A Parametric Family of Subalgebras of the Weyl Algebra I. Structure and Automorphisms

An Ore extension over a polynomial algebra $\mathbb{F}[x]$ is either a quantum plane, a quantum Weyl algebra, or an infinite-dimensional unital associative algebra $\mathsf{A}_h$ generated by elements $x,y$, which satisfy $yx-xy = h$, where $h\in \mathbb{F}[x]$. We investigate the family of algebras $\mathsf{A}_h$ as $h$ ranges over all the polynomials in $\mathbb{F}[x]$. When $h \neq 0$, these algebras are subalgebras of the Weyl algebra $\mathsf{A}_1$ and can be viewed as differential operators with polynomial coefficients. We give an exact description of the automorphisms of $\mathsf{A}_h$ over arbitrary fields $\mathbb{F}$ and describe the invariants in $\mathsf{A}_h$ under the automorphisms. We determine the center, normal elements, and height one prime ideals of $\mathsf{A}_h$, localizations and Ore sets for $\mathsf{A}_h$, and the Lie ideal $[\mathsf{A}_h,\mathsf{A}_h]$. We also show that $\mathsf{A}_h$ cannot be realized as a generalized Weyl algebra over $\mathbb{F}[x]$, except when $h \in \mathbb{F}$. In two sequels to this work, we completely describe the derivations and irreducible modules of $\mathsf{A}_h$ over any field.


Introduction
The focus of this paper is on a family of infinite-dimensional unital associative algebras A h parametrized by a polynomial h = h(x) ∈ F[x], where F is an arbitrary field.The algebra A h has generators x, y, which satisfy the defining relation yx = xy + h, or equivalently, [y, x] = h, where [y, x] = yx − xy.The Ore extensions whose underlying ring is F[x] fall into three specific types.They are quantum planes, quantum Weyl algebras, or one of the algebras A h (compare Lemma 2.2 below).Quantum planes and quantum Weyl algebras are examples of generalized Weyl algebras, and as such, have been studied extensively.It is the aim of our work to investigate the family of algebras A h as h ranges over all the polynomials in F[x].The algebras A h are left and right Noetherian domains.As modules over F[x], they are free with basis {y n | n ∈ Z ≥0 }.Each algebra A h with h = 0 can be viewed as a subalgebra of the Weyl algebra A 1 and thus has a representation as differential operators on F[x], where x acts by multiplication and y by h d dx , so that [h d dx , x] = h holds.
There are several widely-studied examples of algebras in this family.The algebra A 0 is the polynomial algebra F[x, y]; A 1 is the Weyl algebra; and A x is the universal enveloping algebra of the two-dimensional non-abelian Lie algebra (there is only one such Lie algebra up to isomorphism).The algebra A x 2 is often referred to as the Jordan plane.It arises in noncommutative algebraic geometry (see for example, [SZ] and [AS]) and exhibits many interesting features such as being Artin-Schelter regular of dimension 2. In a series of articles [S1]- [S3], Shirikov has undertaken an extensive study of the automorphisms, derivations, prime ideals, and modules of the algebra A x 2 .These investigations have been extended by Iyudu in recent work [I] to include results on varieties of finite-dimensional modules of A x 2 over algebraically closed fields of characteristic zero.Cibils, Lauve, and Witherspoon [CLW] have used quotients of the algebra A x 2 and cyclic subgroups of their automorphism groups to construct new examples of finite-dimensional Hopf algebras in prime characteristic which are Nichols algebras.
There are striking similarities in the behavior of the algebras A h as h ranges over the polynomials in F [x].For that reason, we believe that studying them as one family provides much insight into their structure, derivations, automorphisms, and modules.In this paper, we determine the following: • embeddings of A g into A f (Section 3) • localizations and Ore sets for A h (Section 4) • the center of A h (Section 5) • the Lie ideal [A h , A h ] of A h (Section 6) • the normal elements and the prime ideals of A h (Section 7) • the automorphism group A = Aut F (A h ) and its center, and the subalgebra A A h of A-invariants in A h (Section 8) • the relationship of A h to generalized Weyl algebras (Section 9).
In the sequel [BLO1], we determine the irreducible modules and the primitive ideals of A h in arbitrary characteristic and construct indecomposable A h -modules of arbitrarily large dimension.In further work [BLO2], we completely describe the Lie algebra Der F (A h ) of F-linear derivations and the first Hochschild cohomology HH 1 (A h ) = Der F (A h )/Inder F (A h ) of A h over arbitrary fields F. Our investigations extend earlier results of Nowicki [N].In particular, we determine the Lie bracket in HH 1 (A h ) := Der F (A h )/Inder F (A h ), construct a maximal nilpotent ideal of HH 1 (A h ), and explicitly describe the structure of the corresponding quotient in terms of the Witt algebra (centreless Virasoro algebra) of vector fields on the unit circle when char(F) = 0.

Generalities
An Ore extension A = R[y, σ, δ] is built from a unital associative (not necessarily commutative) algebra R over a field F, an F-algebra endomorphism σ of R, and a σ-derivation of R, where by a σ-derivation δ, we mean that δ is F-linear and δ(rs) = δ(r)s + σ(r)δ(s) holds for all r, s ∈ R. Then A = R[y, σ, δ] is the algebra generated by y over R subject to the relation yr = σ(r)y + δ(r) for all r ∈ R.
The endomorphisms σ considered in this paper will be automorphisms of R. The following are standard facts about Ore extensions.
Theorem 2.1.Let A = R[y, σ, δ] be an Ore extension over a unital associative algebra R over a field F such that σ is an automorphism.
(1) A is a free left and right R-module with basis {y n | n ≥ 0}.
(3) If R is a domain, then A is a domain.
(4) The units of A are the units of R.

