′ ∈ C ′ such that f ≤ f ′ ).Proof . This follows at once from Lemma 3.2 and Theorem 3.1.(1). (cid:3) Corollary 3.4
Let K be a field of characteristic zero and s n be the number of ideals of A n . Then − n + P ni =1 ni ) ≤ s n ≤ n .Proof . Let Sub n be the set of all subsets of { , . . . , n } . Sub n is a partially ordered setwith respect to ‘ ⊆ ’. For each f ∈ B n , the subset supp( f ) := { i | f ( i ) = 1 } of { , . . . , n } iscalled the support of f . The map B n → Sub n , f supp( f ), is an isomorphism of posets.Let SSub n be the set of all subsets of Sub n . An element { X , . . . , X s } of SSub n is called incomparable if for all i = j such that 1 ≤ i, j ≤ s neither X i ⊆ X j nor X i ⊇ X j . An emptyset and one element set are called incomparable by definition. Let Inc n be the subset ofSSub n of all incomparable elements of SSub n . Then the map C n → Inc n , { f , . . . , f s } 7→ { supp( f ) , . . . , supp( f s ) } , (28)is a bijection. For each i = 1 , . . . , n , there are precisely (cid:0) ni (cid:1) subsets of { , . . . , n } thatcontain exactly i elements. Any non-empty collection of these is an incomparable set,hence s n ≥ P ni =1 (2( ni ) −
1) = 2 − n + P ni =1 ni ) where 2 ‘represents’ the zero ideal andthe ideal F ⊗ n which corresponds to an empty set. Clearly, s n = | Inc n | ≤ n . (cid:3) The next corollary classifies all the prime ideals of A n .21 orollary 3.5 Let K be a field of characteristic zero. Then the map Sub n → Spec( A n ) , I p I := X i ∈ I p i , ∅ 7→ , is a bijection, i.e. any nonzero prime ideal of A n is a unique sum of height 1 primes; | Spec( A n ) | = 2 n ; the height of p I is | I | .Proof . 0 is the prime ideal since A n is a prime ring. Let I be a nonzero subset of { , . . . , n } that contains, say s , elements. Then A n / p I ≃ A s ⊗ A n − s . The ring A s is acentral, simple K -algebra, hence the map a
7→ A s ⊗ a is a bijection from the set of ideals of A n − s to the set of ideals of A s ⊗ A n − s (this is an easy consequence of the Density Theorem).Then, A s ⊗ F ⊗ ( n − s ) is the smallest nonzero ideal of A s ⊗ A n − s and it is idempotent, and so A s ⊗ A n − s is a prime ring. This means that p I is a prime ideal. By Theorem 3.1.(1), themap Sub n → Spec( A n ), I p I , is an injection. It remains to prove that it is a surjection.Let p be a prime nonzero ideal of A n . Then p = P f ∈ C I f for some C ∈ C n (Theorem3.1.(1)). If | supp( f ) | = n − f ∈ C then I f = p i where i = i ( f ) is a unique elementof the set { , . . . , n } such that f ( i ) = 0, and so p = P p i .Suppose that | supp( f ) | 6 = n − f ∈ C . Let us show that this case is notpossible since then p would not be a prime ideal. One can choose two distinct functions,say g, h ∈ B n , such that g > f , h > f , and gh = f . Then I f = I gh = I g I h . Let a := I g + c and b := I h + c where c := P f = f ′ ∈ C I f ′ . The ideals a and b strictly contain the ideal p and ab = ( I g + c )( I h + c ) = I g I h + c I h + I g c + c = I gh + c I h + I g c + c ⊆ I f + c = p . This contradicts to the fact that p is a prime ideal, and we are done.It is obvious that | Spec( A n ) | = 2 n . The fact that the height of the ideal p I is | I | followsfrom Lemma 3.6.(1). (cid:3) The next criterion of when a prime ideal contains another prime is used in finding theclassical Krull dimension of the algebra A n (Corollary 3.7). Lemma 3.6
