Hilbert’s Basis Theorem for Non-associative and Hom-associative Ore Extensions

We prove a hom-associative version of Hilbert’s basis theorem, which includes as special cases both a non-associative version and the classical Hilbert’s basis theorem for associative Ore extensions. Along the way, we develop hom-module theory. We conclude with some examples of both non-associative and hom-associative Ore extensions which are all noetherian by our theorem.


Introduction
Hom-associative algebras are not necessarily associative algebras, the associativity condition being replaced by α(a) , where α is a linear map referred to as a twisting map, and a, b, c arbitrary elements in the algebra.Both associative algebras and non-associative algebras can thus be seen as hom-associative algebras; in the first case, by taking α equal to the identity map, and in the latter case by taking α equal to the zero map.
Historically hom-associative algebras originate in the development of hom-Lie algebras, the latter introduced by Hartwig, Larsson and Silvestrov as generalizations of Lie algebras, the Jacobi identity now twisted by a vector space homomorphism; the "hom" referring to this homomorphism [6].The introduction of these generalizations of Lie algebras was mainly motivated by an attempt to study so-called q-deformations of the Witt and Virasoro algebras within a common framework.Makhlouf and Silvestrov then introduced hom-associative algebras as the natural counterparts to associative algebras; taking a hom-associative algebra and defining the commutator as a new multiplication gives a hom-Lie algebra, just as with the classical relation between associative algebras and Lie algebras [7].It was later discovered that there existed formally rigid associative algebras that could now be formally deformed when considered as hom-associative algebras [10], this indicating that hom-associative algebras could be useful in studying deformations as well.Since then, many papers have been written in the subject, and other algebraic structures have been discovered to have natural counterparts in the "hom-world" as well, such as e.g.hom-coalgebras, hom-bialgebras, and hom-Hopf algebras [8,9].
A class of algebras that contains many examples of formally rigid associative algebras such as e.g. the Weyl algebras and some universal enveloping algebras of Lie algebras, is that of Ore extensions.Ore extensions were first introduced under the name non-commutative polynomial rings by Ore [14].Non-associative Ore extensions were introduced by Nystedt, Öinert, and Richter in the unital case [12] (see also [13] for a further extension to monoid Ore extensions).The construction was later generalized to non-unital, hom-associative Ore extensions by the authors of the present article and Silvestrov [3].They give examples, including hom-associative versions of the first Weyl algebra, the quantum plane, and a universal enveloping algebra of a Lie algebra, all of which are formal deformations of their associative counterparts [1,2].
In this paper, we prove a hom-associative version of Hilbert's basis theorem (Theorem 4), including as special cases both a non-associative version (Corollary 7) and the classical associative Hilbert's basis theorem for Ore extensions (Remark 8).In order to prove this, we develop hom-module theory and a notion of being hom-Noetherian (Section 3).Whereas the hom-module theory does not require a multiplicative idenity element, the hom-associative Ore extensions in this article are all assumed to be unital.We conclude with some examples of unital, non-associative and hom-associative Ore extensions which are Noetherian as a consequence of our main theorem.In more detail, the article is organized as follows: Section 2 provides preliminaries from the theory of hom-associative algebras, and of unital, hom-associative Ore extensions as developed in [3].
Section 3 deals with hom-modules over non-unital, hom-associative rings.Here, isomorphism theorems and the property of being hom-Noetherian are introduced.Whereas the notion of a hom-module was first introduced in [10], the theory developed here is new as far as we can tell.
Section 4 contains the proof of a hom-associative version of Hilbert's basis theorem, including as special cases a non-associative and the classical associative version.
Section 5 contains examples of unital, non-associative and hom-associative Ore extensions which are all Noetherian by the aforementioned theorem.

Preliminaries
Throughout this paper, by non-associative algebras we mean algebras which are not necessarily associative, including in particular associative algebras.We call a non-associative algebra A unital if there exists an element 1 ∈ A such that for any element a ∈ A, a • 1 = 1 • a = a.By non-unital algebras, we mean algebras which are not necessarily unital, including unital algebras by definition.
2.1.Hom-associative algebras.This section is devoted to restating some basic definitions and general facts concerning hom-associative algebras.
Definition 1 (Hom-associative algebra).A hom-associative algebra over an associative, commutative, and unital ring R, is a triple (M, •, α) consisting of an Rmodule M , a binary operation • : M ×M → M linear over R in both arguments, and an R-linear map α : The map α is referred to as the twisting map.Remark 1.A hom-associative algebra over R is in particular a non-unital, nonassociative R-algebra, and in case α is the identity map, a non-unital, associative Ralgebra.Moreover, any non-unital, non-associative R-algebra is a hom-associative R-algebra with twisting map equal to the zero map.Remark 2. If A is a unital, hom-associative algebra, then α is completely determined by α(1) since α(a) = α(a)(1 • 1) = a • α(1) for any a ∈ A.