Ore Extensions with Polynomial Coefficients
We are concerned with Ore extensions A = R[y, σ, δ] with R = F[x], a polynomial algebra in the indeterminate x, and σ an automorphism of R. In this case, σ has the form σ(x) = αx + β for some α, β ∈ F with α = 0. Hence, A is isomorphic to the unital associative algebra over F with generators x, y subject to the defining relation yx = (αx + β)y + h, where h is the polynomial given by h(x) = δ(x).The next lemma reduces the study of such Ore extensions to three specific types of algebras.This result is essentially contained in Observation 2.1 of the paper [AVV] by Awami,Van den Bergh,and Van Oystaeyen (compare also [AD2,Prop. 3.2]), although the division into cases here is somewhat different from that given in those papers.
Lemma 2.2.Assume A = R[y, σ, δ] is an Ore extension with R = F[x], a polynomial algebra over a field F of arbitrary characteristic, and σ an automorphism of R. Then A is isomorphic to one of the following: Quantum planes and quantum Weyl algebras are generalized Weyl algebras in the sense of [B, 1.1] and their structure and irreducible modules have been studied extensively in that context.
Our aim in this paper is to give a detailed investigation of the algebras that arise in (c) of Lemma 2.2.The algebra A h is the Ore extension R[y, id R , δ] obtained from the polynomial algebra R = F[x] over the field F by taking h ∈ R, σ to be the identity automorphism id R on R, and δ : R → R to be the F-linear derivation with δ(f ) = f ′ h for all f ∈ R, where f ′ denotes the usual derivative of f with respect to x.
It is convenient to regard A h as the unital associative algebra over F with generators x, y and defining relation Theorem 2.1 implies that A h is both a left and right Noetherian domain with units F * 1 and that } is a basis for A h over F, and A h has Gelfand-Kirillov (GK) dimension 2 by [McR,Cor. 8.2.11].

The Embeddings
In order to distinguish generators for the algebras A f and A g , we will assume those for A f are x, y, 1, and those for A g are x, ỹ, 1.
Lemma 3.1.For f, g, ∈ R, suppose that f | g and g = f r.Then the map ψ : gives an embedding of A g into A f .
Proof.This follows directly from the observation that [yr, x] = [y, x]r = f r = g.
Corollary 3.2.For all nonzero h ∈ F[x], there is an embedding of the algebra A h into the Weyl algebra A 1 .
Because we often use the embedding in Corollary 3.2 as a mechanism for proving results, and because the structure of A 0 = F[x, y] is very well understood, for the remainder of this paper we adopt the following conventions: , and the polynomial h ∈ R is nonzero; • the generators of the Weyl algebra A 1 are x, y, 1; • the generators of the algebra A h are x, ŷ, 1; • when A h is viewed as a subalgebra of A 1 , then ŷ = yh.

The following result provides an important tool for recognizing elements of A
Proof.We show that n i=0 ŷi R = n i=0 y i h i R for all n ≥ 0, and from that we can immediately conclude A h = i≥0 y i h i R. Observe for j ∈ Z, (3.5) Also note that yh = ŷ and y 2 h 2 = y ŷh = yh(ŷ + h ′ ) = ŷ(ŷ + h ′ ) hold.It follows easily from (3.5) and induction that For the other containment, we argue that ŷn ∈ n i≥0 y i h i R by induction on n, with the n = 1 case simply being the definition, ŷ = yh.Now from (3.6) with i = n, we have that y n h n = ŷn + a, where a ∈ n−1 j=0 ŷj R. Thus by induction, ŷn = y n h n − a where a ∈ n−1 i=0 y i h i R, and the containment n i=0 ŷi R ⊆ n i=0 y i h i R holds.The anti-automorphism of A 1 with x → x and y → −y sends ŷ to −ŷ +h ′ .Hence, it restricts to an anti-automorphism of A h .When applied to A h = i≥0 y i h i R, it gives A h = i≥0 Rh i y i and shows that (3.7)

Localizations and Ore Sets
The embedding A h ⊆ A 1 suggests that there is a strong relationship between the skew fields of fractions of A h and A 1 .In this section, we will see that in fact these skew fields are identical.To show this result, we describe certain Ore sets in A 1 and A h .Our starting point is a computational lemma.
Repeated application of this gives the claim.
Lemma 4.2.Fix f, h ∈ R, with f = 0. Then the set Σ = {f n | n ≥ 0} is a left and right Ore set of regular elements in A h .
Proof.That Σ consists of regular elements follows from the fact that A h is a domain.Let a ∈ A h and s ∈ Σ.We must show that there exist a 1 ∈ A h and s 1 ∈ Σ such that as 1 = sa 1 .It is enough to consider the case s = f .Write a = k i=0 r i ŷi and set s 1 = f k+1 .By Lemma 4.1, we see that A similar argument shows that Σ is a left Ore set.
Then Σ is a left and right Ore set of regular elements in both A 1 and A h , and the corresponding localizations are equal: Proof.By applying Lemma 4.2 to A 1 with Σ = {h n | n ≥ 0}, and then to A h with f = h, we see that Σ is a left and right Ore set in both A 1 and A h .Clearly Corollary 4.4.The skew field of fractions of A h is isomorphic to the skew field of fractions of the Weyl algebra A 1 (commonly referred to as the Weyl field).
(2) A 1 is a Noetherian (left or right) A h -module.
(3) A 1 is a free (left or right) A h -module.
Proof.If h ∈ F * , then the embedding A h ⊆ A 1 considered in this section is an equality.Thus as an A h -module, A 1 is free of rank one, and it is Noetherian.Now assume h / ∈ F. For each k ≥ 0, consider the right A h -submodule does not terminate.In particular, A 1 is not a Noetherian A h -module.Since A h is a Noetherian ring, it follows that A 1 is not a finitely generated A h -module either.Assume there exist elements 0 = t i ∈ A 1 , i ∈ I, such that This implies that I must be finite, as the decomposition of 1 ∈ B uses only finitely many summands.This contradicts the fact that A 1 is not a finitely generated A h -module.Hence, A 1 is not a free right A h -module.This proves the corollary for when A 1 is considered as a right A h -module.The left-hand version is analogous.