Let K be a field of characteristic zero; p , q ∈ Spec( A n ) ; p = p i + · · · + p i s and q = p j + · · · + p j t be their decompositions as in Corollary 3.5. Then1. p ⊆ q iff { p i , . . . , p i s } ⊆ { p j , . . . , p j t } .2. If p ⊆ q then pq = p .3. The poset (Spec( A n ) , ⊆ ) is an isomorphic to the set Sub n of all subsets of { , . . . n } .Proof . 1. ( ⇒ ) ( q + p ) / p is the ideal of the algebra A n / p ≃ A s ⊗ A n − s . The algebra A s is central and simple. By the Density Theorem, each ideal of the algebra A s ⊗ A n − s is ofthe type A s ⊗ a for some ideal a of A n − s . By Corollary 3.5 (applied to the algebra A n − s ),we have the inclusion { p i , . . . , p i s } ⊆ { p j , . . . , p j t } .( ⇐ ) If { p i , . . . , p i s } ⊆ { p j , . . . , p j t } then p = p i + · · · + p i s ⊆ p j + · · · + p j t = q .2. By statement 1, if p ⊆ q then q = p + r for an ideal r . Then pq = p + pr = p + pr = p .22. Statement 3 follows from Corollary 3.5 and statement 1. (cid:3) For each s = 0 , , . . . , n , there are precisely (cid:0) ns (cid:1) prime ideals of height s , namely, { p i + · · · + p i s | ≤ i < · · · < i s ≤ n } . Corollary 3.7
Let K be a field of characteristic zero. Then the classical Krull dimensionof A n is n .Proof . By Lemma 3.6.(1), 0 ⊂ p ⊂ p + p ⊂ · · · ⊂ p + · · · + p n is a longest chain ofprimes, and so cl . K . dim( A n ) = n . (cid:3) (Spec( A n ) , ⊆ ) is a poset. Two primes p and q are called incomparable if neither p ⊆ q nor p ⊇ q .For each ideal a of A n such that a = A n , let Min( a ) be the set of all minimal primesover a . The set Min( a ) is a non-empty set since the ring A n has only finitely many primes.For each f ∈ B n , the set csupp( f ) := { i | f ( i ) = 0 } is called the co-support of f . Clearly,csupp( f ) = { , . . . , n }\ supp( f ). Theorem 3.8
Let K be a field of characteristic zero. Then1. Each ideal a of A n such that a = A n is a unique product of incomparable primes, i.e. if a = q · · · q s = r · · · r t are two such products then s = t and q = r σ (1) , . . . , q s = r σ ( s ) for a permutation σ of { , . . . , n } .2. Each ideal a of A n such that a = A n is a unique intersection of incomparable primes,i.e. if a = q ∩ · · · ∩ q s = r ∩ · · · ∩ r t are two such intersections then s = t and q = r σ (1) , . . . , q s = r σ ( s ) for a permutation σ of { , . . . , n } .3. For each ideal a of A n such that a = A n , the sets of incomparable primes in statements1 and 2 are the same, a = q · · · q s = q ∩ · · · ∩ q s .4. The ideals q , . . . , q s in statement 3 are the minimal primes of a , and so a = Q p ∈ Min( a ) p = ∩ p ∈ Min( a ) p .Proof . 1. For each ideal a of A n , we have to prove that a is a product of incomparableprimes and that this product is unique. Since the ring A n is prime these two statementsare obvious when a = 0. So, let a = 0. Existence : Let f ∈ B n ; then I f = Q i ∈ csupp( f ) p i . Let b be any ideal of A n . Since b = b ,it follows at once that I f + b = Y i ∈ csupp( f ) ( p i + b ) . (29)By Theorem 3.1.(1), a = I f + · · · + I f s for some f i ∈ B n . Repeating s times (29), we seethat a = Y i ∈ csupp( f ) ,...,i s ∈ csupp( f s ) ( p i + · · · + p i s ) (30)23s the product of primes, by Corollary 3.5. Note that ideals of A n commute; each ideal isan idempotent ideal; and if p ⊆ q is an inclusion of primes then pq = p . Using these threefacts and (30), we see that a is a product of incomparable primes.Uniqueness follows from the next lemma which will be used several times in the proofof this theorem. Lemma 3.9
