MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A
NEGATION MAP

arXiv:1705.01075v1 [math.RA] 2 May 2017

GUY BLACHAR

Abstract. In this article, we present the basic definitions of modules and Lie semialgebras over
semirings with a negation map. Our main example of a semiring with a negation map is ELT
algebras, and some of the results in this article are formulated and proved only in the ELT theory.
When dealing with modules, we focus on linearly independent sets and spanning sets. We
define a notion of lifting a module with a negation map, similarly to the tropicalization process,
and use it to prove several theorems about semirings with a negation map which possess a lift.
In the context of Lie semialgebras over semirings with a negation map, we first give basic
definitions, and provide parallel constructions to the classical Lie algebras. We prove an ELT
version of Cartan’s criterion for semisimplicity, and provide a counterexample for the naive version
of the PBW Theorem.

Contents
0. Introduction
0.1. Semirings with a Negation Map
0.2. Modules Over Semirings with a Negation Map
0.3. Supertropical Algebras
0.4. Exploded Layered Tropical Algebras
1. Modules over Semirings with a Negation Map
1.1. The Surpassing Relation for Modules
1.2. Basic Definitions for Modules
1.3. -morphisms
1.4. Lifting a Module Over a Semiring with a Negation Map
1.5. Linearly Independent Sets
1.6. d-bases and s-bases
1.7. Free Modules over Semirings with a Negation Map
2. Symmetrized Versions of Important Algebraic Structures
2.1. Semialgebras over Symmetrized Semirings
2.2. Lie Semialgebras with a Negation Map
2.3. Free Lie Algebras with a Negation Map
2.4. Symmetrized Versions of the Classical Lie Algebras
3. Solvable and Nilpotent Symmetrized Lie Algebras
3.1. Basic Definitions
3.2. Lifts of Lie Semialgebras with a Negation Map
3.3. Cartan’s Criterion for Lie Algebras Over ELT Algebras
4. PBW Theorem
4.1. Tensor Power and Tensor Algebra of Modules over a Semirings
4.2. The Universal Enveloping Algebra of a Lie Algebra with a Negation Map
4.3. Counterexample to the Naive PBW Theorem

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Date: 14th March 2018.
This article contains work from the author’s M.Sc. Thesis, which was submitted to the Math Department at BarIlan University. The work was carried under the supervision of Prof. Louis Rowen from Bar-Ilan University, to whom
the author thanks deeply for his help and guidance.
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GUY BLACHAR

References

38

0. Introduction
This paper’s objective is to form an algebraic basis in the context of semirings with negation maps.
Semirings need not have additive inverses to all of the elements. While some of the theory of rings
can be copied “as-is” to semirings, there are many facts about rings which use the additive inverses
of the elements. The idea of negation maps on semirings is to imitate the additive inverse map. In
order to do that, we assume that we are given a map a 7→ (−) a, which satisfies all of the properties
of the usual inverse, except for the fact that a + (−) a = 0. This allows us to use the concept of
additive inverses, even if they do not exist in the usual definition. Semirings with a negation map
are discussed in [1, 8, 9, 2, 3, 24].
During this paper, we will first deal with modules which have a negation map over a semiring,
and then study Lie semialgebras with a negation map over semirings.
In the context of modules, our main interest would be to study linearly independent sets and
spanning sets. We will study maximal linearly independent sets, minimal spanning sets and the
connection between them.
As the introduction demonstrates, the tropical world provides us with several nontrivial examples
of semirings with a negation map – supertropical algebras and Exploded Layered Tropical (ELT)
algebras. In the study of tropical algebra and tropical geometry, one tool is using Puiseux series –
which allow us to “lift” a tropical problem to the classical algebra, and using the classical tools to
solve it. We will give a general definition of what a lift is, and study some of its properties.
We will move to considering Lie semialgebras with a negation map. We will study at first the
basic definitions related to Lie algebras when working over semirings. We will consider on Lie semialgebras which are free as modules, such that their basis contains 1, 2 or 3 elements, and also consider
nilpotent, solvable and semisimple Lie semialgebras.
Cartan’s criterion regarding semisimple Lie algebras states that a Lie algebra is semisimple if and
only if its Killing form is nondegenerate. In this paper, we prove an ELT version of this theorem for
Lie semialgebras over ELT algebras.
Another point of interest is Poincaré-Birkhoff-Witt (PBW) Theorem. In the classical theory of
Lie algebras, the PBW Theorem states that every Lie algebra can be embedded into an associative
algebra (its universal enveloping algebra), where the Lie bracket on this algebra is the commutator.
We will construct the universal enveloping algebra of a Lie semialgebra, and give a counterexample
to the naive version of the PBW theorem.
Throughout this paper, addition and multiplication will always be denoted as + and ·, in order
to emphasize the algebraic structure. Negation maps will always be denoted by (−). N stands for
the set of natural numbers, whereas N0 = N ∪ {0}. Z stands for the ring of integers, and Z /nZ
denotes the quotient ring of Z by nZ.
0.1. Semirings with a Negation Map.
Definition 0.1. Let R be a semiring. A map (−) : R → R is a negation map (or a symmetry)
on R if the following properties hold:
(1) ∀a, b ∈ R : (−) (a + b) = (−) a + (−) b.
(2) (−) 0R = 0R .

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

3

(3) ∀a, b ∈ R : (−) (a · b) = a · ((−) b) = ((−) a) · b.
(4) ∀a ∈ R : (−) ((−) a) = a.
We say that (R, (−)) is a semiring with a negation map. If (−) is clear from the context, we
will not mention it.
For example, the map (−) a = a defines a trivial negation map on every semiring R. If R is a
ring, it has a negation map given by (−) a = −a.
Throughout this article, we use the following notations:
• a (−) a is denoted a◦ .
• R◦ = {a◦ |a ∈ R}.
• R∨ = R \ R◦ ∪ {0R }.
• We define two partial orders on R:
– The relation  (which we call the surpassing relation) defined by
a  b ⇔ ∃c ∈ R◦ : a = b + c
– The relation ∇ defined by
a∇b ⇔ a (−) b ∈ R◦
0.2. Modules Over Semirings with a Negation Map. We now consider modules over semirings
with a negation map.
Definition 0.2. Let R be a semiring with a negation map. A (left) R-module is a commutative
monoid (M, +, 0M ), equipped with an operation of “scalar multiplication”, R × M → M , where we
denote (α, x) 7→ αx, such that the following properties hold:
(1) ∀α ∈ R ∀x, y ∈ M : α (x + y) = αx + αy.
(2) ∀α, β ∈ R ∀x ∈ M : (α + β) x = αx + βx.
(3) ∀α, β ∈ R ∀x ∈ M : α (βx) = (α · β) x.
(4) ∀α ∈ R : α0M = 0M .
(5) ∀x ∈ M : 0R x = 0M .
(6) ∀x ∈ M : 1R x = x.
A right R-module is defined similarly.
Example 0.3. The following are examples of R-modules, where R is a semiring:
(1) R is an R-module, where the scalar multiplication is the multiplication of R. The submodules
of R are its left ideals, meaning sets I ⊆ R that are closed under addition and satisfy
RI ⊆ I.
(2) Taking the direct sum of Mi = R for i ∈ I, we obtain the free R-module:
M
RI =
R
i∈I

Two important special cases are:
(3) Mm×n (R) is an R-module with the standard scalar multiplication, (αA)ij = α · (A)ij .
Y
M
(4) Let {Mi }i∈I be R-modules. Then
Mi and
Mi are R-modules with the usual comi∈I

i∈I

ponentwise sum and scalar multiplication.

As in [24], we consider negation maps on modules.
Definition 0.4. Let R be a semiring, and let M be an R-module. A map (−) : M → M is a
negation map (or a symmetry) on M if the following properties hold:
(1) ∀x, y ∈ M : (−) (x + y) = (−) x + (−) y.
(2) (−) 0M = 0M .
(3) ∀α ∈ R ∀x ∈ M : (−) (αx) = α ((−) x).
(4) ∀x ∈ M : (−) ((−) x) = x.