Definition 2 (Morphism of hom-associative algebras). A morphism from a homassociative R-algebra
Definition 3 (Hom-associative subalgebra).Let A := (M, •, α) be a hom-associative algebra and N a submodule of M that is closed under the multiplication • and invariant under α.The hom-associative algebra (N, •, α| N ) is said to be a hom-subalgebra of A.
Definition 4 (Hom-ideal).A right (left) hom-ideal of a hom-associative R-algebra A is an R-submodule I of A such that α(I) ⊆ I, and for all a ∈ A, i ∈ I, i • a ∈ I (a • i ∈ I).If I is both a left and a right hom-ideal, it is simply called a hom-ideal.
Note that a hom-ideal of a hom-associative algebra A is in particular a homsubalgebra of A.
Remark 3. In case a hom-associative algebra has twisting map equal to the identity map or the zero map, a right (left) hom-ideal is simply a right (left) ideal.
Definition 5 (Hom-associative ring).A hom-associative ring is a hom-associative algebra over the ring of integers.

2.2.
Unital, non-associative Ore extensions.In this section, we recall from [3] some basic definitions and results concerning unital, non-associative Ore extensions.We denote by N the set of all non-negative integers, and by N >0 the set of all positive integers.Let R be a unital, non-associative ring, δ : R → R and σ : R → R additive maps such that σ(1) = 1 and δ(1) = 0.As a set, a unital, non-associative Ore extension of R, written R[X; σ, δ], consists of all formal sums i∈N a i X i , called polynomials, where only finitely many a i ∈ R are nonzero.We endow R[X; σ, δ] with the following addition and multiplication, holding for any m, n ∈ N and a i , b i ∈ R: Here, π m i , referred to as a π function, denotes the sum of all m i possible compositions of i copies of σ and m − i copies of δ in arbitrary order.For instance, We also define π 0 0 := id R and π m i ≡ 0 whenever i < 0, or i > m.The identity element 1 in R also becomes an identity element in R[X; σ, δ] upon identification with 1X 0 .We also think of X as an element of R[X; σ, δ] by identifying it with the monomial 1X.At last, defining two polynomials to be equal if and only if their corresponding coefficients are equal and imposing distributivity of the multiplication over addition make R[X; σ, δ] a unital, non-associative, and non-commutative ring.By identifying any a ∈ R with aX Definition 7 (σ-derivation).Let R be a unital, non-associative ring where σ is a unital endomorphism and δ an additive map on R.
Lemma 1 (Properties of π functions).Let R be a unital, non-associative ring, σ a unital endomorphism and δ a σ-derivation on R.Then, in R[X; σ, δ], the following hold for all a, b ∈ R and l, m, n ∈ N: A proof of (i) in the associative setting can be found in [11].However, the proof makes no use of associativity, so we can conclude that (i) holds in the non-associative setting as well.
Regarding (ii), we first recall that π m+1 l consists of the sum of all m+1 l possible compositions of l copies of σ and m + 1 − l copies of δ.Therefore, we can split the sum into a part containing σ innermost (outermost) and a part containing δ innermost (outermost).When l = 0, we immediately see that the result holds as π m −1 := 0. When l > m, π m l := 0, and in case also , so we can conclude that (ii) holds when l = 0 and when l > m.For the remaining case 1 ≤ l ≤ m, we use the recursive formula for binomial coefficients m+1 l = m l−1 + m l and simply count the terms in the two parts of the sum.
For any unital, hom-associative ring R with twisting map α, we extend α homogeneously to an additive map on R[X; σ, δ] by putting α i∈N a i X i := i∈N α(a i )X i for any a i ∈ R. The next proposition makes use of this construction.

Hom-module theory
In this section, we develop the theory of hom-modules over non-unital, homassociative rings.Most of the proofs are nearly identical to the classical proofs of the associative case.We have, however, provided them for completeness.