Let { q , . . . , q s } and { r , . . . , r t } be two sets of incomparable ideals of a ringsuch that each ideal from the first set contains an ideal from the second and each ideal fromthe second set contains an ideal from the first. Then s = t and q = r σ (1) , . . . , q s = r σ ( s ) fora permutation σ of { , . . . , n } .Proof Lemma 3.9 . For each q i , there are ideals r j and r k such that r j ⊆ q i ⊆ r k , hence q i = r j = r k since the ideals r j and r k are incomparable if distinct. This proves that foreach ideal q i there exists a unique ideal, say r σ ( i ) , such that q i = r σ ( i ) . By symmetry, foreach ideal r j there exists a unique ideal, say q τ ( j ) , such that r j = q τ ( j ) . Then, s = t and q = r σ (1) , . . . , q s = r σ ( s ) for the permutation σ of { , . . . , n } . (cid:3) Uniqueness : Let a = q · · · q s = r · · · r t be two products of incomparable primes. Eachideal q i contains an ideal r j , and each ideal r k contains an ideal q l . By Lemma 3.9, s = t and q = r σ (1) , . . . , q s = r σ ( s ) for a permutation σ of { , . . . , n } .2. Uniqueness : Suppose that an ideal a has two presentations a = q ∩ · · · ∩ q s = r ∩ · · · ∩ r t of incomparable primes. The sets { q , . . . , q s } and { r , . . . , r t } of incomparableprimes satisfy the conditions of Lemma 3.9, and so uniqueness follows. Existence : Let I be the set of all the ideals of A n , and I ′ be the set of ideals of A n thatare intersection of incomparable primes. Then I ′ ⊆ I . The map I → I ′ , q · · · q s q ∩ · · · ∩ q s , is a bijection since |I| < ∞ and by uniqueness of presentations q · · · q s (statement 1) and q ∩ · · · ∩ q s (see above) where q , . . . , q s are incomparable primes. Then I = I ′ . Thisproves that each ideal a of A n is an intersection of incomparable primes.3. Let a be an ideal of A n and a = q · · · q s = r ∩ · · · ∩ r t where S := { q , . . . , q s } and T := { r , . . . , r t } are sets of incomparable primes. The sets S and T satisfy the conditionsof Lemma 3.9, and so s = t and q = r σ (1) , . . . , q s = r σ ( s ) for a permutation σ of { , . . . , n } .This means that a = q · · · q s = q ∩ · · · ∩ q s .4. Let a = q · · · q s = q ∩ · · · ∩ q s be as in statement 3 and let Min( a ) = { r , . . . , r t } be the set of minimal primes over a . Then Min( a ) ⊆ S := { q , . . . q s } ( a = q · · · q s ⊆ r i implies q j ⊆ r i for some j , and so q j = r i by the minimality of r i ). Up to order, let r = q , . . . , r t = q t . It remains to show that t = s . Suppose that t < s , we seek acontradiction. This means that each prime q i , i = t + 1 , . . . , s , contains a and is not minimal over a . Hence, q i contains a minimal prime, say q τ ( i ) , a contradiction (the ideal q i and q τ ( i ) are incomparable). (cid:3) Corollary 3.10
Let K be a field of characteristic zero, a and b be ideals of A n distinctfrom A n in statement 1, 2 and 5. Then . a = b iff Min( a ) = Min( b ) .2. Min( a ∩ b ) = Min( ab ) = the set of minimal elements (with respect to inclusion) ofthe set Min( a ) ∪ Min( b ) .3. a ∩ b = ab .4. If a ⊆ b then ab = a .5. a ⊆ b iff Min( a ) Min( b ) (the means that and each q ∈ Min( b ) contains some p ∈ Min( a ) ).Proof . 1. Statement 1 is obvious due to Theorem 3.8.(4).2. Let M be the set of minimal elements of the union Min( a ) ∪ Min( b ). The elementsof M are incomparable, and (by Theorem 3.8.(4)) a ∩ b = ∩ p ∈ Min( a ) ∩ ∩ q ∈ Min( b ) q = ∩ r ∈M r . By Theorem 3.8.(2), Min( a ∩ b ) = M . By Lemma 3.6.(2), ab = Y p ∈ Min( a ) p · Y q ∈ Min( b ) q = Y r ∈M r = a ∩ b .
3. The result is obvious if one the ideals is equal to A n . So, let the ideals are distinctfrom A n . By statement 2, Min( a ∩ b ) = Min( ab ), then, by statement 1, a ∩ b = ab .4. If a ⊆ b then, by statement 3, ab = a ∩ b = a .5. ( ⇒ ) If a ⊆ b then Min( a ) Min( b ) since a = Q p ∈ Min( a ) p ⊆ Q q ∈ Min( b ) q = b .( ⇐ ) Suppose that Min( a ) Min( b ). For each q ∈ Min( b ), let S ( q ) be the set (nec-essarily nonempty) of p ∈ Min( a ) such that p ⊆ q . Then Min( a ) ⊇ S := ∪ q ∈ Min( b ) S ( q )and a = ∩ p ∈ Min( a ) p ⊆ ∩ p ∈ S p ⊆ ∩ q ∈ Min( a ) q = b . (cid:3) Theorem 3.11