The Center of A h
In this section, we describe the center Z(A h ) of A h and show in Proposition 5.9 that A h is free over Z(A h ).In the case of the Weyl algebra, the center is F1 when char(F) = 0.When char(F) = p > 0, the center has been described by Revoy in [R] (see also [ML]) as follows: Lemma 5.1.Suppose char(F) = p > 0. Then the center of A 1 is the unital subalgebra generated by the elements x p and y p .
In determining Z(A h ) for arbitrary h, we will use the following result which can be shown by a straightforward inductive argument.
(1) If char(F) = 0, then the center of A h is F1.
(2) If char(F) = p > 0, then the center of A h is isomorphic to the polynomial algebra F[x p , h p y p ], where (5.7) If char(F) = 0 then Z(A h ) = F1.Now suppose that char(F) = p > 0. Then x p , h p y p ∈ Z(A 1 ) ∩ A h .For every k ≥ 0, h kp y kp = (h p ) k (y p ) k = (h p y p ) k , thus the elements x p and h p y p are algebraically independent, and it follows that F[x p , h p y p ] ⊆ Z(A h ).Let z ∈ Z(A h ).By (5.7), Lemma 3.4, and Lemma 5.1, we can write z = i≡0 mod p r i y i with The relation h ŷ, use Lemma 3.4 to write h p y p = p n=0 f n ŷn , where f n ∈ F[x] for all n and f p = 1.Then Since δ(x) = h = 0, we see that f p−1 = 0. Then the above gives Proceeding in this way, we obtain f n = 0 for all n = p − 1, p − 2, . . ., 2. As a result, we have 0 and it follows that [ŷ, and hence f 0 ∈ ŷA h .The only way that can happen is if f 0 = 0 and h p y p = ŷp − δ p (x) h ŷ.
Example 5.8.Assume char(F) = p > 0 and h(x) = x n for some n ≥ 1.Then it is easy to verify that Hence, if n ≡ 1 mod p, we can find 1 ≤ k < p with k(n − 1) ≡ −1 mod p so that δ p (x) = 0.This implies that when h(x) = x n , if n ≡ 1 mod p.
In particular, Z(A h ) = F[x p , ŷp ] whenever h(x) = x n and n ≡ 1 mod p.When n = 2, this was shown by Shirikov in [S3].
Proposition 5.9.Assume char(F) = p > 0 and regard A h ⊆ A 1 as in Conventions 3.3.Then A h is a free module over Z(A h ), and the set {x i h j y j | 0 ≤ i, j < p} is a basis.
Proof.Suppose that 0 = 0≤i,j<p c i,j x i h j y j , (5.10) Thus, (5.10) and Theorem 2.1 imply that 0≤i<p c i,j x i h j y j = 0.As h = 0, it follows that 0≤i<p c i,j x i = 0 for every 0 ≤ j < p.The direct sum decomposition F[x, h p y p ] = p−1 i=0 F[x p , h p y p ]x i then implies c i,j = 0 for all i, j.It remains to show that {x i h j y j | 0 ≤ i, j < p} generates A h over Z(A h ).Let a, b ≥ 0 and write a = ãp + i, b = bp + j, for some nonnegative integers ã, b and 0 ≤ i, j < p.Then, Remark 5.11.
(i) The algebra anti-automorphism x → x, y → −y of A 1 can be applied to the basis above to show that {y j h j x i | 0 ≤ i, j < p} is a basis for A h over Z(A h ).
(ii) A standard inductive argument can be used to prove that {x i y j h j | 0 ≤ i, j < p} is also a basis for A h over Z(A h ).

The Lie Ideal
Proof.Recall that A h is spanned by elements of the form aŷ ℓ for ℓ ≥ 0 and a ∈ R.
Thus it suffices to show that [aŷ ℓ , bŷ m ] ∈ hA h for all ℓ, m ≥ 0 and This follows directly from (5.3) as δ j (f ) ∈ hR for all j ≥ 1.
We have the following simple description of [A h , A h ] for fields of characteristic 0.
Proof.By Lemma 6.1, it suffices to prove that hA h ⊆ [ŷ, A h ].Note that hA h = h i≥0 Rŷ i , and by the linearity of the adjoint map ad ŷ (where It will be more convenient to work inside A 1 , where hA h = h i≥0 Rh i y i .Then, for i ≥ 0 and g ∈ R we have 1 i+1 gh i+1 y i+1 ∈ A h and In the next result, we determine the centralizer Proof.We first determine the centralizer Since i = 0 in F as long as i ≡ 0 mod p, we see that im(ad x ) is i ≡−1 mod p hRh i y i , and this sum is evidently direct.The fact that and thus im(ad

The Normal Elements and Prime Ideals of
Similarly, the normal elements of the Weyl algebra A 1 are precisely the central elements (compare Theorem 7.3).In general, for h / ∈ F, there are non-central normal elements in A h .In this section, we determine the normal elements of A h for arbitrary h = 0. Our starting point is i≥0 Rŷ i , it follows that A h g ⊆ gA h and gA h ⊆ A h g, and so gA h = A h g.
Since the product of two normal elements is normal, it is clear at this stage that products of powers of the prime factors of h are normal elements of A h .