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GUY BLACHAR

If the underlying semiring has a negation map (−), every R-module M has an induced negation
map given by (−) x = ((−) 1R ) x. Unless otherwise written, when working with a module over a
semiring with a negation map, the negation map will be the induced one. We note that if it is not
mentioned that the semiring has a negation map, the negation map on the module can be arbitrary
(although the main interest is with the induced negation map).
Definition 0.5. Let R be a semiring, and let M be an R-module with a negation map. A submodule with a negation map of M is a submodule N of M which is closed under the negation
map. Since we are in the context of negation maps, and every module in this paper has a negation
map, we will write “submodule” for a “submodule with a negation map”, where we understand that
every submodule should be closed under the negation map.
A main example of semirings and modules with a negation map uses the process of symmetrization
defined in [24]. Briefly, If R is an arbitrary semiring, we may define R̂ = R × R with componentwise
addition, and with multiplication given by
(r1 , r2 ) · (r1′ , r2′ ) = (r1 r1′ + r2 r2′ , r1 r2′ + r2 r1′ ) .
This is indeed a semiring with a negation map given by (r1 , r2 ) 7→ (r2 , r1 ). Of course, there is a
natural injection R ֒→ R̂ by r 7→ (r, 0).
The idea of this construction is to imitate the way Z is constructed from N. One should think of
the element (r1 , r2 ) as r1 + (−) r2 .
Now, if M is any R-module, we may define M̂ = M × M . We want to define it as an R̂-module;
so we use componentwise addition, and define multiplication by
(r1 , r2 ) (x1 , x2 ) = (r1 x1 + r2 x2 , r1 x2 + r2 x1 ) .
This multiplication is called the twist action on M̂ over R̂. It endows M̂ with a R̂-module structure,
and the induced negation map is
(−) (x1 , x2 ) = (0, 1) (x1 , x2 ) = (x2 , x1 )
For the remainder of the introduction, we will present our two main examples of semirings with
a negation map – supertropical algebras and ELT algebras.
0.3. Supertropical Algebras. Supertropical algebras are a refinement of the usual max-plus algebra. This is a refinement of the max-plus algebra, in which one adds “ghost elements” to replace
the role of the classical zero. Supertropical algebras are discussed in several articles, including
[13, 16, 17, 14, 18, 15].
Definition 0.6. A supertropical semiring is a quadruple R := (R, T , G, ν), where R is a semiring,
T ⊆ R is a multiplicative submonoid, and G0 = G ∪ {0} ⊆ R is an ordered semiring ideal, together
with a map ν : R → G0 , satisfying ν 2 = ν as well as the conditions:

a,
ν (a) > ν (b)
a+b=
ν (a) ν (a) = ν (b)
The monoid T is called the monoid of tangible elements, while the elements of G are called
ghost elements, and ν : R → G0 is called the ghost map. Intuitively, the tangible elements
correspond to the original max-plus algebra, although now a + a = ν (a) instead of a + a = a.
Any supertropical semiring R is a semiring with a negation map, when the negation map is (−) a =
a. Endowed with this negation map, R◦ = G0 .
0.4. Exploded Layered Tropical Algebras. ELT algebras are a more refined degeneration of
the classical algebra than the classical max-plus algebra. The main idea which stands behind this
structure is not only to remember the element, but rather remember also another information – its
layer – which tells us “how many times we added this element to itself”. ELT algebras originate in
[22], were formally defined in [25] and are discussed in [5, 6].

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

5

Definition 0.7. Let L be a semiring, and F a totally ordered semigroup. An Exploded Layered
Tropical algebra (or, in short, an ELT algebra) is the pair R = R (L , F ), whose elements are
denoted [ℓ]a for a ∈ F and ℓ ∈ L , together with the semiring (without zero) structure:

[ℓ ]

a1 > a2
 1 a1
[ℓ
]
[ℓ
]
[ℓ
]
1
2
2
(1)
a1 + a2 :=
a2
a1 < a2 .

[ℓ1 +L ℓ2 ]
a1 a1 = a2
(2) [ℓ1 ]a1 · [ℓ2 ]a2 := [ℓ1 ·L ℓ2 ] (a1 +F a2 ).
We write . For [ℓ]a, ℓ is called the layer, whereas a is called the tangible value.
Let R be an ELT algebra. We write s : R → L for the projection on the first component (the
sorting map):

s [ℓ]a = ℓ
We also write τ : R → F for the projection on the second component:

τ [ℓ]a = a
We denote the zero-layer subset

R◦ = {α ∈ R|s (α) = 0}
and
R∗ = {α ∈ R|s (α) 6= 0} = R \ R◦
Definition 0.8. An ELT algebra R = R (F , L ) in which F is a totally ordered group and L is
a ring (with 1) is called an ELT ring.
Definition 0.9. For any ELT ring R, we define (−) [ℓ]a = [−ℓ]a.
Lemma 0.10. a 7→ (−) a is a negation map on any ELT ring R.
Proof. Straightforward from the definition.



Remark 0.11. When dealing with ELT algebra, the relation  is denoted .
We point out some important elements in any ELT ring R:
(1) [1]0, which is the multiplicative identity of R.
(2) [0]0, which is idempotent to both operations of R.
(3) [−1]0, which has the role of “−1” in our theory.
Since ELT algebras lack an additive identity element, we adjoin one, denoted 0R (denoted −∞
in [5, 6]), and we denote R = R ∪ {0R }. This endows R with an antiring structure.
1. Modules over Semirings with a Negation Map
Modules over rings have an important role in the classical theory. In the classical theory, a module M over a ring R is an abelian group, which has a scalar multiplication operation by elements
from R. If one takes R to be a semiring rather than a ring, then a semimodule M is a commutative
monoid, which has a scalar multiplication operation.
Since semirings with a negation map are not rings, modules over them lack some basic properties
of modules over rings. The study of modules can be found in [10, Chapters 14-18].
However, knowing that we are working over a semiring with a negation map instead of a general
semiring, we note some unique properties of the module. For example, over a semiring with a negation map we have the notion of quasi-zeros, i.e. elements of the form x + (−) x, which play the role
of the classical zero.
In this part, we look into modules over semirings with a negation map. The main issue which we
will deal with is different definitions of base. Following [14], we will give four definitions:
(1) d-base – a maximal linearly independent set, Definition 1.45.

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GUY BLACHAR

(2) s-base – a minimal spanning set, Definition 1.54.
(3) d,s-base – an s-base which is also a d-base, Definition 1.62.
(4) base – defined as the classical base, Definition 1.66.
We will point out the properties satisfied by modules possessing these bases, and we will examine
the differences between these definitions.
1.1. The Surpassing Relation for Modules. We will now give analogous definitions of some
concepts defined previously for semirings with a negation map.
Definition 1.1. Let R be a semiring, and let M be a module with a negation map. An element of
the form x + (−) x for some x ∈ M is called a quasi-zero (or a balanced element, as in [2]). We
denote x◦ = x + (−) x. The submodule of quasi-zeros is
M ◦ = {x◦ | x ∈ M }
If R has a negation map, and the negation map of M coincides with the induced negation map,
then M ◦ = 1◦R M .
Definition 1.2. Let R be a semiring, and let M be an R-module with a negation map. We define
a relation  on M in the following way:
x  y ⇐⇒ ∃z ∈ M ◦ : x = y + z
If x  y, we say that x surpasses y.
Example 1.3. We refer to some of the examples of modules given here, and demonstrate the
meaning of the surpassing relation in these modules.
(1) In R, the surpassing relation of the module coincides with the surpassing relation of the
semiring with a negation map.
(2) In RI , the surpassing relation means surpassing componentwise.
(3) As a special case of the free module, in Mm×n (R) the surpassing relation is equivalent to
surpassing componentwise.
(4) As another special case of the free module, in R [λ], the surpassing relation means surpassing
of each coefficient of the polynomials.
Lemma 1.4.  is a reflexive and transitive relation on any R-module with a negation map. However, it may not be antisymmetric, as demonstrated in [2, Example 4.12].
Proof. Let M be an R-module with a negation map.
(1) If x ∈ M , then x = x + 0M , and 0M ∈ M ◦ ; so x  x.
(2) If x, y, z ∈ M satisfy x  y  z, then x = y + z1 , y = z + z2 , where z1 , z2 ∈ M ◦ . So
x = y + z1 = z + z1 + z2
◦

and z1 + z2 ∈ M , implying x  z.

Although in general  may not be a partial order relation, we give a necessary and sufficient
condition for it to be one, and we point out a specific case in which it is a partial order relation.
Lemma 1.5. The following are equivalent for an R-module M with a negation map:
(1) ∀x ∈ M ∀z1 , z2 ∈ M ◦ : x = x + z1 + z2 ⇒ x = x + z1 ∧ x = x + z2 .
(2) ∀x ∈ M ∀z1 , z2 ∈ M ◦ : x = x + z1 + z2 ⇒ x = x + z1 ∨ x = x + z2 .
(3)  is a partial order relation on M .
Proof. 1 ⇒ 2 is trivial.
2 ⇒ 3: Assume that condition 2 holds. By Lemma 1.4, it is enough to show that  is antisymmetric. Let x, y ∈ M such that x  y and y  x. Then there are z1 , z2 ∈ M ◦ such that x = y + z1
and y = x + z2 . Therefore, x = x + z1 + z2 . By condition 2, either x = x + z2 = y (in which case
we are done) or x = x + z1 . In the latter case,
x = x + z1 + z2 = x + z2 = y

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

7

which proves that  is a partial order relation on M .
3 ⇒ 1: Assume that  is a partial order relation on M , and let x ∈ M and z1 , z2 ∈ M ◦ such that
x = x + z1 + z2 . Write y = x + z1 . By definition, y  x. On the other hand,
x = x + z1 + z2 = y + z2  y
Since  is antisymmetric, x = y = x + z1 , and thus also x = x + z1 + z2 = x + z2 .