Let K be a field of characteristic zero. Then the lattice of ideals of thealgebra A n is distributive, i.e. ( a ∩ b ) c = ac ∩ bc for all ideals a , b , and c .Proof . By Corollary 3.10.(3), ( a ∩ b ) c = a ∩ b ∩ c = ( a ∩ c ) ∩ ( b ∩ c ) = ac ∩ bc . (cid:3) Theorem 3.12
Let K be a field of characteristic zero, a be an ideal of A n , and M be theminimal elements with respect to inclusion of a set of ideals a , . . . , a k of A n . Then1. a = a · · · a k iff Min( a ) = M .2. a = a ∩ · · · ∩ a k iff Min( a ) = M . roof . By Corollary 3.10.(3), it suffices to prove, say, the first statement.( ⇒ ) Suppose that a = a · · · a k then, by Theorem 3.8.(4) and Corollary 3.10.(4), a = k Y i =1 Y q ij ∈ Min( a i ) q ij = Y q ∈M q , and so Min( a ) = M , by Theorem 3.8.(4).( ⇐ ) If Min( a ) = M then, by Corollary 3.10.(4), a = a · · · a k . (cid:3) The involution c . Let K be a field of characteristic zero and I ( A n ) be the set ofall ideals of the algebra A n . Consider the map C n \{∅} → C n \{∅} , C C + 1, wherefor C = { f , . . . , f s } , C + 1 := { f + 1 , . . . , f s + 1 } . The map is well-defined: C ∈ C n iff { supp( f ) , . . . , supp( f s ) } ∈ Inc n iff { csupp( f ) , . . . , csupp( f s ) } ∈ Inc n where csupp( f i ) := { , . . . , n }\ supp( f i ) iff C + 1 ∈ Inc n . Consider the map c : I ( A n ) → I ( A n ) , I C I C +1 , C ∈ C n , where c (0) := 0. Then, for C ∈ C n , c ( P f ∈ C I f ) = P f ∈ C c ( I f ). Note that c ( A n ) = F ⊗ n , c ( p i ) = Q j = i p j , and c ( a n ) = P ni =1 c ( p i ).Let C, C ′ ∈ C n , we write C (cid:22) C ′ if for each f ∈ C there exists f ′ ∈ C ′ such that f ≤ f ′ ,and for each g ′ ∈ C ′ there exists g ∈ C such that g ≤ g ′ . Lemma 3.13
Let K be a field of characteristic zero. Then1. c : I ( A n ) → I ( A n ) is an involution ( c = id ) such that f ≤ g implies c ( I f ) ⊇ c ( I g ) .2. c ( a ) = a iff a = I C for some C = { f , f + 1 , . . . , f s , f s + 1 } .3. If C, C ′ ∈ C n and C (cid:22) C ′ then c ( I C ) ⊇ c ( I C ′ ) .4. c ( a + b ) ⊆ c ( a ) + c ( b ) for all ideals a and b of A n . If a = I C and b = I C ′ for some C, C ′ ∈ C n such that C ∪ C ′ ∈ C n then c ( a + b ) = c ( a ) + c ( b ) .5. c ( ab ) ⊇ c ( a ) + c ( b ) for all nonzero ideals a and b of A n .Proof . 1. c = id since, for all ∅ 6 = C ∈ C n , C + 1 + 1 = C . The rest is obvious.2. c ( I C ) = I C iff C + 1 = C iff C = { f , f + 1 , . . . , f s , f s + 1 } .3. If C (cid:22) C ′ then C ′ + 1 (cid:22) C + 1, and so c ( I C ) ⊇ c ( I C ′ ).4. The second statement is obvious: If C ∪ C ′ ∈ C n then c ( a + b ) = c ( I C ∪ C ′ ) = X f ∈ C ∪ C ′ I f +1 = X f ∈ C I f +1 + X f ∈ C ′ I f ′ +1 = a + b . For arbitrary a and b , let a = I C and b = I D , and a + b = I E for some C, D, E ∈ C n .Then, for each e ∈ E , either e ≥ f for some f ∈ C or e ≥ g for some g ∈ D . Then, either c ( I e ) ⊆ c ( I f ) or c ( I e ) ⊆ c ( I g ). Hence, c ( a + b ) ⊆ c ( a ) + c ( b ).26. Let a = I C , b = I D , and ab = I E for some C, D, E ∈ C n . Then E (cid:22) C and E (cid:22) D .By statement 3, c ( ab ) ⊇ c ( a ) + c ( b ). (cid:3) The involution τ . Let K be a field of characteristic zero. For each p ∈ Spec( A n ),there exists a unique prime ideal τ ( p ) such that a n = p ⊕ τ ( p ). In more detail, if p = p i + · · · + p i s then τ ( p ) = p j + · · · + p j t where { j , . . . , j t } := { , . . . , n }\{ i , . . . , i s } .The map τ : Spec( A n ) → Spec( A n ) is an order reversion involution, i.e. p ⊆ q implies τ ( p ) ⊇ τ ( q ) for p , q ∈ Spec( A n ); and τ = id. In particular, τ is an anti-automorphismof the poset Spec( A n ). τ ( a n ) = 0 and τ ( p i ) = P j = i p j . Let I n be the set of ideals of A n distinct from A n . The map τ can be extended to the map τ : I n → I n , a = ∩ q ∈ Min( a ) τ ( a ) := ∩ q ∈ Min( a ) τ ( q ) . Lemma 3.14