Suppose
where λ ∈ F * , α i ≥ 1 for all i, and the u i ∈ F[x] are distinct monic prime polynomials.We can assume that the factors have been ordered so that the first ones u i , for i ≤ ℓ ≤ t, are the non-central prime divisors of h.Our aim is to establish the following which generalizes (and includes) the result for the Weyl algebra.
Theorem 7.3.Let u 1 , . . ., u ℓ be the distinct monic prime factors of The proof will use the next lemma.
Lemma 7.4.Let u 1 , . . ., u ℓ be the distinct monic prime factors of h in R that are not central in If char(F) = p > 0, the β i may be chosen so that 0 ≤ β i < p for all i.
where µ ∈ F * , γ i ≥ 1 for all i, and q 1 , . . ., q n are distinct monic prime polynomials in F This implies that q j divides γ j q ′ j h for all j.Then either q j divides γ j q ′ j or q j divides h.If q j divides γ j q ′ j then γ j q ′ j = 0 which forces q Otherwise, q j = u k for some non-central prime factor of h.The last assertion in the lemma follows from the observation that when char(F) = p > 0, then r p ∈ F[x p ] for all r ∈ R.
Proof of Theorem 7.3.Assume v = 0 is normal in A h , and write v = n i=0 f i h i y i , where f i ∈ R and f n = 0. Then there exists a ∈ A h so that vx = av, and from considering the coefficient of y n , we see that a ∈ R, and in fact a = x.Thus vx = xv, and v ∈ C A h (x).Since hy ∈ A h by Lemma 3.4, there exists b ∈ A h so that v(hy) = bv and, as above, we conclude that b = hy − r, for some r ∈ F[x].The latter implies [hy, v] = rv.
Recall that Thus, for the remainder of the proof, we assume that char(F) = p > 0, and because which forces rf i = hf ′ i for all i ≡ 0 mod p.This implies that f i divides hf ′ i for all such i, so by Lemma 7.4, there exist w i ∈ F[x p ] and integers β 1i , . . ., β ℓi ∈ {0, 1, . . ., p − 1} such that where ε k = β ki + β kj for k ∈ {1, . . ., ℓ}.If f i , f j = 0, then w i w j = 0, and as a result we have ℓ k=1 , and thus β ki = β kj for all k and all i, j.Letting β k be that common exponent, we have Several authors have studied the problem of determining simplicity criteria for Ore extensions R[y, id R , δ], and it is possible to address the simplicity of the algebras A h by using the results of [J] or [CF,Thms. 3.2 and 3.2a] for example.Instead, we apply our results on normal and central elements of A h to determine when an algebra A h is simple.
Corollary 7.5.The algebra A h is simple if and only if char(F) = 0 and h ∈ F * .Proof.Suppose A h is simple.If b = 0 is a normal element of A h , then bA h = A h b = A h by simplicity, so b is a unit.Since the units of A h are the elements of F * , we see that h ∈ F * by Lemma 7.1, and also Z(A h ) = F1.But then char(F) = 0 by Lemma 5.5.Conversely, if char(F) = 0 and h ∈ F * , then A h is isomorphic to the Weyl algebra, and it is well known that A 1 is simple.
A (noncommutative) Noetherian domain is said to be a unique factorization ring (Noetherian UFR for short), if every nonzero prime ideal contains a nonzero prime ideal generated by a normal element.The height of a prime ideal is the largest length of a chain of prime ideals contained in it (or is ∞ if no bound exists).A Noetherian UFR is said to be a unique factorization domain (Noetherian UFD for short) if every height one prime factor is a domain.These notions were introduced by Chatters and Jordan in [C, CJ].If a Noetherian domain satisfies the descending chain condition on prime ideals (e.g. if it has finite Gelfand-Kirillov dimension [McR,Cor. 8.3.6]), then it is a Noetherian UFR if and only if every height one prime ideal is generated by a normal element.Recently, Goodearl and Yakimov [GY] have used the properties of noncommutative Noetherian UFDs to construct initial clusters for defining quantum cluster algebra structures on a noncommutative domain.
Since R = F[x] is a principal ideal domain, [CJ,Thm. 5.5] trivially implies the first part of the following observation.The second part follows by [GW,Thm. 9.24].
The algebra A 0 = F[x, y] is a Noetherian UFD for any field F. We will see shortly that A h is not a Noetherian UFD when char(F) = p > 0 and h = 0.
The next result describes the height one prime ideals of A h .It is known that over a field of prime characteristic the Weyl algebra A 1 is Azumaya over its center (see [R,Thé. 2]), so in this case the prime ideals of A 1 are in bijection with the prime ideals of Z(A 1 ).If deg h ≥ 1, there may be prime ideals of A h which are not centrally generated.
Theorem 7.7.Let u 1 , . . ., u t be the distinct monic prime factors of h in R, as in (7.2).For every 1 ≤ i ≤ t, the normal element u i generates a height one prime ideal of A h , and the corresponding quotient algebra is a domain.
(i) If char(F) = 0, these are all the height one prime ideals.
(ii) If char(F) = p > 0, then any nonzero irreducible polynomial in Z(A h ) that (up to associates) is not of the form u p i for any 1 ≤ i ≤ t generates a height one prime ideal.These, along with the ideals generated by some u i , constitute all the height one prime ideals.
Proof.First notice that each u i generates a prime ideal of A h , as the quotient algebra A h /u i A h is isomorphic to the commutative polynomial algebra (R/u i R) [ŷ] over the field R/u i R. In particular, A h /u i A h is a domain, and the prime ideal u i A h has height one by the Principal Ideal Theorem (see [McR,Thm. 4.1.11]).
Let P be a height one prime ideal.Since A h is a Noetherian UFR, it follows that P = vA h for some normal element v = 0.Moreover, the primality of P implies that v is not a (non-trivial) product of normal elements.Thus, Theorem 7.3 implies that either v is an irreducible factor of h or a central element which is irreducible as an element in Z(A h ).When char(F) = 0, then v must be an irreducible factor of h, as Z(A h ) = F1, which proves (i).
For the remainder of the proof assume char(F) = p > 0. Note that if z ∈ Z(A h ) is of the form ξu p i for some i and some ξ ∈ F * , then zA h is not a prime ideal.So it remains to show that if z is an irreducible polynomial in Z(A h ), which is not of the form ξu p i for 1 ≤ i ≤ t and ξ ∈ F * , then zA h is a height one prime ideal.We can further assume z is not an irreducible factor of h, as this case has already been considered.Let P ⊇ zA h be a minimal prime over zA h .By the Principal Ideal Theorem, P has height one, and thus P = vA h for some normal element v.
Suppose first that v is an irreducible factor of h, for all i ≡ 0 mod p. Fix j with j ≡ 0 mod p and r j = 0. Let If the latter holds, then z = u n a implies that a ∈ Z(A h ).The irreducibility of z in Z(A h ) implies that a ∈ F * , and thus z is an irreducible factor of h, which contradicts our previous assumption.So it must be that γ 1 ≡ 0 mod p.As γ 1 ≥ 1, it follows that γ 1 ≥ p and u p n divides u n r j .Since j ≡ 0 mod p was arbitrary subject to the restriction that r j = 0, we deduce that z = u p 1 c for some c ∈ Z(A h ).The irreducibility of z in Z(A h ) again implies that z is a scalar multiple of u p n , which violates our assumptions on z.It follows from the arguments in the preceding paragraph that v is not an irreducible factor of h.Hence v ∈ Z(A h ), and again we deduce that z = va for some a ∈ Z(A h ).Thus, as z is irreducible in Z(A h ), it must be that a ∈ F * and zA h = vA h = P is a height one prime ideal.
Proof.By Theorems 5.5 and 7.7, the element h p y p generates a height one prime ideal of A h , as it is irreducible in Z(A h ) and it is not a power of a factor of h.However, by (5.6) we have h p y p = ŷp−1 − δ p (x) h ŷ.Yet neither one of these two factors is in h p y p A h , by considering the degree in y of an element in h p y p A h .Thus, the prime ring A h /h p y p A h is not a domain.
Remark 7.9.Since A h has Gelfand-Kirillov dimension 2, it follows from [McR,Cor. 8.3.6] that the possible values for the height of a prime ideal P of A h are 0, 1, and 2. The zero ideal is prime and is thus the unique prime ideal of height zero.The height one prime ideals are given in Theorem 7.7.The height two prime ideals of A h must be maximal, and no height one prime ideal of A h can be maximal.Indeed, for the height one prime ideals of the form u i A h , 1 ≤ i ≤ t, the quotient A h /u i A h is a commutative polynomial algebra.When char(F) = p > 0, the center Z(A h ) is a polynomial algebra in two variables, so if v is an irreducible polynomial in Z(A h ) as in Theorem 7.7 (ii) above, it follows that any maximal ideal of Z(A h ) containing v induces a maximal ideal of A h strictly containing vA h .
Hence, the height two prime ideals of A h are precisely the maximal ideals of A h , and can be identified with the maximal ideals of A h /P, as P ranges through the height one prime ideals.In particular, if char(F) = 0 and the prime factors of h in F[x] are linear, then the height two prime ideals of A h are the ideals generated by x − λ and q(ŷ), where λ ∈ F is a root of h and q(ŷ) ∈ F[ŷ] is an irreducible polynomial.