Corollary 1.6. Let R be a semiring with a negation map, and let M be an R-module (with the
induced negation map). The following are equivalent:
(1) 1◦R is idempotent.
(2) R◦ is idempotent.
(3) M ◦ is idempotent for every R-module M .
In this case,  is a partial order relation on every R-module M .
Proof. 1 ⇒ 2: Let α◦ ∈ R◦ . Then
α◦ + α◦ = 1◦R · α + 1◦R · α = (1◦R + 1◦R ) α = 1◦R α = α◦
2 ⇒ 3: Assume that R◦ is idempotent, let M be an R-module, and let x◦ ∈ M ◦ . Then
x◦ + x◦ = 1◦R x + 1◦R x = (1◦R + 1◦R ) x = 1◦R x = x◦
3 ⇒ 1: Trivial.
It is left to prove that in this case, if M is an R-module then  is a partial order relation on M .
We prove condition 2 in Lemma 1.5. Indeed, if x ∈ M and z1 , z2 ∈ M ◦ satisfy x = x + z1 + z2 , then
x = x + z1 + z2 = x + z1 + z2 + z2 = x + z2

Example 1.7. If R is a ring with (−) a = −a, a supertropical algebra with (−) a = a or an ELT
ring with (−) a = [−1]0a, then  is a partial order relation on every R-module.
Example 1.8. If  is a partial order relation on R, the induced surpassing relation on an R-module
M might not be a partial order relation on M . For example, consider R = N0 with the usual addition
and multiplication and the negation map (−) a = a. Then N◦0 = 2N0 , and thus for m, n ∈ N0 ,
m  n ⇔ ∃k ∈ N0 : m = n + 2k
This is clearly a partial order relation.
However, consider the R-module M = Z. The induced negation map is (−) a = a, and thus
Z◦ = 2Z. Thus, for m, n ∈ Z,
m  n ⇔ ∃k ∈ Z : m = n + 2k
which is not only not a partial order relation, but an equivalence relation.
We will later see that if  is not a partial order, we can “enforce” it to be one, by taking the
module modulo the congruence it generates (see Definition 1.21).
We note another property of M ◦ :
Lemma 1.9. If x ∈ M ◦ and y  x, then y ∈ M ◦ .
Proof. Since y  x, there exists z ∈ M ◦ such that y = x + z. The result follows, since M ◦ is a
submodule of M .

1.2. Basic Definitions for Modules.

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GUY BLACHAR

1.2.1. Spanning Sets.
Definition 1.10. Let R be a semiring with a negation map, let M be an R-module, and let S ⊆ M
be a subset of M . The R-module spanned by S, denoted Span (S), is
)
( k
X
αi xi ∈ M k ∈ N, α1 , . . . , αk ∈ R, x1 , . . . , xk ∈ S
Span (S) =
i=1

It is easy to see that S ⊆ Span (S), and that Span (S) ⊆ M is also an R-module with respect to
the induced operations.

Definition 1.11. Let R be a semiring with a negation map, and let M be an R-module. We say
that a subset S ⊆ M is a spanning set of M , if Span (S) = M .
Definition 1.12. A module with a finite spanning set is called finitely generated.
1.2.2. Congruences and the First Isomorphism Theorem. Since the usual construction of quotient
modules using ideals fails over semirings, we use the language of congruences. In this subsubsection,
we study congruences of modules over semirings with a negation map. Since this is a special case of
congruences in the context of universal algebra, as written in [10], we mainly cite some known facts.
Definition 1.13. Let R be a semiring, and let M1 , M2 be two R-modules with a negation map. An
R-module homomorphism is a function ϕ : M1 → M2 , which satisfies:
(1) ∀x, y ∈ M1 : ϕ (x + y) = ϕ (x) + ϕ (y).
(2) ∀α ∈ R, ∀x ∈ M1 : ϕ (αx) = αϕ (x).
(3) ∀x ∈ M1 : ϕ ((−) x) = (−) ϕ (x).
We note that if R has a negation map, and the negation maps on M1 and M2 are the induced
negation maps, then condition 3 follows from condition 2, since
ϕ ((−) x) = ϕ (((−) 1R ) x) = ((−) 1R ) ϕ (x) = (−) ϕ (x) .
Remark 1.14. ϕ (M1 ) ⊆ M2 ◦ , since
◦

ϕ (x◦ ) = ϕ (x + (−) x) = ϕ (x) + (−) ϕ (x) ∈ M2 ◦
Definition 1.15. Let R be a semiring, and let M be an R-module with a negation map. An
equivalence relation ∼ on M is called a module congruence, if:
(1) ∀x1 , x2 , y1 , y2 ∈ M : x1 ∼ x2 ∧ y1 ∼ y2 ⇒ x1 + y1 ∼ x2 + y2 .
(2) ∀α ∈ R ∀x, y ∈ M : x ∼ y ⇒ αx ∼ αy.
(3) ∀x, y ∈ M : x ∼ y ⇒ (−) x ∼ (−) y.
The kernel of ∼ is
M∼ = {x ∈ M | ∃y ∈ M ◦ : x ∼ y}
Again, if the negation map on M is induced from a negation map from R, condition 3 follows from
condition 2. Therefore, the negation map (−) of M a negation map (−) on M /∼ by (−) [x] = [(−) x].
It is easy to see that this negation map and the induced negation map from R coincide, since
(−) [x] = [(−) x] = [((−) 1R ) x] = ((−) 1R ) [x]
Lemma 1.16. M /∼ is an R-module.
Lemma 1.17. M∼ is a submodule of M .
Proof. If x1 , x2 ∈ M∼ , take y1 , y2 ∈ M ◦ such that x1 ∼ y1 and x2 ∼ y2 ; then x1 + x2 ∼ y1 + y2 ,
and y1 + y2 ∈ M ◦ . Thus, x1 + x2 ∈ M∼ .
Now, if α ∈ R, x ∈ M∼ , take y ∈ M ◦ such that x ∼ y; then αx ∼ αy, and αy ∈ M ◦ . Thus,
αx ∈ M∼ .

Lemma 1.18. Let ρ : M → M /∼ be the canonical map. Then

M /∼ ◦ = ρ (M∼ )

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

Proof.
ρ (x) ∈ M /∼

◦

9

⇐⇒ ρ (x) = 1◦R ρ (x) = ρ (1◦R x) ⇐⇒ x ∼ 1◦R x ⇐⇒ x ∈ M∼


We recall the first isomorphism theorem:
Lemma 1.19. Let ϕ : M1 → M2 be an R-module homomorphism. Then
x ∼ y ⇔ ϕ (x) = ϕ (y)
is a module congruence on M1 .
Theorem 1.20 (The First Isomorphism Theorem). Let R be a semiring with a negation map, and
let M1 , M2 be R-modules. If ϕ : M1 → M2 is an R-module homomorphism, then there exists a
module congruence ∼ on M1 such that
M1 /∼ ∼
= Imϕ
We return to  to demonstrate how we can “enforce”  to be a partial order on M .
Definition 1.21. Let R be a semiring, and let M be an R-module with a negation map. Define a
relation ≡◦ on M as
a ≡◦ b ⇔ a  b ∧ b  a

Remark 1.22. ≡◦ is a congruence on M , and the R-module M ≡◦ is partially ordered by the
induced negation map (−) [x] = [(−) x].
1.3. -morphisms. As we will see more when dealing with Lie algebras with a negation map, we
cannot always construct functions that will preserve every operation of the Lie algebra. We will now
define the notion of -morphisms, as also defined in [24, Section 8.2].
Definition 1.23. Let R be a semiring, and let M1 , M2 be two R-modules with a negation map. A
-morphism is a function ϕ : M1 → M2 , which satisfies:
(1) ∀x, y ∈ M1 : ϕ (x + y)  ϕ (x) + ϕ (y).
(2) ∀α ∈ R, ∀x ∈ M1 : ϕ (αx)  αϕ (x).
(3) ∀x ∈ M1 : ϕ ((−) x) = (−) ϕ (x).
(4) ϕ (M1 ◦ ) ⊆ M2 ◦ .
Our purpose now will be to formulate a version of the First Isomorphism Theorem for morphisms.
Assume that ϕ : M1 → M2 is a surjective -morphism. We define the equivalence relation ∼
(which is not necessarily a congruence) on M1 as
x ∼ y ⇐⇒ ϕ (x) = ϕ (y)
We now wish to define addition and scalar multiplication on M1 /∼ . The usual definition (adding
or scalar multiplying the representatives) will no longer work, because ∼ may not be a congruence;
so we define
[x] + [y] =
α · [x] =

{z ∈ M1 |ϕ (z) = ϕ (x) + ϕ (y)} = ϕ−1 (ϕ (x) + ϕ (y))
{z ∈ M1 |ϕ (z) = α · ϕ (x)} = ϕ−1 (α · ϕ (x))

and the natural negation map (−) [x] = [(−) x].
The usual verifications show that:
Lemma 1.24. M1 /∼ is an R-module with the above operations. The quasi-zeros are equivalence
classes of the form [x], where x ∈ ker ϕ.
We define the usual projection map ρ : M1 → M1 /∼ as ρ (x) = [x].
Lemma 1.25. ρ is a -morphism.
Proof.

10

GUY BLACHAR

(1) Let x, y ∈ M1 . Since ϕ (x + y)  ϕ (x) + ϕ (y), and since ϕ is injective, there is some z ∈ M1
such that
ϕ (x + y) + ϕ (z) = ϕ (x) + ϕ (y)
◦
and ϕ (z) ∈ M2 , that is, z ∈ ker ϕ. Hence
ρ (x + y) = [x + y]  [x + y] + [z] = [x] + [y] = ρ (x) + ρ (y)
(2) Similarly as 1.
(3) Given x ∈ M1 , it is easily seen that
ρ ((−) x) = [(−) x] = (−) [x] = (−) ρ (x)
(4) This is obvious.