Let K be a field of characteristic zero, and a , b ∈ I n . Then1. τ : I n → I n is the involution ( τ = id ).2. τ ( a ) = a iff τ (Min( a )) = Min( a ) .Proof . 1. The elements of the set Min( a ) are incomparable, then so are the elementsof the set τ (Min( a )), hence τ (Min( a )) = Min( τ ( a )). Then τ = id.2. By Corollary 3.10.(1), τ ( a ) = a iff τ (Min( a )) = Min( a ). (cid:3) A prime ideal of a ring R is called a completely prime if R/ p is a domain. Corollary 3.15
Let K be a field of characteristic zero. Then1. a n is the only completely prime ideal of A n .2. a n is the only ideal a of A n such that the factor ring A n / a is Noetherian (resp. leftNoetherian, resp. right Noetherian).Proof . 1. By Corollary 3.5, any prime ideal p of A n is a unique sum p = p i + · · · + p i s .Then A n / p ≃ A s ⊗ A n − s . The ring A s ⊗ A n − s is a domain iff s = n , that is p = a n .2. The factor ring A n / a n ≃ A n is Noetherian. It remains to show that A n / a is notleft and right Noetherian for all ideals a distinct from a n . By Theorem 3.8.(4), if a = a n then a n Min( a ). Choose p ∈ Min( a ). Then A n / p ≃ A s ⊗ A n − s for some s ≥
1. The ring A s ⊗ A n − s is not left or right Noetherian. The ring A s ⊗ A n − s is a factor ring of A n / a , andthe result follows. (cid:3) A ∗ n of A n In this section, K be a field of characteristic zero. Let A ∗ n , A ∗ n and K ∗ be the groups of unitsof the algebras A n , A n and K respectively. Using the Z n -grading of the skew polynomialalgebra A n (see (2)), it follows that A ∗ n = ( S − n P n ) ∗ = { K ∗ Y i ∈ Z n Y j =1 ( H j + i ) n ij | ( n ij ) ∈ ( Z n ) ( Z ) } ≃ K ∗ × ( Z n ) ( Z ) (31)27here the abelian group ( Z n ) ( Z ) is the direct sum of Z copies of Z n . The group K ∗ × H n . Let, for a moment, n = 1. In this case we usually drop thesubscript 1. For each integer i ≥ λ ∈ K ∗ , the element ( H − i ) λ := H − i + λπ i − isa unit of the algebra D and its inverse is equal to( H − i ) − λ := ( ρ + λ − π if i = 1 ,ρ i + P i − j =0 1 j +1 − i π j + λ − π i − if i ≥ . As a function of the discrete argument H , ( H − i ) − λ coincides with H − i but instead ofhaving pole at H = i it takes the value λ − . Consider the following subgroup of D ∗ , H := { Y i ≥ ( H + i ) n i · Y i ≥ ( H − i ) n − i | ( n i ) ∈ Z ( Z ) } ≃ Z ( Z ) . (32)For an arbitrary n ≥
1, recall that A n = ⊗ ni =1 A ( i ) = A ⊗ n . For each tensor multiple A ( i ) = A , let H ( i ) be the corresponding group H . Their product H n := H (1) · · · H ( n )is a subgroup of D ∗ n and H n ≃ H n ≃ ( Z n ) ( Z ) . The natural inclusion H n ≃ ( H n + a n ) / a n ⊂ A n / a n ≃ A n and (31) yield the isomorphism of groups K ∗ × H n → A ∗ n , λ λ, H s + i H s + i, ( H s − j ) H s − j, (33)where λ ∈ K ∗ , 1 ≤ s ≤ n , i ∈ N and 1 ≤ j ∈ N . K ∗ H n is the subgroup of D ∗ n such that K ∗ H n ≃ K ∗ × H n . The group (1 + F ⊗ n ) ∗ of units of the monoid F ⊗ n . We are going to find thegroup (1 + F ⊗ n ) ∗ of units of the multiplicative (noncommutative) monoid 1 + F ⊗ n . Let,for a moment, n = 1. The ring F = ⊕ i,j ∈ N KE ij is the union M ∞ ( K ) := ∪ d ≥ M d ( K ) =lim −→ M d ( K ) of the matrix algebras M d ( K ) := ⊕ ≤ i,j ≤ d − KE ij , i.e. F = M ∞ ( K ).For each d ≥
1, consider the (usual) determinant det d = det : 1 + M d ( K ) → K , u det( u ). These determinants determine the (global) determinantdet : 1 + M ∞ ( K ) = 1 + F → K, u det( u ) , where det( u ) is the common value of all determinants det d ( u ), d ≫