Automorphisms of A h
Extending results of Dixmier [D] on the automorphisms of the Weyl algebra A 1 , Bavula and Jordan [BJ] considered isomorphisms and automorphisms of generalized Weyl algebras over polynomial algebras of characteristic 0. Alev and Dumas [AD2] initiated the study of automorphisms of Ore extensions over the polynomial algebra R = F[x], and the results in [AD2] have been further developed in the recent work [G] of Gaddis.In Theorem 8.2, we summarize results from [AD2] that pertain to the algebras A h studied here, but suitably interpreted in the notation of the present paper.Since one of those results assumes that char(F) = 0, we first prove Lemma 8.1, which can be used to remove that characteristic assumption.This will enable us to prove our main results, Theorems 8.7 and 8.13, which give a complete description of the automorphisms of A h over arbitrary fields.
Proof.Let B h be the ideal of A h minimal with the property that A h /B h is commutative.Then [y, x] = 0 in the quotient A h /B h , so it follows that h ∈ B h .The element h is normal in A h and hA h ⊆ B h , so the minimality of B h , with the fact that A h /hA h is commutative, implies that hA h = B h .Similar reasoning shows that B g = gA g is the ideal of A g minimal with the property that A g /B g is commutative.As B h and B g are obviously characteristic ideals, it follows that θ(B h ) = B g .Since A g is a domain and gA g = B g = θ(B h ) = θ(h)A g , we have that θ(h) = λg for some λ ∈ F * .Now with Lemma 8.1, the argument in the proof [AD2,Prop. 3.6] can be extended to arbitrary fields, and as a result, we have the following.
(i) A h is isomorphic to A g if and only if there exist α, β, ν ∈ F, with αν = 0 such that νg(x) = h(αx + β).In particular, if A h is isomorphic to A g , then g and h have the same degree.