We now get a version of the First Isomorphism Theorem:
Theorem 1.26 (The First Isomorphism Theorem for -morphisms). Let R be a semiring, let
M1 , M2 be two R-modules with a negation map, and let ϕ : M1 → M2 be a -morphism. Let ∼
be the equivalence relation defined above, and rho : M1 → M1 /∼ the projection map. Then there
exists a unique R-isomorphism ϕ̂ : M1 /∼ → M2 such that ϕ = ϕ̂ ◦ ρ.
Proof. We define ϕ̂ : M1 /∼ as ϕ̂ ([x]) = ϕ (x). The usual verifications as in the proof of the First
Isomorphism Theorem prove that ϕ̂ is an R-isomorphism, and that ϕ̂ is uniquely defined.

We define a similar concept for semirings with a negation map:
Definition 1.27. Let R1 and R2 be semirings with a negation map. A -morphism is a function
ϕ : R1 → R2 such that the following properties hold:
(1) ∀α, β ∈ R1 : ϕ (α + β)  ϕ (α) + ϕ (β).
(2) ∀α, β ∈ R1 : ϕ (α · β)  ϕ (α) · ϕ (β).
(3) ∀α ∈ R1 : (−) ϕ (α) = ϕ ((−) α).
(4) ϕ (0R1 ) = 0R2 .
(5) ϕ (1R1 ) = 1R2 .
1.4. Lifting a Module Over a Semiring with a Negation Map.
1.4.1. Lifting a semiring with a negation map. When dealing with tropical algebra, we had a powerful tool – Puiseux series. This tool enables one to use known results in the classical theory to prove
tropical results. In this section, we attempt to give a similar construction for an arbitrary semiring
with a negation map.
Definition 1.28. Let R be a semiring with a negation map. For any subset A ⊆ R, define
A = {β ∈ R | ∃α ∈ A : β  α}
If A = R, we say that A is -dense in R.
For example, R∨ = R. We note that this defines a topology on R; however, this topology is
usually not even T1 , since, for example, {0R } = R◦ .
b and a map
Definition 1.29. Let R be a semiring with a negation map. A lift of R is a ring R
b → R, such that the following properties hold:
ϕ
b:R
b is (−) α
(1) ϕ
b is a -morphism, where the negation map on R
b = −b
α.
∨
(2) Imϕ
b ⊆ R is -dense in R.
(3) ϕ
b (b
α) = 0R ⇐⇒ α
b = 0Rb
Example 1.30. We give several examples of lifts.
(1) If R is a ring with the negation map (−) a = −a, then the identity map R → R is a lift of R.
b = L {{t}} with the EL tropicalization map is a
(2) Given an ELT algebra R = R (F , L ), R
lift of R, as defined and proved in [5, Lemma 0.4].

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

11

(3) Although the intuition is that lifts should be “very big”, this example proves that this is
not always the case. Consider the semiring (N0 , +, ·) with the negation map (−) a = a.
Then Z /2Z is a lift of N0 , with the map defined by ϕ
b (0̄) = 0 and ϕ
b (1̄) = 1.

We now prove the following theorem:

Theorem 1.31. Let R be an antiring with a negation map. Then R possesses a lift.
Proof. We shall first construct the lift. We denote by A the following set:
A = {aα | α ∈ R∨ }

(We use the notation aα rather than α to distinguish the elements of A and those of R∨ ).
We define a multiplication on the elements of A as follows:

aαβ , αβ ∈ R∨
aα · aβ =
a0R , Otherwise

This multiplication endows A with a monoid structure.

b = Z [A], meaning
We also denote R
)
( k
X
∨
b=
ni aαi ni ∈ Z, αi ∈ R
R
i=1

b is a ring.
Since A is a monoid, R

b → R. We define an action of {−1, 0, 1} on R by
We are left with defining the lifting map ϕ
b:R

1 · α = α, 0 · α = 0R , (−1) · α = (−) α

b → R is defined by ϕ
The map ϕ
b:R
b 0Rb = 0R and
!
k
k
X
X
|ni | (sign (ni ) · αi )
ni aαi =
ϕ
b
i=1

i=1

where

k
X

ni aαi is in its reduced representation, i.e. αi 6= αj for i 6= j, and sign (ni ) · αi is calculated

i=1

by the above action.


b ϕ
We shall now prove that R,
b is a lift of R. We first prove that ϕ
b is a -morphism.
(1) It is easy to see that if αi 6= αj whenever i 6= j,
!
k
k
X
X
ϕ
b
ϕ
b (ni aαi )
ni aαi =
i=1

i=1

Therefore we only have to prove that ϕ
b (maα ) + ϕ
b (naα )  ϕ
b (maα + naα ) where m, n 6= 0.
If sign (m) = sign (n), then

ϕ
b (maα ) + ϕ
b (naα )

= |m| (sign (m) · α) + |n| (sign (n) · α) = |m + n| (sign (m + n) · α) =
= ϕ
b ((m + n) aα ) = ϕ
b (maα + naα )

Now, suppose sign (m) 6= sign (n). Without loss of generality, we assume n < 0 < m and
−n < m (the other cases are proved similarly). Thus,

ϕ
b (maα ) + ϕ
b (naα )

= mα + (−n) ((−) α) = (m + n) α + (−n) α + (−n) ((−) α) =
= (m + n) α + (−n) α◦  (m + n) α = ϕ
b ((m + n) aα ) =
= ϕ
b (maα + naα )

as we needed to prove.

12

GUY BLACHAR

(2) Let us begin by observing that
ϕ
b (aα ) ϕ
b (aβ )  ϕ
b (aαβ )

∨

(Since if αβ ∈ R , there is equality; otherwise, the LHS is αβ ∈ R◦ , whereas the RHS is
0R ). Now,



!  ℓ
!
k
ℓ
k
X
X
X
X
n j βj  =
mi αi · 
b
nj aβ j  =
mi aαi · ϕ
ϕ
b
j=1

i=1

j=1

i=1

X

=

(mi nj ) αi βj +

αi βj ∈R∨

X



αi βj

=
(3)
ϕ
b −

k
X
i=1

ni aαi

!

=

k
X

∈R∨

X

(mi nj ) αi βj 

αi βj ∈R
/ ∨



(mi nj ) αi βj = ϕ
b

X

αi βj

∈R∨


!  ℓ
k
X
X
nj aβj 
mi aαi · 
ϕ
b


(mi nj ) aαi βj 

j=1

i=1

|−ni | (sign (−ni ) · αi ) = (−)

k
X
i=1

i=1



|ni | (sign (ni ) · αi ) = (−) ϕ
b

k
X
i=1

ni aαi

!


(4) ϕ
b 0Rb = 0R by definition.

b 1Rb = 1R .
(5) Noting that 1Rb = 1 · a1R , ϕ
This proves that ϕ
b is a -morphism. Since Imϕ
b = R∨ , we are left to prove that ϕ
b (b
α) = 0R if
and only if α
b = 0Rb . But this follows from the fact that R is an antiring.


b ϕ
Hence R,
b is a lift of R.


1.4.2. Lifting a module. We move towards lifting a module. We use again the concept of -density:

Definition 1.32. Let R be a semiring, let M be an R-module with a negation map. For any subset
S ⊆ M , define
S = {y ∈ M | ∃x ∈ S : y  x}
If S = M , we say that S is -dense in M .


b ϕ
Definition 1.33. Let R be a semiring with a negation map with a lift R,
b , and let M be an
b
c and a map ψb : M
c → M such that the following properties
R-module. A lift of M is an R-module
M
hold:
c : ψb (x1 ) + ψb (x2 )  ψb (x1 + x2 ).
(1) ∀x1 , x2 ∈ M
b ∀x ∈ M
c:ϕ
(2) ∀b
α∈R
b (b
α) ψb (x)  ψb (b
αx
b).

b
(3) ψ 0M
c = 0M .
(4) Imψb is -dense in M .
Theorem 1.34. If R has a lift, then every R-module has a (free) lift. Furthermore, if the module
is generated by µ elements (for some cardinal number µ), it possesses a lift which is a free module
generated by µ elements.


b ϕ
c=R
bS , and define a
Proof. Let R,
b be a lift of R. Let S ⊆ M be a generating set. Define M
c → M by
function ψb : M
 X
ψb (b
rs )
=
ϕ
b (b
rs ) s
s∈S



c, ψb is a lift of M .
We now prove that M

s∈S

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

13

c. Then
(1) Let (b
xs )s∈S , (b
ys )s∈S ∈ M
X
X
X


ψb (b
xs )s∈S + ψb (b
ys )s∈S
=
ϕ
b (b
xs ) s +
ϕ
b (b
ys ) s =
(ϕ
b (b
xs ) + ϕ
b (b
ys )) s 
s∈S



X
s∈S

s∈S

s∈S

ϕ
b (b
xs + ybs ) s = ψb (b
xs )s∈S + (b
ys )s∈S



b and (b
c. Then
(2) Let α
b∈R
xs )s∈S ∈ M
X
X
X


ϕ
b (b
α) ψb (b
xs )s∈S = ϕ
b (b
α)
ϕ
b (b
xs ) s =
(ϕ
b (b
α) ϕ
b (b
xs )) s 
ϕ
b (b
αx
bs ) s = ψb α
b (b
xs )s∈S
s∈S

s∈S

s∈S

(3) Follows immediately.
P
b
b is -dense in R, for each αs there is α
bs ∈ R
(4) Let x ∈ M . Write x = s∈S αs s. Since Imϕ
such that αs  ϕ
b (b
αs ). If αs = 0R , we may choose α
bs = 0Rb . Therefore,
X
X

x=
αs s 
ϕ
b (b
αs ) s = ψb (b
αs )s∈S
s∈S

s∈S




 n 
b ϕ
b , ψb is a lift of Rn , where
Remark 1.35. If R,
b is a lift of R, then
R


b ((b
x)i )
ψb (b
x) = ϕ



i

n
I.e., ψb applies ϕ
b on each entry of the given vector. In this case, Imψb = (R∨ ) .