1. The (global)determinant has usual properties of the determinant. In particular, for all u, v ∈ M ∞ ( K ), det( uv ) = det( u ) · det( v ). It follows from Cramer’s formula that the groupGL ∞ ( K ) := (1 + M ∞ ( K )) ∗ of units of the monoid 1 + M ∞ ( K ) is equal to GL ∞ ( K ) = { u ∈ M ∞ ( K ) | det( u ) = 0 } . (34)Therefore, (1 + F ) ∗ = { u ∈ F | det( u ) = 0 } = GL ∞ ( K ) . (35)The kernel SL ∞ ( K ) := { u ∈ GL ∞ ( K ) | det( u ) = 1 } of the group epimorphism det : GL ∞ ( K ) → K ∗ is a normal subgroup of GL ∞ ( K ).28or any n ≥ F ⊗ n = ⊗ ni =1 F ( i ) = ⊗ ni =1 ( ∪ d i ≥ M d i ( K )) = ∪ d ,...,d n ≥ ⊗ ni =1 M d i ( K )= ∪ d ,...,d n ≥ M d ··· d n ( K ) = M ∞ ( K ) . Consider the determinantdet : 1 + F ⊗ n = 1 + M ∞ ( K ) → K, u det( u ) , as in the case n = 1. Hence,(1 + F ⊗ n ) ∗ = { u ∈ F ⊗ n | det( u ) = 0 } = (1 + M ∞ ( K )) ∗ = GL ∞ ( K ) . (36)For each element u ∈ (1 + F ⊗ n ) ∗ , using Cramer’s formula one can easily find a formula forthe inverse u − , it is Cramer’s formula. Lemma 4.1
Let K be a field of characteristic zero and u ∈ F ⊗ n . The followingstatements are equivalent.1. u ∈ A ∗ n .2. The element u has left inverse in A n ( vu = 1 for some v ∈ A n ).3. The element u has right inverse in A n ( uv = 1 for some v ∈ A n ).4. det( u ) = 0 .Proof . Using the Z n -grading on A n , it is obvious that the first three statements areequivalent to the fourth. (cid:3) Since F ⊗ n is an ideal of A n , the subgroup (1 + F ⊗ n ) ∗ of A ∗ n is a normal subgroup: Forall a ∈ A ∗ n , a (1 + F ⊗ n ) a − = 1 + aF ⊗ n a − ⊆ F ⊗ n , and so a (1 + F ⊗ n ) ∗ a − ⊆ (1 + F ⊗ n ) ∗ . The subgroup K ∗ × ( H n ⋉ (1 + F ⊗ n ) ∗ ) of A ∗ n . Let A ′ n be the subgroup of the group A ∗ n generated by its subgroups K ∗ , H n and (1 + F ⊗ n ) ∗ . Let us prove that A ′ n = K ∗ × ( H n ⋉ (1 + F ⊗ n ) ∗ ) . (37)The subgroup (1 + F ⊗ n ) ∗ of A ∗ n (and of A ′ n ) is normal and the subgroup K ∗ belongs to thecentre of A ∗ n , hence A ′ n = K ∗ H n (1 + F ⊗ n ) ∗ , i.e. each element a of A ′ n is a product a = λαu for some elements λ ∈ K ∗ , α ∈ H n and u ∈ (1 + F ⊗ n ) ∗ . In order to prove (37) it suffices toshow uniqueness of the decomposition a = λαu . Since a + a n = λα + a n ∈ ( A n / a n ) ∗ ≃ A ∗ n ,the uniqueness of λ and α follows from (31) and (33). Then u = ( λα ) − a is unique as well.This finishes the proof of (37). The group A ∗ and its commutants . Theorem 4.2
Let K be a field of characteristic zero. Then . A ∗ = K ∗ × ( H ⋉ (1 + F ) ∗ ) , each unit a of A is a unique product a = λα (1 + f ) forsome elements λ ∈ K ∗ , α ∈ H , and f ∈ F such that det(1 + f ) = 0 .2. A ∗ = K ∗ × ( H ⋉ GL ∞ ( K )) .3. The centre of the group A ∗ is K ∗ .4. The commutant A ∗ (2)1 := [ A ∗ , A ∗ ] of the group A ∗ is equal to SL ∞ ( K ) := { v ∈ (1 + F ) ∗ = M ∞ ( K ) | det( v ) = 1 } , and A / [ A ∗ , A ∗ ] ≃ K ∗ × H × K ∗ .5. All the higher commutants A ∗ ( i )1 := [ A ∗ , A ∗ ( i − ] , i ≥ , are equal to A ∗ (2)1 .Proof . 1. By (37), A ′ = K ∗ × ( H ⋉ (1 + F ) ∗ ) ⊆ A ∗ . It suffices to show the reverseinclusion. By (33), there is the exact sequence of groups1 → (1 + F ) ∗ → A ∗ → ( A /F ) ∗ ≃ A ∗ → A ∗ ⊆ A ′ . The rest of statement 1 follows from(36).2. Statement 2 is equivalent to statement 1 since (1 + F ) ∗ = GL ∞ ( K ), see (35).3. Let Z be the centre of the group A ∗ . Since K ∗ ⊆ Z , we have Z = Z ∩ A ∗ = Z ∩ ( K ∗ H (1 + F ∗ ) = K ∗ ( Z ∩ H (1 + F ) ∗ ) . We have to show that Z ∩ H (1 + F ) ∗ = { } . Let z = αu ∈ Z ∩ H (1 + F ) ∗ where α = α ( H ) ∈ H and u ∈ (1 + F ) ∗ . It remains to show that z = 1. βz = zβ for all β ∈ H iff βu = uβ for all β ∈ H (since H is an abelian group) iff [ β ] u = u [ β ] (the equality of infinitematrices) for all β = β ( H ) ∈ H where [ β ] is the diagonal matrix diag( β (1) , β (2) , . . . ) iff u is a diagonal matrix of (1 + F ) ∗ . The diagonal entries of