Automorphisms of A h Definitions and the Decomposition
If h ∈ F, the automorphism group of A h is known [VDK, D, ML] (see also the discussion in Sec.8.5 below), so in what follows, we assume deg h ≥ 1.In view of Theorem 8.2, we introduce the following definitions.Let It is easy to verify that each pair (α, β) ∈ P determines an automorphism τ α,β of A h whose values on x and ŷ are given by The pair (α −1 , −βα −1 ) belongs to P whenever (α, β) does, and and having inverse φ −f .Furthermore, . One important example is the automorphism φ h ′ with φ h ′ (x) = x and φ h ′ (ŷ) = ŷ + h ′ .The normality of the element h ∈ A h (see Lemma 7.1) implies that this automorphism has the property that for all a ∈ A h (compare (3.5)).
(iv) The abelian subgroup {φ f | f ∈ F[x]}, which we identify with (F[x], +), is a normal subgroup of Aut F (A h ).
Proof.Part (i) is immediate from Theorem 8.2.If τ α,β = φ f for some (α, β) ∈ P and Suppose (α, β), (γ, ε) ∈ P. Then (αγ, βγ + ε) ∈ P, as Direct calculation shows that Since every automorphism is a product of automorphisms in the subgroups F[x] and τ P , we have that the subgroup F[x] is normal in Aut F (A h ).Part (v) follows then, since the two subgroups have trivial intersection by (ii).
The automorphism group Aut F (A h ) will be completely determined once we establish conditions for a pair (α, β) to belong to P. This will of course depend on the polynomial h.
The group F[x] ⋊ τ 1,G may not be all of Aut F (A h ), and in that situation, there exists some (α, β) ∈ P with α = 1 so that τ α,β ∈ Aut F (A h ).The next result draws conclusions in that case.
Theorem 8.13.Assume h has k distinct roots in F for k ≥ 1.
(Case k = 1) Let λ be the unique root of h in F.
where for all f ∈ F[x] and α ∈ F * , (Case k ≥ 2) The group τ P /τ 1,G is a finite cyclic group.In particular, when Proof.Assume (α, β) ∈ P. By the definition of P, the affine bijection σ α,β of F given by σ α,β (λ) = αλ + β permutes the roots of h(x) in such a way that the corresponding multiplicities are preserved.Thus λ + ν is a root of h(x) whenever λ is a root of h(x) and ν ∈ G, so it follows that G = {0} when k = 1.
When h(x) has the form h , we may identify the group τ P with F * in this case.Thus, Aut F (A h ) = F[x] ⋊ F * .The product formula appearing in (a) follows from (8.9).Hence, the theorem holds when k = 1 and λ ∈ F.
Hence, the orbits of this action of σ α,β on the roots of h(x) have size either 1 or ℓ.Let r be the number of orbits of size 1 and q the number of orbits of size ℓ.It follows that k = r + qℓ, so ℓ divides k − r.If the orbits of two roots λ and λ have size 1, then λ = β 1−α = λ, so r ≤ 1.Thus, either r = 0 and ℓ divides k or r = 1 and ℓ divides k − 1.
By (8.8), the projection map ψ : τ P → F * given by ψ(τ µ,ν ) = µ is a group homomorphism with kernel τ 1,G .The image is a finite subgroup of F * , since F * has only finitely many k and k − 1 roots of unity.As finite subgroups of F * are cyclic, we have that τ P /τ 1,G is generated by a coset τ α,β τ 1,G for some (α, β) ∈ P such that α k−1 = 1 or α k = 1 (but not both).The rest of the statements follow from Lemma 8.11 and Theorem 8.7.
In the next result, we will use the notation σ P = ζ,ε | (ζ, ε) ∈ P} for the group of affine maps on F determined by P, and σ 1,G for the subgroup determined by G, along with the fact that these groups act on the set of roots of h in F.
Corollary 8.14.Assume h has k distinct roots in F for k ≥ 1.
(Case k = 1) Let λ be the unique root of h in F.
, and there exist orbit representatives λ i , i ∈ I, for the action of σ 1,G on the roots of h, so that h = γ i∈I h n i i , where γ ∈ F * , n i ≥ 1, and or there exists (α, β) ∈ P, where α is a primitive ℓth root of unity for some ℓ > 1 such that ℓ divides k − 1 or k, and (c) If ℓ divides k, then there are orbit representatives λ i , i ∈ I, for the action of σ P on the roots of h so that h = γ i∈I h n i i for some γ ∈ F * and integers n i ≥ 1, where Proof.We may assume k ≥ 2, since the first case follows directly from Theorem 8.13.
Now if (a) holds, then either are roots of h in F, which are representatives for the orbits of roots of h in F under the affine bijections σ 1,ν for ν ∈ G. Since each orbit is of size p d , we have k = qp d .Then h has the form displayed in (a).When , where α is primitive ℓth root of unity for some ℓ > 1 that divides k or k − 1.
When ℓ divides k − 1, then as we have seen previously, there is one orbit of size one under the action of σ α,β generated by the root λ 0 := β/(1 − α) ∈ F. Either the group G = {0}, or char(F) = p > 0 and G has order p d for some d ≥ 1, and G is invariant under multiplication by the cyclic group generated by α by (3) of Lemma 8.11.Under this action of the group α , there is one orbit of size 1 (namely {0}), and all the other orbits have size ℓ.Thus, rℓ + 1 = p d for some r ≥ 0.
Consider the orbits of roots under the group generated by the maps σ α,β and σ 1,ν as ν ranges over the elements of G.One such orbit is {λ 0 + ν | ν ∈ G}.Assume λ i for i ∈ I are the representatives for the other orbits.Then h has the factorization into linear factors given in (8.15) for some γ ∈ F * , and n i ≥ 1.Counting roots of h in F, we have qℓ+1 = k when G = {0}, and The case when ℓ divides k is similar and follows the same line of reasoning -just omit the factors of h involving λ 0 , and use the fact that σ j α,β Remark 8.18.Suppose α ∈ F is an ℓth root of unity for ℓ > 1.Let G be a finite subgroup of (F, +) invariant under multiplication by α (necessarily G = {0} when char(F) = 0).By choosing λ i for i in some index set I so that λ 0 + ν, α j (λ i + ν) + λ 0 (1 − α j ) are distinct for ν ∈ G, i ∈ I ∪ {0}, and j = 0, 1, . . ., ℓ − 1, and taking arbitrary n i ≥ 1 for i ∈ I ∪ {0}, we can construct h as in (8.15) with τ 1,G ⋊ τ α,λ 0 (1−α) ⊂ Aut F (A h ).Similarly, if we choose β arbitrarily, G as above, and λ i for i ∈ I so that α j (λ i + ν) + β(1 − α j )/(1 − α) are all distinct for ν ∈ G, i ∈ I, and j = 0, 1, . . ., ℓ − 1, and take arbitrary n i ≥ 1, we can construct h as in (8.17) with Example 8.19.In this example, we compute Aut F (A h ) for any monic quadratic polynomial h We conclude that there are two possibilities: either In this calculation, we have tacitly assumed that char(F) = 2.When char(F) = 2, then (α, β) ∈ P if and only if