We will later see theorems which hold over modules which have a lift, such as Corollary 1.49.
 
b ⊆M
c be a submodule. Then ψb N
b is a submodule of M .
Lemma 1.36. Let N

 
b and let α, β ∈ R. Take x
c such that x  ψb (b
b, yb ∈ M
x) and y  ψb (b
y ), and
Proof. Let x, y ∈ ψb N
 
b such that α  ϕ
take α
b, βb ∈ R
b (b
α) and β  ϕ
b βb . Then
 
 


by  ψb α
by
αx + βy  ϕ
b (b
α) ψb (b
x) + ϕ
b βb ψb (b
y )  ψb (b
αx
b) + ψb βb
bx
b + βb
by ∈ N , we are finished.
Since α
bx
b + βb



1.5. Linearly Independent Sets. Up until now, most of our results were formulated and proved
for a module with a negation map, where we didn’t assume that the underlying semiring has a
negation map. However, now that we are going to deal with linear dependency, we need the notion
of quasi-zero scalars; hence, we assume for the rest of this section that our semiring has a negation
map.
We first specialize our underlying semiring, to avoid the problem of “quasi-zero divisors”:
Definition 1.37. A semiring with a negation map R is called entire, if R is commutative in both
of the operations and if
∀α, β ∈ R : αβ ∈ R◦ ⇒ α ∈ R◦ ∨ β ∈ R◦
Definition 1.38. Let R be an entire semiring with a negation map, and let M be an R-module. A
set S ⊆ M is called ◦-linearly dependent, if
∃k ∈ N, ∃x1 , . . . , xk ∈ S, ∃α1 , . . . , αk ∈ R∨ \ {0R } : α1 x1 + · · · + αk vk ∈ M ◦
S is called ◦-linearly independent, if it is not linearly dependent.
Since we are dealing with negation maps, we will omit the ◦ in ◦-linearly dependent and ◦-linearly
independent.
Lemma 1.39. Let R be an entire semiring with a negation map, let M be an R-module, and let
S ⊆ M be a linearly independent set. Then
S ∩ M◦ = ∅

14

GUY BLACHAR

Proof. Assume there exists some x ∈ S such that x ∈ M ◦ . Then the linear combination 1R x is
quasi-zero, contradicting the fact that S is linearly independent.

Lemma 1.40. Let R be an entire semiring with a negation map, let M be an R-module, and let
S = {x1 , . . . , xm } be a linearly dependent subset of M . Assume that S ′ = {y1 , . . . , ym } ⊆ M satisfies
∀i : yi  xi . Then S ′ is also linearly dependent.
X
αi xi ∈ M ◦ , where ∀i : αi ∈ R∨ ; then
Proof. Assume that
i

implying

X

X

αi yi 

αi xi ∈ M ◦

i

i

◦

X

αi yi ∈ M , by Lemma 1.9.



i

Lemma 1.41. Let M be an R-module. If x  y, then x + (−) y ∈ M ◦ .
Proof. By definition, there is some z ∈ M ◦ such that x = y + z. Therefore,
x + (−) y = y + z + (−) y = 1◦R y + z ∈ M ◦

Lemma 1.42. Let R be an entire semiring with a negation map, and let M be an R-module. Assume
k
X
αi xi . Then the set {x1 , . . . , xk , y} is linearly dependent.
that y 
i=1

Proof. Write

I = {i = 1, . . . , n|αi ∈
/ R◦ }
◦
We note that if i ∈
/ I, then αi ∈ R , and thus αi xi ∈ M ◦ . If I = ∅, then y ∈ M ◦ , and thus
{x1 , . . . , xk , y} is linearly dependent. Therefore, we may assume that I 6= ∅, and thus
y

k
X

αi xi 

i=1

By Lemma 1.41,
y+

X

X

αi xi

i∈I

((−) αi ) xi = y + (−)

i∈I

X

αi xi ∈ M ◦ .

i∈I

We found a linear combination of some of the vectors in the set {x1 , . . . , xk , x}, which is quasi-zero,
and the coefficients are not quasi-zero, implying {x1 , . . . , xk , x} is linearly dependent.

Lemma 1.43. Let S ⊆ M be a linearly independent set. Then for all x ∈ S, S\ {x} is not a
spanning set of M .
Proof. Assume that S\ {x} is a spanning set of M . Then
∃k ∈ N, ∃α1 , . . . , αk ∈ R, ∃x1 , . . . , xk ∈ S : α1 x1 + · · · + αk xk = x = 1R x
By Lemma 1.42, (S\ {x}) ∪ {x} = S is linearly dependent, a contradiction. Thus, S\ {x} is not a
spanning set of M .

Corollary 1.44. Suppose A ⊆ M is a linearly independent set, and that B ⊆ M is a spanning set
of M . If B ⊆ A, then A = B.
Proof. If A 6= B, there exists x ∈ A \ B. Therefore, there exists a linear combination
X
x=
αb b
b∈B

By Lemma 1.42, B ∪ {x} is linearly dependent. But B ∪ {x} ⊆ A, contradicting the assumption
that A is linearly independent. Thus A = B.

1.6. d-bases and s-bases.

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

15

1.6.1. d-bases.
Definition 1.45. Let R be an entire semiring with a negation map. A d-base (for dependence
base) of an R-module M is a maximal linearly independent subset of M .
Definition 1.46. Let R be an entire semiring with a negation map, and let M be an R-module.
The rank of M is
rk (M ) = max {|B||B is a d-base of M }
Lemma 1.47. Every linearly independent set is contained in some d-base.
Proof. Let M be an R-module, and let S ⊆ M be a linearly independent set. Consider
{S ′ ⊆ M |S ⊆ S ′ is linearly independent}
This set satisfies Zorn’s condition, and thus, has a maximal element S ′′ , which is a d-base of M . 
1.6.2. d-bases of modules

which possess a lift. For this part, we fix an entire semiring with a negation
b
map R with a lift R, ϕ
b .


c, ψb , and let x
c be vectors. If
Lemma 1.48. Let M be an R-module with a lift M
b1 , . . . , x
bm ∈ M
x
b1 , . . . , x
bm are linearly dependent, then ψb (b
x1 ) , . . . , ψb (b
xm ) are also linearly dependent.
b not all are 0 b , such that
Proof. x
b1 , . . . , x
bm are linearly dependent, so there are α
b1 , . . . , α
bm ∈ R,
R
m
X
i=1

Applying ψb on the equation, we get
m
X
i=1

Using Lemma 1.9, we get

ϕ
b (b
αi ) ψb (b
xi )  ψb
m
X
i=1

α
bi x
bi = 0M
c
m
X
i=1

α
bi x
bi

!


= ψb 0M
c = 0M

ϕ
b (b
αi ) ψb (b
xi ) ∈ M ◦

Since Imϕ
b ⊆ R∨ , we get that for each i, ϕ
b (b
αi ) ∈ R∨ . But we know that not all of the α
bi ’s are 0Rb ,
and thus not all of the ϕ
b (b
αi ) are 0R , as required.


Corollary 1.49. If R has a lift, than any n + 1 vectors in Rn are linearly dependent.
 n 
b , ψb . Let
Proof. Let x1 , . . . , xn+1 ∈ Rn be any vectors. By Remark 1.35), Rn has a lift
R
 n
b
x′1 , . . . , x′n+1 ∈ Imψb such that xi  x′i for each i. Now, let y1 , . . . , yn+1 ∈ R
be some vectors
′
b
such that for each i, ψ (yi ) = x .
i

 n
b is a ring, hence any n + 1 vectors in R
b
R
are linearly dependent; in particular, y1 , . . . , yn+1

are linearly dependent. By Lemma 1.48, x′1 , . . . , x′n+1 are linearly dependent. By Lemma 1.40, we
are finished.

Theorem 1.50. If R has a lift, then rk (Rn ) = n.
Proof. By Corollary 1.49, any n + 1 vectors in Rn are linearly dependent; so rk (Rn ) ≤ n. However,
since {e1 , . . . , en } are linearly dependent, rk (Rn ) ≥ n, which together yield the desired equality. 
Recall the relation ≡◦ from Definition 1.21, defined as


a ≡◦ b ⇔ a  b ∧ b  a

Theorem 1.51. Assume that R ≡◦ has a lift (rather than R). Then any n + 1 vectors in Rn are
linearly dependent.

16

GUY BLACHAR

n
Proof. Let x1 , . . . , xn+1 ∈ Rn , and let [x1 ] , . . . , [xn+1 ] ∈ R ≡◦ be their equivalence classes under ≡◦ . Noting that

 n
Rn ≡ ∼
= R ≡
◦

◦

by Corollary 1.49, there are α1 , . . . , αn+1 ∈ R∨ , not all are 0R , such that
Hence,

◦
n
[α1 x1 + · · · + αn+1 xn+1 ] = α1 [x1 ] + · · · + αn+1 [xn+1 ] ∈ R ≡◦

[α1 x1 + · · · + αn+1 xn+1 ] = 1◦R [α1 x1 + · · · + αn+1 xn+1 ] = [1◦R (α1 x1 + · · · + αn+1 xn+1 )]
By the definition of ≡◦ ,
◦

α1 x1 + · · · + αn+1 xn+1  1◦R (α1 x1 + · · · + αn+1 xn+1 ) ∈ (Rn )
By Lemma 1.9,

◦

α1 x1 + · · · + αn+1 xn+1 ∈ (Rn )
as required.


n

Corollary 1.52. Let R be a ring with some negation map. Then any n+1 vectors in R are linearly
dependent.