the matrix u , say u i , i ∈ N , areelements of K ∗ such that u i = 1 for all i ≥ d for some natural number d = d ( u ). For alldistinct i, j ∈ N , 1 + E ij ∈ (1 + F ) ∗ . Now, it follows from the equalities z + α ( i + 1) u i E ij = z (1 + E ij ) = (1 + E ij ) z = z + E ij α ( j + 1) u j that α ( i + 1) u i = α ( j + 1) u j where α ( i + 1) := α ( H ) | H = i +1 (we have used that αE ij = α ( i + 1) E ij and E ij α = E ij α ( j + 1)). For all distinct natural numbers i and j such that i, j > d , we have α ( i + 1) = α ( j + 1). This means that the function α ( H ) ∈ H is a constant,i.e. α = 1. This gives u i = u j for all i, j ∈ N such that i = j , i.e. all u i = 1. Therefore, z = 1, as required.4. The determinant can be extended from the subgroup (1 + F ) ∗ of A ∗ to the wholegroup by the ruledet : A ∗ → K ∗ × H × K ∗ , λ α u λ α det( u ) := ( λ, α, det( u )) , (39)where λ ∈ K ∗ , α ∈ H , and u ∈ (1 + F ) ∗ . It turns out that det is a group epimorphism . Bythe very definition, det is a surjection. It remains to show that det( aa ′ ) = det( a )det( a ′ )for all a := λαu, a ′ := λ ′ α ′ u ′ ∈ A ∗ . This follows from the following equalitydet( α − uα ) = det( u ) , α ∈ H , u ∈ (1 + F ) ∗ . (40)30ndeed, det( aa ′ ) = det( λλ ′ αα ′ · ( α ′ ) − uα ′ u ′ ) = λλ ′ αα ′ · det(( α ′ ) − uα ′ )det( u ′ )= λλ ′ αα ′ · det( u )det( u ′ ) = det( a )det( a ′ ) . The proof of (40):det( α − uα ) = det([ α − ] u [ α ]) (where [ α ] := diag( α (1) , α (2) , . . . ))= det([ α − ] d u d [ α ] d ) for all d ≫
1= det( u d ) = det( u ) for all d ≫ X = P i,j ∈ N x ij E ij , X d := P ≤ i,j ≤ d − x ij E ij is the sub-matrixof X of size d × d .The kernel of the epimorphism det, (39), is SL ∞ ( K ) := { u ∈ (1+ F ) ∗ = GL ∞ ( K ) | det( u ) =1 } . In particular, SL ∞ ( K ) is a normal subgroup of A ∗ such that the factor group A ∗ / SL ∞ ( K ) ≃ K ∗ × H × K ∗ is abelian. Hence, [ A ∗ , A ∗ ] ⊆ SL ∞ ( K ) and there is the short exact sequenceof groups 1 → SL ∞ ( K ) → A ∗ −→ K ∗ × H × K ∗ → . (41)The group SL ∞ ( K ) is generated by the transvections t ij ( λ ) := 1 + λE ij , λ ∈ K ∗ , i, j ∈ N , i = j . Note that [ H, t ij ( λ )] = t ij ( i − jj + 1 λ ) (42)where [ a, b ] := aba − b − is the commutator of two elements of a group. In more detail,[ H, t ij ( λ )] = Ht ij ( λ ) H − t ij ( λ ) − = t ij ( i + 1 j + 1 λ ) t ij ( − λ ) = t ij ( i − jj + 1 λ ) . Since H ∈ H , we have the inclusion SL ∞ ( K ) ⊆ [ A ∗ , A ∗ ], i.e. [ A ∗ , A ∗ ] = SL ∞ ( K ).5. Statement 5 follows from (42) and the fact that the group SL ∞ ( K ) is generated bythe transvections. (cid:3) An inversion formula for u ∈ A ∗ . Let K be a field of characteristic zero. By Theorem4.2.(1), each element u of A ∗ can written as u = λa (1 + f ). The inverse (1 + f ) − can befound using Cramer’s formula for the inverse of matrix. Then u − = λ − (1 + f ) − a − . Let f ∈ K [ x ] be a given polynomial and y ∈ K [ x ] is an unknown. Then the integro-differentialequation uy = f can be solved explicitly: y = u − f .In contrast to differential operators on an affine line, in general, the space of solutionsfor integro-differential operators is infinite-dimensional: Example. E ij y = 0.For an ideal I of A n such that I = A n , let (1 + I ) ∗ be the group of units of themultiplicative monoid 1 + I . Lemma 4.3