The Aut F (A h ) Invariants
Throughout this section and the next, we let A = Aut F (A h ).In this section, we determine the invariants under A in A h : is clear, since φ r (x) = x for all r ∈ R. We will prove that the reverse inclusion holds as well.
Assume by contradiction that there is a ∈ A , where c i,k = i k if k < i and c i,k = 0 otherwise.Assume first that char(F) = 0. Take g = 1 and k = m − 1 above.Then we get mf m = 0, which is a contradiction.Now suppose char(F) = p > 0, and take g = x np , where n is chosen so that np > max{deg f i | 1 ≤ i ≤ m}, and k = 0. We have m i=1 f i g i = 0.For every i, either This implies that f m g m = 0, so f m = 0, which is a contradiction.Thus A and equality is proved.
The above shows that A A h ⊆ R A ⊆ R P .However, since φ r (x) = x for all r ∈ R, R A = R P , and the rest follows.
Next we determine the invariants under A in R: Lemma 8.21.Suppose R A = F. Then there exists a unique monic polynomial s of minimal degree in R A \ F with zero constant term such that R A = F[s].
Proof.Let s be a monic polynomial of minimal degree in R P \ F. We may assume that s has zero constant term.Now for every r ∈ R P , r = sf + g for some f, g ∈ R with deg g < deg s.Applying τ ζ,ε to that relation gives and subtracting that from the above gives 0 = s(f − τ ζ,ε (f )) + g − τ ζ,ε (g).Since this is true for all (ζ, ε) ∈ P, and since τ ζ,ε preserves degree, we have that f ∈ R P and g ∈ F. Thus R P = sR P ⊕ F.
Clearly F[s] ⊆ R A = R P .For the other direction, we proceed by induction on the degree of an element of R A ; the case of degree 0 being obvious.Assuming the result for degree < n, we suppose r ∈ R A has degree n where n ≥ 1.Then there exist f ∈ R A and ξ r ∈ F such that r = sf + ξ r .By induction, f ∈ F[s].Hence so is r, and The uniqueness of such an s is clear.
where the polynomial t ∈ R can be taken as follows: , where α is a primitive ℓth root of unity for some Proof.Assume r ∈ R A and deg r ≥ 1, and let Λ be the set of roots of r in F. Since every automorphism of the form φ f leaves R pointwise fixed, the first part of (i) is clear.We will assume we have nontrivial automorphisms in τ P .For any τ 1,ν ∈ τ 1,G , the equality r(x + ν) = τ 1,ν (r) = r(x) implies that µ + ν ∈ Λ for all µ ∈ Λ.Thus G acts faithfully on Λ, and roots of r in the same G-orbit have the same multiplicity.This implies that deg r is divisible by |G|.
In particular, if τ P = τ 1,G , then we claim that the polynomial s in Lemma 8.21 is given by s(x) = t(x) − t(0), where t(x) = ν∈G (x + ν).Indeed, it is easy to see that the polynomial t belongs to R A in case (a) of (ii).Moreover, t(x) − t(0) is a monic polynomial of degree |G| in R A with zero constant term.Since every r ∈ R A \ F has deg r ≥ |G|, t(x) − t(0) is the polynomial s in Lemma 8.21.Finally, In all the remaining possibilities for A = Aut F (A h ), coming from Theorem 8.13, there exists an automorphism of the form τ α,β , with (α, β) ∈ P and α = 1.Since deg r ≥ 1, it follows from considering the leading coefficient of r = τ α,β (r) that α deg r = 1, and thus when r ∈ F, deg r is at least the multiplicative order of any α ∈ F * with (α, β) ∈ P for some β ∈ F. Now when A = F[x] ⋊ F * in Theorem 8.13, F * is identified with τ P = {τ α,(1−α)λ | α ∈ F * }, where λ ∈ F is the unique root of h.If r ∈ R A with deg r ≥ 1, then by the previous paragraph deg r is greater than or equal to the multiplicative order of every α ∈ F * .If F is infinite, there is no upper bound on the order of elements of F * , so no such r can exist.Hence, we have the second part of (i).
Finally, note that |G| and ℓ = |τ P /τ 1,G | are coprime by Remark 8.12.Therefore, in case (ii)(b) the degree of the polynomial r is divisible by the coprime integers |G| and ℓ, so deg r ≥ ℓ|G|.Observe that From Lemma 8.11, we know that αG = G, hence α −1 ν ∈ G. Thus the polynomial In particular, Fh ′ is a subgroup of Z(A) (under our usual identification of r ∈ F[x] with the automorphism φ r ).
Proof.We first argue that the centralizer of the normal subgroup F Then by (8.9), This shows that the centralizer and the other containment is trivial, so we have equality.Now ω = φ r ∈ Z(A) if and only if φ r commutes with τ ζ,ε , for every (ζ, ε) ∈ P. Equation (8.9) gives that τ we multiply both sides of this equation by an arbitrary λ ∈ F, we see that For the other containment, assume f ∈ R Z , and use the division algorithm to write f = qr + g with r, g ∈ F[x] and deg g < deg q.Then for (ζ, ε) ∈ P, we have , and g = 0 by the minimality of deg q.Thus, we have f ∈ qR A .
Combining these results with the description of the invariants R A in Theorem 8.22, we obtain the main result of this section -a description of the center of Aut and Z(A) and R Z are as follows: (1) (3) If A = F[x] ⋊ F * and |F| = ∞, then h = γ(x − λ) n for some γ ∈ F * and some λ ∈ F, and R (4) If A = F[x] ⋊ τ P , where τ P = τ 1,G ⋊ τ α,β and α is a primitive ℓth root of unity for some ℓ > 1, then R Z = qF[t], where Proof.It will be seen in the course of the proof that in all cases R Z = {0}, so from Lemma 8.24, we know that R Z = qR A , where q is the monic polynomial of minimal degree in R Z .Since we have determined R A in Theorem 8.22, we need to find the polynomial q.For all (ζ, ε) ∈ P we have from q Let's consider the various cases arising from Theorem 8.13 and Corollary 8.14: where |F| = ∞ and F * is identified with the group {τ α,(1−α)λ | α ∈ F * }, then by the above, α deg q = α deg h−1 for all α ∈ F * , which forces deg q = deg h − 1. Recall that this case occurs when h(x) = γ(x − λ) n for some γ ∈ F * , λ ∈ F, and n ≥ 1.The monic polynomial (x − λ) n−1 has degree equal to deg h − 1, and it is in R Z .Thus, q(x) = (x − λ) n−1 , and R Z = (x − λ) n−1 R A .
We have shown that Hence, there exists a polynomial f (t) ∈ F[t] so that u = qf (t).However, since the degree of t in x is ℓ|G| and the degree of u in x is n|G| and n < ℓ, it must be that f (t) ∈ F. But since both q and u are monic, this says q = u.Remark 8.27.In the case of the Weyl algebra, the center of Aut F (A 1 ) is trivial by [KA,Prop. 3].However, when h ∈ F * , we can have the opposite extreme.For example, if h = x 2 (x − 1), then P = {(1, 0)}, as any permutation of the roots of h has to fix 0 and 1 (since they have different multiplicities), and the affine permutations determined by elements of P can have at most 1 fixed point, except for the identity map.So Aut F (A h ) = F[x] is commutative, and its center is the entire automorphism group in this case.

