Two-dimensional perfect evolution algebras over domains

We will study evolution algebras A that are free modules of dimension two over domains. We start by making some general considerations about algebras over domains: They are sandwiched between a certain essential D-submodule and its scalar extension over the field of fractions of the domain. We introduce the notion of quasiperfect algebras and we characterize the perfect and quasiperfect evolution algebras in terms of the determinant of its structure matrix. We classify the two-dimensional perfect evolution algebras over domains parametrizing the isomorphism classes by a convenient moduli set.


Introduction
There is a large number of publications studying 2-dimensional evolution algebras ( [6], [9], [8], [5], [1], [4], [2], etc.), among them we could highlight those works that deal with their classification.So, in the works of [8] and [6] the evolution algebras of dimension two over the reals and complex respectively are classified, the paper [5] contains the classification of the evolution algebras of dimension two and three over the field K , where K = R or C. The work [1] is addressed to the case in which the ground field K is algebraically closed.In [2] the classification of the evolution algebras of dimension two over arbitrary fields is provided.We can find more information about the evolution of the research in the field of Tiam's evolution algebras in [3].We are contributing our grain of sand by studying the perfect two-dimensional evolution algebras over domains.So, in this paper, the word domain will stand for a commutative ring such that xy = 0 implies x = 0 or y = 0.If D is a domain, an evolution algebra over D is a D-algebra which is free as D-module and has a basis such that the product of any two different elements is zero.Such basis is called a natural basis.Of course, any one-dimensional D-algebra is an evolution algebra.Evolution algebras over domains are much more involved than evolution algebras over fields.For instance, there are two one-dimensional evolution algebras over fields up to isomorphism: the ground field with zero multiplication and the ground field with its usual multiplication.But if you consider a domain D and a one-dimensional evolution Dalgebra there are more isomorphic classes of evolution algebras.Ruling out the trivial one, we have define in D a multiplication with One of the elements of this study is the use of moduli sets from certain classes of algebras: the idea is to parametrize the algebras of a class by tuples of parameters ranging in a given space.It turns out that in some cases the tuples range in curves or surfaces or other varieties.The different algebras in the same isomorphic class may happen to be in a curve of an affine plane and the different curves fill the space modulo the restrictions on the parameters imposed by the class of algebras.This may be seen as a bundle in the category of sets.
This paper is organized as follows.In Section 2 we prove some results on algebras over domains in Proposition 1 and Corollary 1.We introduce the class of quasiperfect algebras and give necessary and sufficient conditions for an evolution algebra to be perfect or quasiperfect in terms of natural bases (in Proposition 2).Next, we define the required terminology to be able to classify our D-algebras using moduli sets.Finally, in Section 4 we associate a colored directed graph to a quasiperfect evolution D-algebra and give the classification theorem of perfect two-dimensional evolution algebras over a domain (Theorem 1).

Preliminaries and previous results
If D is a (commutative) domain we will denote by Q := Q(D) the field of fractions of D. For any D-module M we will construct with the usual operations of sum, product and product by elements of Q.To be more specific we consider the set of couples (m, d) with m ∈ M, 0 = d ∈ D modulo the equivalence relation Then we denote the equivalence class of (m, d) in the usual way: m d and M Q is a Q-vector space relative to the usual sum of fractions and 1 .Usually we will denote m 1 = m so that the elements of M and their images in M Q will be identified.Unless otherwise stated, we will work allthrough this paper with torsion-free D-modules.
A well known property of M is that a set of vectors {e i } ⊂ M is linearly independent if and only if its image in M Q is D-linearly independent.As well if {u i } i∈I is linearly independent in the Q-vector space M Q and u i = m i d i for any i ∈ I, then the set of numerators {m i } i∈I is linearly independent in M. We also have: Proposition 1.Let D be a domain, Q its field of fractions, M a torsion-free D-module and M Q the Q-module of fractions of M.
(a) If {u i } i∈I is a basis of M Q (as a Q-vector space) and u i = m i d i for i ∈ I, then {m i } is also a basis of M Q and a maximal linearly independent subset of M.
Proof.For the first assertion, take into account that the set of the m i 's is linearly independent also in M Q because if we have Let us prove now that {m i } is a maximal among the linearly independent subsets of M: if {m i } T for a linearly independent subset T ⊂ M, then T contains properly a basis of M Q and is linearly independent, a contradiction.For the second assertion, take a maximal linearly independent set {m i } ⊂ M. We know that {m i } is linearly independent also in M Q .If this set is not a basis, there is some x ∈ M Q such that {m i } ∪ {x} is again linearly independent.If x = z d then {m i } ∪ {z} is a linearly independent subset of M contradicting the maximality of {m i }.Reciprocally, if {m i } ⊂ M is a basis of M Q we know that {m i } is linearly independent in M. To prove the third assertion take a nonzero submodule N of M. We must prove that N ∩ (⊕ i Dm i ) = 0. Take 0 = n ∈ N. Since {m i } is a basis of M Q we have dn = i d i m i for some d, d i ∈ D (and d = 0).Thus 0 = dn ∈ N ∩ (⊕ i Dm i ).Finally we prove the fourth assertion.We have u i = a j i e j (using Einstein summation convention) for any i ∈ I.If det[(a j i )] ∈ D × , then there are scalars b j i ∈ D such that e i = b j i e j for any i.Whence {u i } i∈I is a basis of M. Reciprocally, if {u i } i∈I turns out to be a basis we may write e i = b j i e j for suitable scalars b j i ∈ D. But then the matrices (a j i ) and (b j i ) have product 1, that is, (a j i )(b j i ) = 1.This implies that the determinant of each such matrix is an invertible element of D.
Corollary 1.Let A be a D-algebra, then there is a maximal linearly independent subset {a i } of A (in fact a basis of the Q-vector space A Q ) such that A is contained as D-module in a sandwich and ⊕Da i is an essential D-module of A.
Proof.Take a basis {u i } of A Q as a Q-vector space.If u i = a i /d i then {a i } is a maximal linearly independent subset of A by Proposition 1.Also ⊕Da i is essential as a D-submodule of A by Lemma 1.Note that A is a torsion-free D-module: if da = 0 for some nonzero d ∈ D, write a = i q i u i as a linear combination of the u i 's.Then we have 0 = d i q i u i whence dq i = 0 for any i.Since d = 0 we have q i = 0.