Let K be a commutative Q -algebra, I and J be ideals of A n which are distinctfrom A n . Then1. A ∗ n ∩ (1 + I ) = (1 + I ) ∗ . . (1 + I ) ∗ is a normal subgroup of A ∗ n .Proof . 1. The inclusion A ∗ n ∩ (1+ I ) ⊇ (1+ I ) ∗ is obvious. To prove the reverse inclusion,let 1 + a ∈ A ∗ n ∩ (1 + I ) where a ∈ I , and let (1 + a ) − = 1 + b for some b ∈ A n . Theequality 1 = (1 + a )(1 + b ) can be written as b = − a (1 + b ) ∈ I , i.e. 1 + a ∈ (1 + I ) ∗ . Thisproves the reverse inclusion.2. For all a ∈ A ∗ n , a (1 + I ) a − = 1 + aIa − = 1 + I , and so a (1 + I ) ∗ a − = a ( A ∗ n ∩ (1 + I )) a − = a A ∗ n a − ∩ a (1 + I ) a − = A ∗ n ∩ (1 + I ) = (1 + I ) ∗ . Therefore, (1 + I ) ∗ is a normalsubgroup of A ∗ n . (cid:3) Let K be a field of characteristic zero. By (33), the group homomorphism A ∗ n → ( A n / a n ) ∗ ≃ A ∗ n is an epimorphism. By Lemma 4.3.(1), its kernel is A ∗ n ∩ (1+ a n ) = (1+ a n ) ∗ ,and we have the short exact sequence of groups1 → (1 + a n ) ∗ → A ∗ n → A ∗ n → Theorem 4.4
Let K be a field of characteristic zero. Then1. A ∗ n = K ∗ × ( H n ⋉ (1 + a n ) ∗ ) .2. The centre of the group A ∗ n is K ∗ .Proof . 2. Let Z be the centre of the group A ∗ n . Then K ∗ ⊆ Z and, by statement 1, Z = Z ∩ A ∗ n = Z ∩ ( K ∗ H n (1 + a n ) ∗ ) = K ∗ ( Z ∩ H n (1 + a n ) ∗ ) . It remains to show that Z ′ := Z ∩ H n (1 + a n ) ∗ = { } .Let us show first that Z ′ = Z ∩ (1 + a n ) ∗ . Let z = ϕu ∈ Z ′ for some ϕ ∈ H n and u ∈ (1 + a n ) ∗ . It suffices to show that ϕ = 1. Note that, for each element a ∈ a n , thereexists a natural number c = c ( a ) such that aE αβ = E αβ a = 0 for all α, β ∈ N n such that all α i , β i ≥ c . For short, we write α, β ≫
0. So, uE α,β = E αβ u = E αβ for all α, β ≫
0. Notethat E αβ = 0 for all α = β , and so 1 + E αβ ∈ A ∗ n . Now, for all α, β ≫ α = β , z (1 + E αβ ) = (1 + E αβ ) z ⇔ z + ϕ ( α + 1) E αβ = z + E αβ ϕ ( β + 1) ⇔ ϕ ( α + 1) = ϕ ( β + 1) ⇔ ϕ = 1, as required, where ϕ ( α + 1) is the value of the function ϕ = ϕ ( H , . . . , H n ) at H = α + 1 , . . . , H n = α n + 1. This proves the claim.So, it remains to show that Z ′ := Z ∩ (1 + a n ) ∗ = { } . The result is true for n = 1(Theorem 4.2.(3)). So, we may assume that n ≥
2. Consider the descending chain f ⊃· · · ⊃ f i ⊃ · · · ⊃ f n ⊃ f n +1 := 0 of ideals f i := X ≤ j < ··· 32n the proof of the above series of ring isomorphisms without 1 we have used the facts that F = M ∞ ( K ) and F ⊗ i ≃ F . f i / f i +1 ≃ Y ≤ j < ··· Let K be a commutative Q -algebra, a ∈ a n . Then a ∈ (1 + a n ) ∗ iff1. a + f ∈ (1 + a n / f ) ∗ ( ≃ GL ∞ ( A n − ) n ) , and2. a + (1 + a ) c ∈ (1 + a ) f and a + c (1 + a ) ∈ f (1 + a ) where c := s ((1 + a + f ) − − (the value of the section s : a n / f → a n at the element (1 + a + f ) − − ∈ a n / f ). uppose that conditions 1 and 2 hold and a +(1+ a ) c = (1+ a ) r (resp. a + c (1+ a ) = l (1+ a ) )for some r ∈ f (resp. l ∈ f ) then a − = 1 + c − r (resp. a − = 1 + c − l ).Proof . ( ⇒ ) Suppose that (1 + a ) ∈ (1 + a n ) ∗ . Due to the group homomorphism(1 + a n ) ∗ → (1 + a n / f ) ∗ , we have 1 + a a + f ∈ (1 + a n / f ) ∗ , i.e. the first conditionholds. (1 + a n ) ∗ ∋ (1 + a ) − = 1 + b for some element b ∈ a n = im( s ) ⊕ f which can bewritten as b = c + d where c := s ((1 + a + f ) − − ∈ im( s ) and d := b − c ∈ f . Theequalities (1 + a )(1 + c + d ) = 1 and (1 + c + d )(1 + a ) = 1 can be rewritten as follows a + (1 + a ) c = − (1 + a ) d ∈ (1 + a ) f and a + c (1 + a ) = − d (1 + a ) ∈ f (1 + a ), and so thesecond statement holds.( ⇐ ) Suppose that conditions 1 and 2 hold, we have to show that 1 + a ∈ (1 + a n ) ∗ ,i.e. the element 1 + a has a left and a right inverse. Condition 2 can be written as follows a + (1 + a ) c = (1 + a ) r and a + c (1 + a ) = l (1 + a ) for some elements r, l ∈ f . These twoequalities can we rewritten as (1 + a )(1 + c − r ) = 1 and (1 + c − l )(1 + a ) = 1. This meansthat 1 + a ∈ (1 + a n ) ∗ and a − = 1 + c − r = 1 + c − l . (cid:3) An inversion formula for u ∈ K ∗ × ( H n ⋉ (1 + F ⊗ n ) ∗ ). Let K be a field of charac-teristic zero. 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