Basic definitions and theorems.
Definition 8 (Hom-module).Let R be a non-unital, hom-associative ring with twisting map α R , multiplication written with juxtaposition.Let M be an additive group with a group homomorphism α M : M → M , also called a twisting map.A right R-hom-module M R consists of M and an operation • : M × R → M , called scalar multiplication, such that for all r 1 , r 2 ∈ R and m 1 , m 2 ∈ M , the following hold:

associativity). (M3)
A left R-hom-module is defined analogously and written R M .
For the sake of brevity, we also allow ourselves to write M in case it does not matter whether it is a right or a left R-hom-module, and simply call it an Rhom-module.Furthermore, any two right (left) R-hom-modules are assumed to be equipped with the same twisting map α R on R.
Definition 10 (Hom-submodule).Let M be a right (left) R-hom-module.An R-hom-submodule, or just hom-submodule, is an additive subgroup N of M that is closed under scalar multiplication and invariant under α M .N is then a right (left) R-hom-module with twisting maps α R and α N , the latter being given by the restriction of α M to N .We denote that N is a hom-submodule of M by N ≤ M or M ≥ N , and in case N is a proper subgroup of M , by N < M or M > N .
Proposition 2 (Image and preimage under hom-module morphism).Let f : M → M ′ be a morphism of right (left) R-hom-modules, N ≤ M and N ′ ≤ M ′ .Then f (N ) and f −1 (N ′ ) are hom-submodules of M ′ and M , respectively.
Proof.We see that f (N ) and f −1 (N ′ ) are additive subgroups when considering f as a group homomorphism.Let r ∈ R and a ′ ∈ f (N ) be arbitrary.Then there is some a ∈ N such that a ′ = f (a), so The left case is analogous.

Proposition 3 (Intersection of hom-submodules). The intersection of any set of hom-submodules of a right (left) R-hom-module is a hom-submodule.
Proof.We show the case of right R-hom-modules; the case of left R-hom-modules is analogous.Let N = ∩ i∈I N i be an intersection of hom-submodules N i of a right R-hom-module M , where I is some index set.Take any a, b ∈ N and j ∈ I. Since a, b ∈ N j and N j is an additive subgroup, (a − b) ∈ N j , and therefore (a Definition 11 (Generating set of hom-submodule).Let S be a nonempty subset of a right (left) R-hom-module M .The intersection N of all hom-submodules of M that contain S is called the hom-submodule generated by S, and S is called a generating set of N .If there is a finite generating set of N , then N is called finitely generated.Remark 6.The hom-submodule N of a right (left) R-hom-module M generated by a nonempty subset S is the smallest hom-submodule of M that contains S in the sense that any other hom-submodule of M that contains S also contains N .
Proposition 4 (Union of hom-submodules in an ascending chain).Let M be a right (left) R-hom-module, and consider an ascending chain i=1 N i by N , and let a, b ∈ N .Then a ∈ N j and b ∈ N k for some j, k ∈ N >0 , and since so a • r ∈ N for the right case, and analogously for the left case.Finally, Proposition 5 (Sum of hom-submodules).Let M be a right (left) R-hom-module and N 1 , N 2 , . . ., N k any finite number of hom-submodules of M .Then Proof.We prove the right case; the left case is analogous.
Corollary 1 (The modular law for hom-modules).Let M be a right (left) R-hommodule, and M 1 , M 2 , and Proof.The modular law holds for M 1 , M 2 and M 3 when considered as additive groups.By Proposition 3 and Proposition 5, the intersection and sum of any two hom-submodules of M are also hom-submodules of M , and hence the modular law holds for M 1 , M 2 and M 3 as hom-modules as well.
Proposition 6 (Direct sum of hom-modules).Let M 1 , M 2 , . . ., M k be any finite number of right R-hom-modules.Endowing the (external) direct sum M := with the following scalar multiplication and twisting map on M , makes it a right R-hom-module: Here, (m 1 , m 2 , . . ., m k ) ∈ M , r ∈ R, and α Mi is the twisting map on M i for Proof.Since M is an additive group, what is left to check is that α M is a group homomorphism, i.e. an additive map, and that (M1), (M2) and (M3) in Definition 8 holds.This is readily verified by routine calculations.
An analogous result holds for left R-hom-modules.
Corollary 2 (Associativity of the direct sum).For any right (left) R-hom-modules M 1 , M 2 , and Proof.We prove the right case of the first isomorphism.The proof of the second isomorphism is similar, as are all the left cases.Considered as additive groups, Proposition 7 (Quotient hom-module).Let M R be a right R-hom-module with twisting map α M on M .Let N R ≤ M R and consider the additive groups M and N of M R and N R , respectively.Form the quotient group M/N with elements of the form m + N for m ∈ M .Then M/N becomes a right R-hom-module when endowed with the following twisting map and scalar multiplication for m ∈ M and r ∈ R: Proof.First, let us make sure that the scalar multiplication and twisting map are both well-defined.To this end, take two arbitrary elements of M/N .They are of the form By straightforward calculations, one readily verifies that the three hom-module axioms (M1), (M2), and (M3) in Definition 8 hold.
Again, an analogous result holds for left R-hom-modules as well.Proof.We know that π is a surjective group homomorphism, and for any m • r for the right case, and analogously for the left case.We also have that π(α Corollary 4 (Hom-submodules of quotient hom-modules).Let M be a right (left) R-hom-module with N ≤ M .If L is a hom-submodule of M/N , then L = K/N for some hom-submodule K of M that contains N .
Proof.Let L be a hom-submodule of M/N .Using the natural projection π : M → M/N from Corollary 3, we know that K = π −1 (L) is a hom-submodule of M since it is the preimage of a morphism of hom-submodules (cf.Proposition 2).By the surjectivity of π, π Theorem 1 (The first isomorphism theorem for hom-modules).Let f : M → M ′ be a morphism of right (left) R-hom-modules.Then ker f is a hom-submodule of M , im f is a hom-submodule of M ′ , and M/ ker f ∼ = im f .
Proof.We prove the case of right R-hom-modules; the case of left R-hom-modules is analogous.By definition, ker f is the preimage of the hom-submodule 0 of M ′ , and hence it is a hom-submodule of M by Proposition 2. Now, im f = f (M ), so by the same proposition, im f is a hom-submodule of M ′ .The map g : M/ ker f → im f defined by g(m We also see that ker f = N/L, so using Theorem 1, (M/L)/ ker f = (M/L)/(N/L) ∼ = im f = M/N .