In the situation above
An example of the situation described in Corollary 1 is given by taking D = Z and A = {( x 2 , y) : x, y ∈ Z} = Z( 1 2 , 0) ⊕ Z(0, 1).Then A is a two-dimensional free Z-module and Definition 1.An algebra A over a domain D will be termed quasiper- Let D be a domain and E a free D-module such that E is an evolution algebra with natural basis {e 1 , e 2 }.We will need the following proposition whose first item is exactly the same as in the case of evolution algebras over fields: Proposition 2. Assume that E is an evolution algebra over a domain D with a finite natural basis {e i } i∈I .Let ω j i ∈ D be the structure constants, that is, e 2 i = ω j i e j (using Einstein summation convention).Then we have: (1) E 2 = E if and only if the matrix (ω j i ) is invertible.Moreover, for any other natural basis {f i } i∈I there is a permutation σ of (2) E is quasiperfect if and only if the determinant of (ω j i ) is nonzero.As in the previous case, for any other natural basis {f i } i∈I there is a permutation σ of I such that Proof.From E = E 2 we deduce that e i = x j i e 2 j for any i ∈ I. Then i is an isomorphism whence E 2 = E .Assuming the perfection of E , if {f i } is another natural basis and we write f i = a j i e j , then for i = j we have 0 = f i f j = a k i e k a q j e q = a k i a q j e k e q = a k i a q j δ kq ω s k e s = a k i a k j ω s k e s whence a k i a k j ω s k = 0 for any s and any couple (i, j) with i = j.Since the matrix (ω j i ) is invertible, we consider its inverse matrix (ω j i ), so we have Thus a q i a q j = 0 for any q provided i = j.So in each column and each row of the matrix (a j i ) there is a unique nonzero element.Consequently, f i = k i e σ(i) for a certain permutation σ of I. Now, the coefficients k i are invertible in D since the determinant of the matrix of basis change is invertible.Let us prove now the second assertion.If we have

Corollary 2. Any perfect evolution algebra over a domain is quasiperfect.
Next, we prove a result which generalizes the situation of perfect evolution algebras over fields.
Lemma 1. Assume that E is a quasiperfect evolution algebra over a domain D with a finite natural basis {e i } i∈I .Then the following numbers do not depend on the natural basis chosen: (1) The number of nonzero entries in the structure matrix (ω j i ).
(2) The number of nonzero entries in the diagonal of (ω j i ).
(3) The number of invertible elements in (ω j i ).(4) The number of invertible elements in the diagonal of (ω j i ).Proof.For our original natural basis we have e 2 i = ω j i e j .Take now any other natural basis {f i }.There is a permutation σ of I such that So, the number of nonzero (respectively invertible) elements in the matrix (ω j i ) coincides with the number of nonzero (respectively invertible) elements in the matrix (τ j i ).Similarly for the diagonal elements.We can associate a colored graph to any quasiperfect evolution algebra over a domain.The key idea is of course that of [7] but slightly modified.We fix a natural basis {e i } of an evolution D-algebra E .Let ω j i be the structure constants.Then, we consider the graph whose vertices are in bijection with {e i } and we draw a black edge from vertex i to vertex j if ω j i ∈ D × .We draw a blue edge from vertex i to vertex j if ω j i ∈ D * \ D × .The (isomorphic class) of the associated graph does not depend on the chosen natural basis because of Lemma 1. So, the graph associated to an evolution algebra over a domain is a colored directed graph.Proposition 3. Let D be a domain and Q its field of fractions.If E is a quasiperfect evolution algebra over D of finite dimension and {m i } is maximal linearly independent subset of E such that m i m j = 0 for i = j, then {m i } is a natural basis of E .
Proof.Fix a natural basis {e i } of E .We know that {m i } is a basis of E Q by Proposition 1 and it is a natural basis.Then m i = k i e σ(i) for some k i ∈ Q × and a permutation σ.On the other hand, there are expressions m i = a j i e j where a j i ∈ D. Consequently a