Proof. In this case, R ≡◦ is a ring, and the induced negation map is (−) [a] = − [a]; therefore,

R ≡ has a lift, and Theorem 1.51 can be applied.

◦
In the above cases, we have another corollary:

Corollary 1.53. If M ⊆ Rn is a submodule, then rk (M ) ≤ n.
Proof. Any d-base of M is contained in a d-base of Rn , due to Lemma 1.47, whose order is at
most n.

1.6.3. s-bases.
Definition 1.54. Let R be an entire semiring with a negation map. An s-base (for spanning base)
of an R-module M is a minimal spanning subset of M .
An easy corollary from the First Isomorphism Theorem is the following:
Corollary 1.55. Let M be an R-module with a finite spanning set S, |S| = n. Then there exists a
congruence ∼ on Rn , such that
n
M∼
= R /∼
Corollary 1.56. Any spanning set of an R-module M must have at least rk (M ) elements.
n
Proof. By Corollary 1.55, if M = Span {x1 , . . . , xn }, then M ∼
= R /∼ for some congruence ∼.
n

′
]} of R /∼ .
Thus, any d-base {y1 , . . . , ym } of M maps to a linearly independent set {[y1′ ] , . . . , [ym
′
′
n

Then also y1 , . . . , ym are linearly independent in R , and thus m ≤ n.

1.6.4. Critical elements versus s-bases. We define a similar concept to the concept of critical elements
presented in [14]:
Definition 1.57. Let R be an entire semiring with a negation map, and let M be an R-module.
(1) We define an equivalence relation ∼ on M , called projective equivalence, as the transitive
closure of the following relation: we say that x ∼ y if there is an invertible α ∈ R∨ such that
x = αy.
(2) We define the equivalence class of x to be
[x]∼ = {y ∈ M |y ∼ x}
◦

(3) We say that x ∈ M \M is critical, if there is no linear combination x = x1 + x2 where
x1 , x2 ∈ M \ [x]∼ .
(4) A critical set is a set of representatives of all the equivalence classes of ∼.
Remark 1.58. Any critical set is projectively unique.

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

17

Lemma 1.59. Suppose x ∈ M is critical. Then it is not spanned by M \ [x]∼ .
Proof. Assume that
x=

n
X

αi yi

i=1

for yi ∈ M \ [x]∼ . x is critical, thus n > 1. Write x1 = α1 y1 and x2 =
of criticality, x2 ∈ [x]∼ . We get a contradiction by induction on n.

Pn

i=2

αi yi . By the definition


Lemma 1.60. Suppose S is an s-base of M . Then S contains a critical set of M .
Proof. Suppose x ∈ M is critical. S is a s-base, thus x is spanned by S. So, by Lemma 1.59, it must
be an element of S (up to projective equivalence).

In the classical theory (meaning when working with modules over rings), this definition is meaningless; there are no critical vectors. However, In [14, Theorem 5.24] it is proven that in the
supertropical theory, any s-base (if it exists) is a critical set. One could hope that over any entire
antiring with a negation map these definition will coincide – but even in the ELT theory there is a
counterexample.
Example 1.61. Consider R = R (C, R), and
  
 

[1]0
−∞
 [1]0

M = Span [1]0 , −∞ ,  [1]0 
 [0]
[1] (−1) 
−∞
0
{z
}
|
S

The set is a spanning set of M . It is straightforward to prove that any vector in S cannot be
presented as a linear combination of the others. However,
[1] 
[1]  
 
 [1]  

[1]0
−∞
−∞
0
0
0
[1]0 = [0]0 [1]0 + −∞ +  [1]0  = [0]0 +  [1]0 
[1] (−1)
[1] (−1)
[0]0
[0]0
[0]0
−∞
[1] 
0
implying [1]0 is not critical. Since this vector is not critical, we get an example of an ELT module
[0]0
with two s-bases which are not projectively equivalent:
  
  
 

 

[1]0
[1]0
−∞
−∞
 [1]0

 [1]0

M = Span [1]0 , −∞ ,  [1]0  = Span [0]0 , −∞ ,  [1]0 
 [0]
 [0]
[1] (−1) 
[1] (−1) 
−∞
−∞
0
0

1.6.5. d,s-bases.

Definition 1.62. Let R be an entire semiring with a negation map, and let M be an R-module. A
d,s-base of M is an s-base of M , which is also a d-base.
Lemma 1.63. If S is a d,s-base of M , with M finitely generated, then |S| = rk (M ).
Proof. By Corollary 1.56, |S| ≥ rk (M ); but, since S is also a d-base, we get equality.



Lemma 1.64. Any linearly independent spanning set is a d,s-base.
Proof. Assume that S is a linearly independent spanning set of M . If S ′ ⊆ S is an s-base, we get
that S ′ = S, by Corollary 1.44.
Now, assume that S ⊆ S ′′ is a d-base, using the same corollary we get S = S ′′ . So S is a d-base.
To sum up, S is an s-base and a d-base, and thus S is a d,s-base.



Note that not every s-base is a d,s-base.
Example 1.65. Any submodule spanned by zero-layered elements has no d,s-base, since the only
linearly independent subset of this submodule is the empty set, which is not a spanning set.

18

GUY BLACHAR

1.7. Free Modules over Semirings with a Negation Map.
1.7.1. Definitions and examples.
Definition 1.66. Let R be an entire semiring with a negation map. An R-module M is called free,
if there exists a set B ⊆ M such that for all x ∈ M \ {0M } there exists a unique choice of n ∈ N,
α1 , . . . , αn ∈ R and x1 , . . . , xn ∈ B such that
x = α1 x1 + · · · + αn xn
Such a set B is called a base of M .
Example 1.67. Let R be an entire semiring with a negation map. Then Rn is a free R-module,
with the base {e1 , . . . , en }. Indeed, the linear combination is determined uniquely, each component
separately.
Definition 1.68. Let R be an entire semiring with a negation map, and let M be a free Rmodule with a base B (|B| = n). For every x ∈ M there exists a unique representation x =
αi1 vi1 + · · · + αik vik . We define the coordinate vector of x by the base B, [x]B ∈ Rn , to be:
(
0R j ∈
/ {i1 , . . . , ik }
([x]B )j =
, j = 1, . . . , n
αij j ∈ {i1 , . . . , ik }
It is easy to verify that [x + y]B = [x]B + [y]B , and [αx]B = α [x]B .
Lemma 1.69. Let R be an entire semiring with a negation map, and let M be an R-module. M is
free with a base B of size n if and only if M ∼
= Rn .
Proof. If M ∼
= Rn , then M is free with a base of size n.
Now, assume that M is free with a base B of size n. Denote B = {x1 , . . . , xn }, and define a
function ϕ : M → Rn by ϕ (x) = [x]B .
(1) ϕ is an R-module homomorphism: since [x + y]B = [x]B + [y]B and [αx]B = α [x]B .
(2) ϕ is injective: If ϕ (x) = ϕ (y), then the representations of x and y with respect to B are
the same. But we also know that these representations
are unique, and thus x = y.
!
n
X
αi xi = (αi ).
(3) ϕ is surjective: If (αi ) ∈ Rn , then ϕ
i=1

Thus, M ∼
= Rn .



We now prove that any base is a d,s-base.
Lemma 1.70. If B is a base of M , and if
n
X

αi bi ∈ M ◦

i=1

for α1 , . . . , αn ∈ R, b1 , . . . , bn ∈ B, then α1 , . . . , αn ∈ R◦ .
Proof. Since

n
X

αi bi ∈ M ◦ , there is some x ∈ M such that

αi bi = x◦ . Writing x =

i=1

i=1

have

n
X

n
X
i=1

αi bi = x◦ =

X

b∈B

βb b

!

=

X

X

βb b, we

b∈B

βb◦ b

b∈B

But B is a base, and thus α1 , . . . , αn ∈ R◦ (since all of the coefficients in the right side are quasizeros).

Lemma 1.71. If B is a base of M , then it is a linearly independent set.

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

Proof. Assume that

X

19

αb b ∈ M ◦

b∈B
◦

By Lemma 1.70, ∀b ∈ B : αb ∈ R , and thus B is linearly independent.



Corollary 1.72. Any base is a d,s-base.
Proof. Let M be a free R-module with base B. By definition, B is a spanning set of M . In addition,
by Lemma 1.71, B is linearly dependent. We are finished, by Lemma 1.64.