3.2.
The hom-Noetherian conditions.Recall that a family F of subsets of a set S satisfies the ascending chain condition if there is no properly ascending infinite chain S 1 ⊂ S 2 ⊂ . . . of subsets from F .Furthermore, an element in F is called a maximal element of F provided there is no element of F that properly contains that element.Proof.The following proof is an adaptation of a proof that can be found in [5] to the hom-associative setting.
(NM1) =⇒ (NM2): Let F be a nonempty family of hom-submodules of M that does not have a maximal element and pick an arbitrary hom-submodule S 1 in F .Since S 1 is not a maximal element, there exists some S 2 ∈ F such that S 1 < S 2 .Now, S 2 is not a maximal element either, so there exists some S 3 ∈ F such that S 2 < S 3 .Continuing in this manner we get an infinite chain of hom-submodules S 1 < S 2 < . . ., which proves the contrapositive statement.
(NM2) =⇒ (NM3): Assume (NM2) holds, let N be an arbitrary hom-submodule of M , and G the family of all finitely generated hom-submodules of N .Since the zero module is a hom-submodule of N that is finitely generated, G is clearly nonempty and thus contains a maximal element L by assumption.If N = L, we are done, so assume the opposite and take some n ∈ N \L.Now, let K the hom-submodule of N generated by the set L ∪ {n}.Then K is finitely generated as well, so K ∈ G.Moreover, L < K, which is a contradiction since L is a maximal element in G. Therefore, N = L, and N is finitely generated.
(NM3) =⇒ (NM1): Assume (NM3) holds, let T 1 ≤ T 2 ≤ . . .be an ascending chain of hom-submodules of M , and T = ∪ ∞ i=1 T i .By Proposition 4, T is a hom-submodule of M , and hence it is finitely generated by some set S which by Definition 11 is contained in T .Moreover, since S is finite, it needs to be contained in T j for some j ∈ N >0 .However, T j = T by Remark 6, so T k = T j for all k ≥ j, and hence the ascending chain condition holds.
Definition 12 (Hom-Noetherian module).A right (left) R-hom-module is called hom-Noetherian if it satisfies the three equivalent conditions of Proposition 8 on its hom-submodules.
Appealing to Remark 5, all properties that hold for right (left) hom-modules necessarily also hold for hom-associative rings, replacing "hom-submodule" by "right (left) hom-ideal".Hence we have the following:

Corollary 5 (The hom-Noetherian conditions for hom-associative rings). Let R be a non-unital, hom-associative ring. Then the following conditions are equivalent: (NR1) R satisfies the ascending chain condition on its right (left) hom-ideals. (NR2) Any nonempty family of right (left) hom-ideals of R has a maximal element. (NR3) Any right (left) hom-ideal of R is finitely generated.
Definition 13 (Hom-Noetherian ring).A non-unital, hom-associative ring R is called right (left) hom-Noetherian if it satisfies the three equivalent conditions of Corollary 5 on its right (left) hom-ideals.If R satisfies the conditions on both its right and its left hom-ideals, it is called hom-Noetherian.Remark 7. If the twisting map is either the identity map or the zero map, a right (left) hom-Noetherian ring is simply a right (left) Noetherian ring (cf.Remark 3).If R is a unital, hom-associative ring, then all right (left) ideals of R are actually right (left) hom-ideals; if I is a right ideal of R and i ∈ I, then α(i) = α(i) • (1 • 1) = i • α(1) ∈ I, and similarly for the left case.In particular, R is right (left) hom-Noetherian if and only if R is right (left) Noetherian.
Proof.It is sufficient to prove this for any of the three equivalent conditions (NM1), (NM2), or (NM3) in Proposition 8, so let us choose (NM2).To this end, let f : M → M ′ be a surjective morphism of right (left) R-hom-modules where M is hom-Noetherian.Let F ′ be a nonempty family of right (left) hom-submodules of M ′ .Now, consider the corresponding family in M , By the surjectivity of f , this family is nonempty, and since M is Noetherian, it has a maximal element f −1 (N ′ 0 ) for some N ′ 0 ∈ F ′ .We would like to show that N ′ 0 is a maximal element of F ′ .Assume there exists an element N ′ ∈ F ′ such that N ′ 0 < N ′ .We know that the operation of taking preimages under any function preserves inclusions sets.We also know that the preimage of any hom-submodule is again a hom-submodule by Proposition 2, so taking preimages under a hommorphism preserves the inclusions on the hom-submodules, and therefore Proof.This is again an adaptation of a proof that can be found in [5] to the homassociative setting.
(=⇒) : Assume M is hom-Noetherian and N ≤ M .Then any hom-submodule of N is also a hom-submodule of M , and hence it is finitely generated, and N therefore also hom-Noetherian.If L 1 ≤ L 2 ≤ . . . is an ascending chain of hom-submodules of M/N , then from Corollary 4, each (⇐=) : Assume M/N and N are hom-Noetherian.Let S 1 ≤ S 2 ≤ . . .be an ascending chain of hom-submodules of M .By Proposition 3, S i ∩ N is a homsubmodule of N for every i ∈ N >0 , and furthermore S i ∩ N ≤ S i+1 ∩ N .We thus have an ascending chain S 1 ∩ N ≤ S 2 ∩ N ≤ . . . of hom-submodules of N .By Proposition 5, S i + N is a hom-submodule of M , and moreover, N = 0 + N is a hom-submodule of S i + N , so we can consider (S i + N ) /N .Now, (S i + N ) /N ≤ (S i+1 + N ) /N by Corollary 4, so we have an ascending chain (S 1 + N )/N ≤ (S 2 + N )/N ≤ . . . of hom-submodules of M/N .Since both N and M/N are hom-Noetherian, there is some k such that S j ∩N = S k ∩N and (S j +N )/N = (S k +N )/N for all j ≥ k.The latter equation implies that for any s j ∈ S j and n ∈ N , there are s k ∈ S k and n ′ ∈ N such that (s j + n) + N = (s k + n ′ ) + N .Hence x := ((s j + n) − (s k + n ′ )) ∈ N , and therefore s j + n = (s k + (x + n ′ )) ∈ (S k + N ), so that (S j + N ) ≤ (S k + N ), and by a similar argument, (S k + N ) ≤ (S j + N ), so S j + N = S k + N for all j ≥ k.Using this and the modular law for hom-modules for all j ≥ k, and hence M is hom-Noetherian.
Corollary 6 (Finite direct sum of hom-Noetherian modules).Any finite direct sum of hom-Noetherian modules is hom-Noetherian.
Proof.We prove this by induction.
Induction step (∀k ∈ N >1 (P(k by Corollary 2. The latter of the two is hom-Noetherian by the base case, and by Proposition 9 the former as well.