The moduli set.
We have to introduce some terminology in order to present what we will call the moduli set for the different classes of algebras.Given a class C of algebras we will say that a set S is a moduli for C (or a moduli set) if there is a one-to-one correspondence between the isomorphic classes of algebras of C and the elements of S. In some occasions the moduli of a class will have an additional algebraic or geometric structure.In the category of sets, we recall that a bundle is an epimorphism π : E → B where E is called total set, B is the base set and for every b ∈ B, p −1 (b) is the fiber over b.By a cross section of π we understand a right inverse s : B → E. We have an example of bundle if we consider as the total set a class of algebras C and as the base set the set C/ ∼ = of isomorphic classes of the algebras in C.Then, π maps any algebra to its isomorphic class.The fiber of an element represents an isomorphic class of algebras.If we specify a cross section of the bundle, then this is equivalent to give a representative of the isomorphic class of any element of C. Thus, the classification problem of C under isomorphism consists just in giving a cross section of the corresponding bundle.The moduli set of the classification is the base set of the corresponding bundle.
3.1.Direct limits.Some of the cases of our classification are based on direct limits.If M is an abelian monoid and n ∈ Z we consider the direct system where (•) n is the homomorphism such that g → g n , then we will denote the direct limit of such system by lim →n M. Recall that this monoid can be described as follows: consider the sequence of monoids {M i } i∈N such that M i := M for any i.Then, in the disjoint union ⊔ i M i we define an equivalence relation: if x ∈ M i and y ∈ M j we say that x ∼ y if and only if y n k = x n h for some naturals k, h.We denote the equivalence class of g ∈ M by [g].So lim →n M is the quotient of ⊔ i M i modulo ∼.A particular case of this arises if we take M := D × /(D × ) [q] where (D × ) [q] := {x q : x ∈ D × } with q ∈ N * .This specific M is a group and its elements are equivalence classes λ with λ ∈ D × .We have λ = μ if and only if λ = µ r q for some r ∈ D × .We can consider lim →n M whose elements are the equivalence classes [ λ] with λ ∈ M.
3.2.Algebraic sets.Assume that we have an action D × × X → X where X is some subset of D n .So, for t ∈ D × and a = (a 1 , . . ., a n ) ∈ X we might have t • (a 1 , . . ., a n ) = (x 1 , . . ., x n ) where each x i is a polynomial in t with coefficients in D. Consequently, we may write x i = p i (t) where p i is the mentioned polynomial.The orbit of a under the above action is in a curve x i = p i (t).More precisely, the orbit is contained in the image of the map c : Q → Q n such that t → (p 1 (t), . . ., p n (t)) where Q is the field of fractions of D. Then the Zariski closure of the image of c is an algebraic set V ⊂ Q n and the orbit of a ∈ X is just V ∩ X.This setting will appear in our classification of evolution algebras.
(I) There is a family C of two-dimensional evolution algebras depending on two parameters, namely those algebras A(1, λ, µ) = D × D where D is a fixed domain and the product in the algebra being e 2 1 = e 2 , e 2 2 = λe 1 + µe 2 with λ ∈ D × and µ ∈ D * \ D × .As we will see in the next section the isomorphism conditions for algebras of C is A(1, λ, µ) ∼ = A(1, λ ′ , µ ′ ) if and only if there is some k ∈ D × such that λ ′ = k 3 λ and µ ′ = k 2 µ.This induces an action D × × X 0 → X 0 where X 0 = D × × (D * \ D × ) given by k • (λ, µ) = (k 3 λ, k 2 µ).Note that the isotropy subgroup of any (λ, µ) ∈ X 0 is trivial.This implies that the cardinal of each orbit agrees with that of D × .
For a fixed (λ, µ) ∈ X 0 we can consider c : The Zariski closure of the image of c is V (I), the algebraic set of zeros of the ideal I ⊳ Q[x, y] generated by the polynomial µ 3 x 2 − λ 2 y 3 .So, it is a curve c λ,µ of Q 2 .Thus the cardinal of the orbit of (λ, µ) agrees with that of the set of points of c λ,µ lying on X 0 .Any point of X 0 is in some curve c λ,µ , in fact, So, X 0 is the disjoint union of all sections c * λ,µ and we have a bundle (in the category of sets) p : X 0 → X 0 /D × in which p(λ, µ) = orb(λ, µ) can be identified with c * λ,µ .The fibers of this bundle are the points in one specific curve so the fibers represent classes of isomorphic algebras.Lemma 3. The cardinal of each orbit of X 0 /D × is |D × | and agrees with that of the set of points of the curve c λ,µ ≡ µ3 x 2 − λ 2 y 3 = 0 in D × × (D * \ D × ).The orbit set X 0 /D × is the base space of a bundle p : X 0 → X 0 /D × where the fibers represent classes of isomorphic algebras.So X is a disjoint union of sections c λ,µ ∩ X 0 .
The previous bundle can be "lifted" to specific fields, for instance, in the real case we may consider the plane with the axis removed: Γ := R × × R × .Denote by O the origin O = (0, 0).Consider also the curves c λ,µ each one of which, is the zero set of µ 3 x 2 − λ 2 y 3 .Then Γ = ⊔ λ,µ c * λ,µ is the disjoint union of the perforated curves c * λ,µ := c λ,µ \ {O} with λ, µ = 0.Each such curve c λ,µ cuts the line x = 1 in an unique point: ).Then we consider π : R × × R × → R × where π(λ, µ) can be defined as the intersection of c λ,µ with the vertical line x = 1.In other words π(λ, µ) where c * t := ct \ {0}.We could paraphrase this by saying that the axis-less plane is a disjoint union (indexed in R × ) of perforated curves.
If we consider for instance the domain in one-to-one correspondence with the monoid M .Consequently, the isomorphic classes of algebras of type A(1, λ, µ) are in one-to-one correspondence with M .