Corollary 1.73. The cardinality of a base of a free module is uniquely determined.
Example 1.74. We give an example of a submodule of a free module, which has a d,s-base which
is not a base. Take R = R (C, R), and consider

  [1]   [1] 
0
1
 [1]1

M = Span  [1]0  ,  [1]1  ,  [−1]0 
 [1]

[1]1
[1]0
0

  [1]   [1] 
0
1
 [1]1

The set  [1]0  ,  [1]1  ,  [−1]0  is a d,s-base of M (to prove this set is linearly inde [1]

[1]1
[1]0
0
pendent, we use [25, Theorem 3.3.5]; their determinant is [2]2, which is not zero-layered, and thus
this set is linearly dependent). However, it is not a base, because there is a vector which can be
written as a linear combination of its elements in two different ways:
 [1]   [1]   [1]   [1]   [1] 
1
0
1
0
1
 [1]0  +  [1]1  =  [1]1  =  [1]1  +  [−1]0 
[1]0

[1]1

[1]1

[1]1

[1]0

However, if we know that a module is free, we also know all of its d,s-bases:

Lemma 1.75. Let M be a free R-module, and let S be a d,s-base of M . Then S is a base of M .
Proof. By Lemma 1.69, we can assume M = Rn . Since S is an s-base of M , by Lemma 1.60, S
contains some critical set, A. Since the critical elements in Rn are precisely αi ei for some αi ∈ R∨ ,
we may write
A = {α1 e1 , . . . , αn en }
We now prove that each αi is invertible. Otherwise, assume without loss of generality that α1 is
not invertible. Since S is an s-base of Rn , and since e1 is critical, there are β1 e1 , . . . , βk e1 ∈ S such
that
∃γ1 , . . . , γk ∈ R : γ1 β1 e1 + · · · + γk βk e1 = e1
Since S is linearly independent, we must have k = 1, and thus βe1 ∈ S such that β is invertible.
Again, since S is linearly independent, α1 = β is invertible, as required.

1.7.2. Free modules over entire antirings with a negation map. For the remainder of this part, we
consider free modules over entire antirings with a negation map.
Definition 1.76. Let R be an entire semiring with a negation map, and let M be a free R-module
with two bases B = {v1 , . . . , vn } and C = {w1 , . . . , wn }. We define the transformation matrix
B
from B to C, [I]C , by:
B

[I]C = ([v1 ]C , . . . , [vn ]C )
Notice that [I]B
C ∈ Mn (R).
B

Lemma 1.77. Under the conditions of Definition 1.76, ∀x ∈ M : [I]C [x]B = [x]C .

20

GUY BLACHAR

Proof. Denote

x=

B
[I]C

n
X

k=1

= (αi,j ), [x]B = (βk ). That means x =

βk vk =

n
X

k=1

βk

n
X
ℓ=1

αℓ,k wℓ

!

n
X

βk vk , and vk =

k=1

=

n
n
X
X
ℓ=1

k=1

αℓ,k βk

!

n
X

αℓ,k wℓ . We get

ℓ=1

wℓ =

n 
X



And by the definition, ∀ℓ = 1, . . . , n : [I]B
[x]
= ([x]C )ℓ , as needed.
C
B

B

[I]C [x]B

ℓ=1



ℓ

wℓ


ℓ

Lemma 1.78. Under the conditions of Definition 1.76,

B
[I]C
B · [I]C = In

Proof. We will show equality of columns. For all i = 1, . . . , n,


Lemma 1.77
B
C
B
C
B
·
[I]
=
[I]C
Ci [I]C
B
C = [I]B · [I]C · ei = [I]B · [I]C · [vi ]B
B · [vi ]C

Lemma

=

1.77

as needed.

[vi ]B = ei


Corollary 1.79. A base of a free module over an entire antiring with a negation map is unique, up
to invertible scalar multiplication of each element and permutation. In other words, any two bases
of a free ELT module are projectively equivalent.
B

Proof. If B and C are two bases of a free R-module M , then [I]C is invertible. We also know (by
B
Corollary 1.73) that [I]C is a square matrix. By [7, Corollary 3], it is a generalized permutation
matrix, namely C is B up to scalar multiplication of each element (ci ’s) and change of order (σ). 
2. Symmetrized Versions of Important Algebraic Structures
We now turn to study Lie semialgebras over semirings with a negation map. In the classical
theory, a Lie algebra is a vector space endowed with a bilinear alternating multiplication, which
satisfies Jacobi’s identity. In the context of negation maps, our definition will be quite similar to
the classical one. We will mostly follow the approach of [12].
2.1. Semialgebras over Symmetrized Semirings.
Definition 2.1. A nonassociative semialgebra with a negation map over a semiring R is an
R-module with a negation map A, together with an R-bilinear multiplication, A × A → A, that is
distributive over addition and satisfies the axioms:
(1) ∀α ∈ R ∀x, y ∈ A : α (xy) = (αx) y = x (αy).
(2) ∀x, y ∈ A : (−) (xy) = ((−) x) y = x ((−) y).
Note that “nonassociative” means “not necessarily associative”.
As in the case of homomorphisms and congruences, if the negation map on A is induced from a
negation map on R, the last condition is follows from the other.
Example 2.2. Let A, B be nonassociative semialgebras with a negation map over a semiring R.
Then A × B is also a nonassociative semialgebra with a negation map (a module as in Example 0.3),
where the multiplication and the negation map are defined componentwise.
Definition 2.3. An associative semialgebra is a nonassociative semialgebra, whose multiplication is associative.
Definition 2.4. Let A, B be nonassociative semialgebras with a negation map over a semiring. A
function ϕ : A → B is called an R-homomorphism, if it satisfies the following properties:
(1) ϕ is an R-module homomorphism (as defined in Definition 1.13).
(2) ∀x, y ∈ A : ϕ (xy) = ϕ (x) ϕ (y).
The set of all R-homomorphisms ϕ : A → B is denoted Hom (A, B). We also say that:
• ϕ is an R-monomorphism, if ϕ is injective.
• ϕ is an R-epimorphism, if ϕ is surjective.
• ϕ is an R-isomorphism, if ϕ is bijective. In this case we denote A ∼
= B.

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

21

In addition, if A is a semialgebra with negation over a semiring R, then End (A) = Hom (A, A) is
also semialgebra with a negation map over R, with the usual addition and scalar multiplication,
composition as multiplication and negation map
((−) ϕ) (x) = (−) ϕ (x)
One may also define -morphism of algebras with a negation map similarly.
Definition 2.5. Let A, B be nonassociative semialgebras with a negation map over a semiring R,
and let ϕ : A → B be an R-homomorphism. The kernel of ϕ is
ker ϕ = {x ∈ A | ϕ (x) ∈ B ◦ } = ϕ−1 (B ◦ )
By Remark 1.14, A◦ ⊆ ker ϕ.
We recall the definition of an ideal:
Definition 2.6. Let A be a nonassociative semialgebra over a semiring R. A subalgebra I ⊆ A
which satisfies IA, AI ⊆ I is called an ideal of A, denoted I ⊳ A.
Lemma 2.7. ker ϕ is an ideal of A.
Proof. We prove it directly:
(1) If x, y ∈ ker ϕ, then ϕ (x + y) = ϕ (x) + ϕ (y) ∈ B ◦ . Hence, x + y ∈ ker ϕ.
(2) If α ∈ R and x ∈ ker ϕ, then ϕ (αx) = αϕ (x) ∈ B ◦ , and thus αx ∈ ker ϕ.
(3) If x ∈ ker ϕ, then ϕ ((−) x) = (−) ϕ (x) ∈ B ◦ , so (−) x ∈ ker ϕ.
(4) If x ∈ ker ϕ and y ∈ A, then ϕ (x) ∈ B ◦ . Hence,
ϕ (xy) = ϕ (x) ϕ (y) = (1◦R ϕ (x)) ϕ (y) = 1◦R (ϕ (x) ϕ (y))
Therefore, ϕ (xy) ∈ B ◦ , implying xy ∈ ker ϕ. Similarly, yx ∈ ker ϕ.

2.2. Lie Semialgebras with a Negation Map.
2.2.1. Basic definitions. We first restrict our base semiring to be a semifield with a negation
map, i.e. a commutative semiring with a negation map in which every element which is not quasizero is invertible.
Definition 2.8. Let R be a semifield. A Lie semialgebra with a negation map over R is a
nonassociative semialgebra with a negation map over R, whose multiplication [·, ·] : L × L → L,
called a negated Lie bracket, satisfies the following axioms:
(1) Commutes with the negation map: ∀x, y ∈ L : [(−) x, y] = (−) [x, y].
(2) Alternating on L: ∀x ∈ L : [x, x] ∈ L◦ .
(3) Anticommutativity: ∀x, y ∈ L : [y, x] = (−) [x, y].
(4) Jacobi’s identity: ∀x, y, z ∈ L : [x, [y, z]] + [z, [x, y]] + [y, [z, x]] ∈ L◦ .
Remark 2.9.
(1) If charR 6= 2, then 2 ⇒ 1.
(2) If R has a negation map and the negation map of L is the induced negation map from R,
then condition 1 follows from the bilinearity of the negated Lie bracket, since
[(−) x, y] = [((−) 1R ) x, y] = ((−) 1R ) [x, y] = (−) [x, y] .
Our definition is a bit different than Rowen’s definition [24]. The difference is reflected in Jacobi’s
identity; Rowen’s definition requires that
∀x, y, z ∈ L : [x, [y, z]] (−) [y, [x, z]]  [[x, y] , z]
We call this axiom the strong Jacobi’s identity. Note that while the strong Jacobi’s identity
implies our version of Jacobi’s identity, the converse does not hold. An example is given in Example 4.6.
Lemma 2.10. ∀x, y ∈ L : [x, y] + [y, x] ∈ L◦ .