Hilbert's basis theorem for hom-associative Ore extensions
In this section, we consider unital, non-associative Ore extensions R[X; σ, δ] over some unital, non-associative ring R. First, recall from Proposition 1 that if R is hom-associative, then a sufficient condition for R[X; σ, δ] to be hom-associative is that σ is a unital endomorphism and δ a σ-derivation that both commute with the twisting map α of R, the latter extended homogeneously to R[X; σ, δ].Moreover, from Remark 7, for unital, hom-associative rings, being right (left) hom-Noetherian is the same as being right (left) Noetherian.Also recall that the associator is the map (•, •, •) : R×R×R → R defined by (r, s, t) = (r •s)•t−r •(s•t) for any r, s, t ∈ R. The left, middle, and right nucleus of R are denoted by N l (R), N m (R), and N r (R), respectively.As sets, they are defined as ), and hence also N (R), are all associative subrings of R.
Proposition 11 (Associator of X k ).Let R[X; σ, δ] be a unital, non-associative Ore extension of a unital, non-associative ring R. Assume σ is a unital endomorphism and δ a σ-derivation on R. Then X k ∈ N (R[X; σ, δ]) for any k ∈ N.
Proof.By identifying X 0 with 1 ∈ R, X 0 ∈ N (R[X; σ, δ]).We now wish to show that X ∈ N (R[X; σ, δ]).In order to do so, we must show that X associates with all polynomials in R[X; σ, δ].Due to distributivity, it is however sufficient to prove that X associates with arbitrary monomials aX m and bX n in R[X; σ, δ].To this end, first note that aX m • X = i∈N (a • π m i (1)) X i+1 = aX m+1 since σ is unital by assumption, and δ(1) = 0 by Remark 4.Then, ).Also, by using (ii) in Lemma 1, ) is a ring it also contains all powers of X, so Proposition 12 (Hom-modules of R[X; σ, δ]).Let R be a unital, Noetherian, homassociative ring with twisting map α, σ a unital endomorphism and δ a σ-derivation that both commute with α.Extend α homogeneously to R[X; σ, δ].Then, for any m ∈ N, Proof.Let us prove the right case; the left case is similar, but slightly simpler.Put M = m i=0 X i R. First note that M really is a subset of R[X; σ, δ], where the elements are of the form m i=0 1X i • r i X 0 with r i ∈ R. When identifying 1X i with X i and r i with r i X 0 , this gives us elements of the form m i=0 X i • r i .Using this identification also allows us to write the multiplication in R, which in Definition 8 is done by juxtaposition, by "•" instead.The purpose of this is do be consistent with our previous notation.
Since distributivity follows from that in R[X; σ, δ], it suffices to show that the multiplication in R[X; σ, δ] is a scalar multiplication, and that we have twisting maps α M and α R that give us hom-associativity.To this end, for any r ∈ R and any element in M (which is of the form described above), by using Proposition 11, and the latter is clearly an element of M .Now, we claim that M is invariant under the homogeneously extended twisting map on R[X; σ, δ].To follow the notation in Definition 8, let us denote this map when restricted to M by α M , and that of R by α R .Then, by using the additivity of α M and α R , as well as the fact that the latter commutes with δ and σ, we get which again is an element of M .At last, let r, s ∈ R be arbitrary.Then, We see that f is additive, and for any r ∈ R, we have f , which shows that f is a morphism of two right R-hom-modules.Moreover, f is surjective, and so by Proposition 9, M is hom-Noetherian.
Proof.That σ −1 is an automorphism and −δ • σ −1 a σ −1 -derivation on R op is an exercise in [5] that can be solved without any use of associativity.Now, since α commutes with δ and σ, for any ), which proves the first and second statement.
For the third statement, let us start by putting We claim that f is an isomorphism of hom-associative rings.First, note that an arbitrary element of S ′ by definition is of the form p := m i=0 a i X i for some m ∈ N and a i ∈ R op .Then, The second last step also shows that m i=0 RX i ⊆ m i=0 X i R as sets, and a similar calculation shows that m i=0 X i R ⊆ m i=0 RX i , so that as sets, where the implication comes from comparing coefficients with the left-hand side, which is equal to zero.Let us prove by induction that r j = r ′ j for 0 ≤ j ≤ m.Put k = m − j, where m is fixed, and consider the statement P(k) : Base case (P(0)) : k = 0 ⇐⇒ j = m, so using that σ is an automorphism, 0 Induction step (For 0 ≤ k ≤ m : (P(k) → P(k + 1))): By putting j = m − (k + 1) and then using the induction hypothesis, 0 which implies 0 = r m−(k+1) = r ′ m−(k+1) .Hence r j = r ′ j for 0 ≤ j ≤ m, so j=0 r ′ j X j , proving that f is injective.Additivity of f follows immediately from the definition by using distributivity.Using additivity also makes it sufficient to consider only two arbitrary monomials aX m and bX n in S when proving that f is multiplicative.To this end, let us use the following notation for multiplication in S: aX m • bX n := i∈N (a • op πm i (b)) X i+n , and then use induction over n and m; Base case (P(0, 0)) : Proposition 11, and so Induction step over m (∀(m, n) ∈ N×N (P(m, n) → P(m+1, n))): We know that X ∈ N (S ′op ) ∩ N (S) by Proposition 11, and so by a straightforward calculation, f aX m+1 • bX n = f aX m+1 • op f (bX n ).Now, according to Definition 2 with R[X; σ, δ] considered as a hom-associative algebra over the integers, we are done if we can prove that f • α = α • f for the homogeneously extended map α.Since both α and f are additive, it again suffices to prove that f ((α(aX m )) = α (f (aX m )) for some arbitrary monomial aX m in R[X; σ, δ].This is verified by a simple computation.
Theorem 4 (Hilbert's basis theorem for hom-associative rings).Let R be a unital, hom-associative ring with twisting map α, σ an automorphism and δ a σ-derivation that both commute with α.
Proof.This proof is an adaptation of a proof in [5] to the hom-associative setting.Let us begin with the right case, and therefore assume that R is right Noetherian.We wish to show that any right ideal of R[X; σ, δ] is finitely generated.Since the zero ideal is finitely generated, it is sufficient to show that any nonzero right ideal . ., r 0 ∈ R}, i.e.J consists of the zero element and all leading coefficients of polynomials in I.We claim that J is a right ideal of R. First, one readily verifies that J is an additive subgroup of R. Now, let r ∈ J and a ∈ R be arbitrary.Then there is some polynomial p = rX d + [lower order terms] in I.
Since R is right Noetherian and J is a right ideal of R, J is finitely generated, say by {r 1 , . . ., r k } ⊆ J.All the elements r 1 , . . ., r k are assumed to be nonzero, and moreover, each of them is a leading coefficient of some polynomial p i ∈ I of degree n i .Put n = max(n 1 , . . ., n k ).Then each r i is the leading coefficient of p i • X n−ni = r i X ni • X n−ni + [lower order terms] = r i X n + [lower order terms], which is an element of I of degree n.
Let N := n−1 i=0 RX i .Then calculations similar to those in the proof of the third statement of Lemma 2 show that as sets, N , N is then a hom-Noetherian right R-hom-module.Now, since I is a right ideal of the ring R[X; σ, δ] which contains R, in particular, it is also a right R-hom-module.By Proposition 3, I ∩ N is then a hom-submodule of N , and since N is a hom-Noetherian right R-hom-module, I ∩ N is finitely generated, say by the set {q 1 , q 2 , . . ., q t }.
Let I 0 be the right ideal of R[X; σ, δ] generated by Since all the elements in this set belong to I, we have that I 0 ⊆ I.We claim that I ⊆ I 0 .In order to prove this, pick any element p ′ ∈ I.