The perfect case
In the previous section we have analyzed some of the different moduli sets that we will use now in our classification task.Consider a 2-dimensional evolution algebra E over the domain D with a natural basis {e 1 , e 2 } and assume that E is perfect.We analyse several cases.the change of basis and only if there is a k ∈ D × such that λ ′ = k 3 λ and µ ′ = k 2 µ.Summarizing: the algebras in this case fall into two mutually non-isomorphic classes: those of the form A(λ, 1, 1) with λ ∈ D × and those of the form The algebras of the form A(1, λ, µ) only exist over domains which are not fields.The isomorphic classes of algebras of type A(1, λ, µ) are in one-to-one correspondence with the monoid M defined in equation (1).
The isomorphism condition in this class of algebras is given in 3.2 (II) and the moduli set is that of the curves ∂ ξ,ν,ρ .
(2) β, γ / ∈ D × .So, we have the algebras B(α, β, γ, δ) where the four scalars are nonzero and noninvertible but αδ − βγ ∈ D × .We will denote Ω 4 as the set of all (α, β, γ, δ) The isomorphic classes of algebra of this type are in one-to-one correspondence with the elements of the orbit set Ω 4 /(D × × D × ).Furthermore, this orbit set is in one-to-one correspondence with the set of sections ω described in 3.
1 2 = d where d ∈ D * := D \ {0}.Denote this algebra by D d .The product in D d is x • y = xyd for any x, y ∈ D. If f : D d → D e is an isomorphism, it is a D-module isomorphism hence f (a) = af (1) for any a.So f is the multiplication times an element x := f (1).Consequently x ∈ D × (the group of invertibles of D).But also f (uvd) = f (u)f (v)e whence f (d) = f (1) 2 e or d = xe.Thus the isomorphic condition is D d ∼ = D e if and only if there exists x ∈ D × such that d = xe.So the isomorphic classes of nontrivial one dimensional evolution algebras over domains with nonzero product is in one to one correspondence with the set D * /D × , that is, the set of equivalence classes of D * modulo the action D × × D * → D * such that x • d = xd for any x ∈ D × , d ∈ D * .For instance, if D = Z we have Z * /{±1} ∼ = N * = {1, 2, . ..}.Thus there are countably many isomorphic classes of one-dimensional evolution algebras over certain domains.If we consider the domain D := K[x] of polynomials in one indeterminate over the field K then the isomorphic classes of one-dimensional evolution D-algebras are in one to one correspondence with the set of monic polynomials of K[x].