Automorphisms of the Weyl Algebra
In this section we contrast the previous results on automorphisms of A h for h ∈ F, with known results on the automorphisms of the Weyl algebra A 1 .The Weyl algebra has more automorphisms because of its high degree of symmetry.
Remark 8.29.Unlike the situation for A h , with deg h ≥ 1, the subgroup F[x] fails to be normal in Aut F (A 1 ), which can be seen from the above calculation.
The following provide generating sets of automorphisms for Aut F (A 1 ).(Compare [ML] and [S], and see also [KA] for part (iii).)Theorem 8.30.Each of the following sets gives a generating set for the automorphism group Aut F (A 1 ):

Dixmier's Conjecture
In [D,Problem 1], Dixmier asked if every algebra endomorphism of the nth Weyl algebra must be an automorphism when char(F) = 0.This conjecture was shown to be equivalent to the longstanding Jacobian conjecture (see [T] and [BK]).In this section, we explore whether monomorphisms for the algebra A h with deg h ≥ 1 necessarily are automorphisms.
Proposition 8.31.Assume h = x n for some n ≥ 1, and fix k ≥ 1.When char(F) = p > 0 assume further that p does not divide k.Then there is an algebra monomorphism η k : A h → A h such that η k (x) = x k and η k (ŷ) = 1 k x (k−1)(n−1) ŷ.If k ≥ 2, then η k is not an automorphism.
homogeneous with respect to the Z-grading.Assume v has degree n < 0. Then we can write v = cd −n and b = cu 1−n , for some c, c ∈ D. We have: The above equation implies that du = a is a unit in D, which is a contradiction.Hence, v has degree n ≥ 0. Similarly, assuming that d ∈ vD(σ, a), we conclude that v has degree n ≤ 0. It follows that if vD(σ, a) contains both u and d, then v ∈ D. But then the equation vcu = u, for c ∈ D, implies that vD(σ, a) = D(σ, a).
Theorem 9.5.Assume h ∈ F. Then the algebra A h is not a generalized Weyl algebra over a polynomial ring in one variable.
Proof.Assume h = 0 and A h ∼ = D(σ, a), for D = F[t].First, notice that a / ∈ F, as otherwise we would have ud = 0 = du, and A h would not be a domain, or else u = d −1 and A h would have nontrivial units.By [RS,Prop. 2.1.1]we need only consider three possibilities for σ: (A) σ is the identity automorphism;  Notice that if σ is the identity then D(σ, a) must be commutative and thus h = 0, so case (A) above does not occur.Cases (B) and (C) are usually referred to as the classical and quantum cases, respectively.
Let Frac(A h ) be the skew field of fractions of A h .By Corollary 4.4, Frac(A h ) is the (first) Weyl field, i.e., the field of fractions of the Weyl algebra.Thus, it follows by [RS,Prop. 2.1.1]and [AD1,Thé. 3.10] that D(σ, a) must be of classical type, i.e., σ(t) = t − 1.
Let the ideal B h of A h (resp.J of D(σ, a)) be minimal with the property that A h /B h (resp.D(σ, a)/J) is commutative.Then, by the defining relations of A h and the fact that h is normal, we have B h = hA h .In particular, B h is a principal ideal, and it follows that J is also principal.In D(σ, a), the relations u = [t, u] and d = [d, t] show that u, d ∈ J.But Lemma 9.4 implies that J = D(σ, a), and thus hA h = A h , so h ∈ F * .
(a) a quantum plane (b) a quantum Weyl algebra (c) a unital associative algebra A h with generators x, y and defining relation yx = xy + h for some polynomial h = h(x) ∈ F[x].

Example 8. 26 .
Assume h(x) = x n for some n ≥ 1.Then by Theorem 8.13,A = Aut F (A h ) = F[x] ⋊ F * , where F * is identified with the automorphisms {τ α,0 | α ∈ F * }.If F is infinite, the monic polynomial generator of R Z is q(x) = x n−1 by Theorem 8.25, and according to Theorem 8.22, the invariants are given by R A = F. Thus, in this case R Z = Fx n−1 and Z(A) = {φ f | f ∈ Fx n−1 }.If |F * | = ℓ < ∞,then part (4) of Theorem 8.25 shows that the monic polynomial generator of R Z is q(x) = x m , where 0 ≤ m < ℓ and m ≡ n − 1 mod ℓ.Now Theorem 8.22 asserts that RA = F[t], where t(x) = x ℓ , thus R Z = x m F[x ℓ ] and Z(A) = {φ f | f ∈ x m F[x ℓ ]}.
and also under τ α,β , so t(x) is invariant under A. As above, since deg t = ℓ|G| and any non-constant r ∈ R A has deg r ≥ ℓ|G|, we deduce that R A = F[t].8.4The Center of AutF (A h )The explicit description of the automorphism group Aut F (A h ) in Theorem 8.7 enables us to determine the center of this group.Proposition 8.23.Assume deg h ≥ 1.Then the center of A and Fh ′ is clearly a subgroup under addition.Lemma 8.24.Assume deg h ≥ 1 and R