Base case (P(n)
On the other hand, the generating set of I ∩ N is a subset of the generating set of I 0 , so I ∩ N ⊆ I 0 , and therefore p ′ ∈ I 0 .
Induction step (∀m ≥ n (P(m) → P(m + 1))): Assume deg p ′ = m ≥ n and that I 0 contains all elements of I with deg < m.Does I 0 contain all elements of I with deg < m + 1 as well?Let r ′ be the leading coefficient of p ′ , so that we have p ′ = r ′ X m + [lower order terms].Since p ′ ∈ I by assumption, r ′ ∈ J.We then claim that r and some a ij1 , a ij2 , . . ., a ijk ′′ ∈ R. First, we note that since J is generated by {r 1 , r 2 , . . ., r k }, it is necessary that J contains all elements of that form.Secondly, we see that subtracting any two such elements or multiplying any such element from the right with one from R again yields such an element, and hence the set of all elements of this form is not only a right ideal containing {r 1 , r 2 , . . ., r k }, but also the smallest such to do so.
Recalling that p and by iterating this multiplication from the right, we set is a generator of I 0 , c ij is an element of I 0 as well, and therefore also q : lower order terms].However, as I 0 ⊆ I, we also have that q ∈ I, and since p ′ ∈ I, (p ′ −q) ∈ I. Now, p ′ = r ′ X m +[lower order terms], so deg(p ′ − q) < m, and therefore (p ′ − q) ∈ I 0 .This shows that p ′ = (p ′ − q) + q is an element of I 0 as well, and thus I = I 0 , which is finitely generated.
For the left case, first note that any hom-associative ring S is right (left) Noetherian if and only if S op is left (right) Noetherian, due to the fact that any right (left) ideal of S is a left (right) ideal of S op , and vice versa.Now, assume that R is left Noetherian.Then, R op is right Noetherian, and using (i) and (ii) in Lemma 2, σ −1 is an automorphism and −δ • σ −1 a σ −1 -derivation on R op that both commute with α.Hence, by the previously proved right case One verifies that surjective morphisms between hom-associative rings preserve the Noetherian conditions (NR1), (NR2), and (NR3) in Corollary 5 by examining the proof of Proposition 9, changing the module morphism to that between rings instead, and "submodule" to "ideal".Therefore, R[X; σ, δ] op is right Noetherian, so R[X; σ, δ] is left Noetherian.Remark 8.By putting α = id R in Theorem 4, we recover the classical Hilbert's basis theorem for Ore extensions.
Corollary 7 (Hilbert's basis theorem for non-associative rings).Let R be a unital, ring, σ an automorphism and δ a σ-derivation on R. If R is right (left) Noetherian, then so is R[X; σ, δ].

Examples
Here we provide some examples of unital, non-associative and hom-associative Ore extensions which are all Noetherian by the above theorem.First, recall that there are, up to isomorphism, only four normed, unital division algebras over the real numbers: the real numbers themselves, the complex numbers, the quaternions (H), and the octonions (O) [16].The largest of the four are the octonions, and while sharing the property of not being commutative with the quaternions, the octonions are the only ones that are not associative.All of the four algebras above are Noetherian, and hence also all iterated Ore extensions of them: let D be any unital division algebra, and I any nonzero right ideal of D. If a ∈ D is an arbitrary nonzero element, then 1 = a • a −1 ∈ I, so I = D, and analogously for the left case.As an ideal of itself, D is finitely generated (by 1, for instance), as is the zero ideal.
The derivations on any normed division algebra D is a linear combination of derivations x ∈ D [15].These derivations are called inner, and in particular, all derivations on H are of the form [a, •] for some a ∈ H.
Given a unital and associative algebra A with product • over a field of characteristic different from two, one may define a unital and non-associative algebra A + by using the Jordan product {•, •} : A + → A + .This is given by {a, b} := 1 2 (a • b + b • a) for any a, b ∈ A. A + is then a Jordan algebra, i.e. a commutative algebra where any two elements a and b satisfy the Jordan identity, {{a, b}{a, a}} = {a, {b, {a, a}}}.Since inverses on A extend to inverses on A + , one may infer that if A = H, then A + is also Noetherian.Using the standard notation i, j, k for the quaternion units in H with defining relation i 2 = j 2 = k 2 = ijk = −1, one can see that H + is not associative as e.g.(i, i, j) H + := {{i, i}, j} − {i, {i, j}} = −j.
Example 2. Let [j, •] H be the inner derivation on H induced by j.Any derivation on H is also a derivation on H + , and so we may form the unital, non-associative Ore extension H + [X; id H , [j, •] H ] which is Noetherian due to Corollary 7.

Example 3.
From the Jordan identity one may infer that a map δ a,b : J → J defined by δ a,b (x) := (a, x, b) J for any a, b, x ∈ J where J is a Jordan algebra, is a derivation, called an inner derivation.On H + one could for instance take a = i and b = j, resulting in δ i,j (x) = {{i, x}, j} − {i, {x, j}} for any x ∈ H + .Then H + [X; id H , δ i,j ] is a unital, non-associative Ore extension which is Noetherian by Corollary 7.
Example 4. Take any derivation on O, e.g.δ i,j defined by δ i,j (x) := [[i, j], x] − 3(i, j, x) for any x ∈ O. Then O[X; id O , δ i,j ] is a unital, non-associative Ore extension which is Noetherian by Corollary 7.
Example 5.One may define an octonionic Weyl algebra A(O) as the tensor product of the usual Weyl algebra A(R) over R, and O. A(O) is then a free module of finite rank over A(R), and hence it is Noetherian.One may also define an octonionic Weyl algebra as an iterated Ore extension of O. Using this latter approach, let us first mention that for any unital, non-associative ring R, the non-associative Weyl algebra over R was introduced in [ Example 6.For any q ∈ R\{0, 1}, one may define an octonionic q-Weyl algebra A q (O) as the tensor product of the usual q-Weyl algebra A q (R) over R, and O.In particular, A q (O) is a free module of finite rank over A q (R), and hence it is Noetherian.Alternatively, one may see that A q (O) is Noetherian by noting that it both δ and σ, and therefore U [X; σ, δ] is Noetherian.Here, one could e.g.take R to be any base ring together with σ and δ from the previous examples.

Corollary 3 (
The natural projection).Let M be a right (left) R-hom-module with N ≤ M .Then the natural projection π : M → M/N defined by π(m) = m + N for any m ∈ M is a surjective morphism of hom-modules.

Proposition 8 (
The hom-Noetherian conditions for hom-modules).Let M be a right (left) R-hom-module.Then the following conditions are equivalent: (NM1) M satisfies the ascending chain condition on its hom-submodules.(NM2) Any nonempty family of hom-submodules of M has a maximal element.(NM3) Any hom-submodule of M is finitely generated.
and M ′ is hom-Noetherian.Proposition 10 (Hom-Noetherian condition on quotient hom-module).Let M be a right (left) R-hom-module, and N ≤ M .Then M is hom-Noetherian if and only if M/N and N are hom-Noetherian.