22

GUY BLACHAR

Proof. For all x, y ∈ L,
[x, y] + [y, x] = [x, y] + (−) [x, y] ∈ L◦

Definition 2.11. A Lie semialgebra with a negation map L is called abelian, if
∀x, y ∈ L : [x, y] ∈ L◦
Example 2.12. Given a semifield R, let A be an associative semialgebra with a negation map
over R. We define an operation as follows: [x, y] = xy + (−) yx, which is called the negated
commutator. We will now show that it forms a structure of a Lie semialgebra with a negation
map over R, which satisfied the strong Jacobi’s identity:
(1) Bilinear: Let x, y, z ∈ A and α, β ∈ R. Then,
[αx + βy, z] = (αx + βy) z + (−) (z (αx + βy)) =
= α (xz + (−) zx) + β (yz + (−) zy) = α [x, z] + β [y, z]
Linearity in the second component is proved similarly.
(2) [·, ·] commutes with (−): For all x, y ∈ L,
[(−) x, y] = ((−) x) y + (−) (y ((−) x)) = (−) (xy + (−) yx) = (−) [x, y]
(3) Alternating on A: For all x ∈ A,
[x, x] = x2 + (−) x2 ∈ L◦
(4) Anticommutativity: For all x, y ∈ A,
[y, x] = yx + (−) xy = (−) (xy + (−) yx) = (−) [x, y]
(5) Strong Jacobi’s identity: a proof can be found in [24], which uses the strong transfer principle.
Lemma 2.13.
∀x, y ∈ L : x ∈ L◦ ∨ y ∈ L◦ =⇒ [x, y] ∈ L◦
Proof. Assume x ∈ L◦ ; write x = z + (−) z for some z ∈ L. Then
[x, y] = [z + (−) z, y] = [z, y] + (−) [z, y] ∈ L◦

Definition 2.14. Let R be a semifield with a negation map, and let L be a Lie semialgebra with a
negation map over R. A subset L1 ⊆ L is called a subalgebra of L, if it is a Lie semialgebra with
a negation map over R with the restrictions of the operations of L.
Definition 2.15. Let R be a semifield with a negation map, and let L be a Lie semialgebra with a
negation map over R. The center of L is
Z (L) = {x ∈ L | ∀y ∈ L : [x, y] ∈ L◦ }
Lemma 2.16. L◦ ⊆ Z (L).
Proof. By Lemma 2.13.



Definition 2.17. Let R be a semifield with a negation map, let L be a Lie semialgebra with a
negation map over R, and let L1 , L2 ⊆ L be subalgebras of L. Define
[L1 , L2 ] = Span {[x1 , x2 ]|x1 ∈ L1 , x2 ∈ L2 }
Lemma 2.18.
[Z (L) , L] ⊆ L◦ ⊆ Z (L)
Proof.
∀x ∈ Z (L) ∀y ∈ L : [x, y] ∈ L◦ ⊆ Z (L)


MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

23

2.2.2. Homomorphisms and Ideals. We use the general definitions for homomorphisms and ideals of
nonassociative algebras given in section 2.1.
Lemma 2.19. L◦ ⊳ L and Z (L) ⊳ L.
Proof. L◦ is immediate by Lemma 2.13, and Z (L) by Lemma 2.18.



Lemma 2.20. If I, J ⊳ L, then I ∩ J ⊳ L.
Proof. Let x ∈ I ∩ J, y ∈ L. x ∈ I, so [x, y] ∈ I. Also x ∈ J, so [x, y] ∈ J. In conclusion,
[x, y] ∈ I ∩ J.

Definition 2.21. Let R be a semifield, let L be a Lie semialgebra with a negation map over R, and
let I, J ⊳ L be two ideals of L. Then their sum is defined as
I + J = {x + y|x ∈ I, y ∈ J}
Lemma 2.22. If I, J ⊳ L, then I + J ⊳ L.
Proof. Obviously, I + J is a submodule of L. Now, let x ∈ I + J, y ∈ L. x ∈ I + J, so there exists
x′ ∈ I and x′′ ∈ J such that x = x′ + x′′ . We get
[x, y] = [x′ + x′′ , y] = [x′ , y] + [x′′ , y] ∈ I + J

2.2.3. The Adjoint Algebra.
Definition 2.23. Let R be a semifield, let L be a Lie semialgebra with a negation map over R, and
let x ∈ L. We define a homomorphism adx : L → L by
adx (y) = [x, y]
Definition 2.24. Let R be a semifield, and let L be a Lie semialgebra with a negation map over R.
The adjoint algebra of L is the set
AdL = {adx |x ∈ L}
endowed with the following negated Lie bracket:
∀adx , ady ∈ AdL : [adx , ady ] = ad[x,y]
and the induced negation map from End (L): (−) adx = ad(−)x .
Jacobi’s identity can be rewritten now as
∀x, y, z ∈ L : adx (ady (z)) + adz (adx (y)) + ady (adz (x)) ∈ L◦
and the strong Jacobi’s identity can be rewritten now as
adx ady + (−) ady adx  ad[x,y]
Lemma 2.25. AdL is a Lie semialgebra with a negation map over R.
Proof. We first check that AdL is a submodule of End (L). But this is obvious, since adx + ady =
adx+y ∈ AdL, αadx = adαx ∈ AdL and (−) adx = ad(−)x ∈ AdL.
Now, we need to check that it is a Lie semialgebra with a negation map.
(1) ∀α, β ∈ R ∀adx , ady , adz ∈ AdL : [αadx + βady , adz ] = [adαx+βy , adz ] = ad[αx+βy,z] =
adα[x,z]+β[y,z] = αad[x,z] + βad[y,z] =
 α [adx , adz] + β [ady , adz ]
(2) ∀adx , ady ∈ AdL : [(−) adx , ady ] = ad(−)x , ady = ad[(−)x,y] = (−) ad[x,y] = (−) [adx , ady ].
(3) ∀adx ∈ AdL : [adx , adx ] = ad[x,x] , which is a quasi-zero.
(4) ∀adx , ady ∈ AdL : [ady , adx ] = ad[y,x] = ad(−)[x,y] = (−) ad[x,y] = (−) [adx , ady ].
(5) ∀adx , ady , adz ∈ AdL : [adx [ady , adz ]]+[adz [adx , ady ]]+[ady [adz , adx ]] = ad[x[y,z]]+[z[x,y]]+[y[z,x]] ,
which is a quasi-zero.


24

GUY BLACHAR

Lemma 2.26. There is a homomorphism of Lie semialgebras with a negation map L → AdL given
by
x 7→ adx
Therefore, given the congruence ≡ on L defined by x ≡ y ⇔ adx = ady , one has
L/ ≡ ∼
= AdL
Proof. Using the definition of the operation on AdL, this is a homomorphism. By Theorem 1.20,
we get the conclusion.

A word of caution: with the definition given above to the negated Lie bracket, AdL is not a
subalgebra of End (L). Generally, there is no obvious way of fixing this problem; however, one can
use the strong Jacobi’s identity.
If L satisfies the strong Jacobi’s identity, then one can define
adL = {adx | x ∈ L} ⊆ End (L)
as in [24], and this is also a Lie semialgebra with a negation map over R (under the negated
commutator). Nevertheless, we now have only a -morphism L → adL given by x 7→ adx .
2.3. Free Lie Algebras with a Negation Map. In this subsection we turn to study Lie semialgebras with a negation map which are free as modules and have a base consisting of one, two or
three elements.
2.3.1. 1-dimensional Lie Algebras with a negation map.
Lemma 2.27. Any Lie semialgebra L over a semifield with a negation map R with a base B = {x}
is abelian.
Proof. Given y, z ∈ L, there exist α, β ∈ R such that y = αx, z = βx. Therefore,
[y, z] = [αx, βx] = αβ [x, x] ∈ L◦
Thus, L is abelian.



2.3.2. 2-dimensional Lie Algebras with a negation map.
Lemma 2.28. Let R be a semifield with a negation map, and let L be a free R-module with base
B = {x1 , x2 }. Define
[x1 , x1 ] = α1 x1 + α2 x2 , [x2 , x2 ] = β1 x1 + β2 x2 , [x1 , x2 ] = (−) [x2 , x1 ] = γ1 x1 + γ2 x2
where α1 , α2 , β1 , β2 , γ1 , γ2 ∈ R and α1 , α2 , β1 , β2 ∈ R◦ . Now, extend [·, ·] to [·, ·] : L × L → L by
bilinearity.
Then L equipped with [·, ·] is a Lie semialgebra over R.
Proof. By bilinearity, it is enough to check Jacobi’s identity in two cases:
(1) [x1 , [x1 , x2 ]] + [x1 , [x2 , x1 ]] + [x2 , [x1 , x1 ]] ∈ L◦ .
(2) [x2 , [x2 , x1 ]] + [x2 , [x1 , x2 ]] + [x1 , [x2 , x2 ]] ∈ L◦ .
We shall prove the first part; the second is proved similarly. Indeed,

◦
◦
[x1 , [x1 , x2 ]] + [x1 , [x2 , x1 ]] + [x2 , [x1 , x1 ]] = x1 , [x1 , x2 ] + [x2 , [x1 , x1 ]] = [x1 , [x1 , x2 ]] + [x2 , [x1 , x1 ]]
We note that [x1 , x1 ] ∈ L◦ , and thus [x1 , [x1 , x2 ]] + [x1 , [x2 , x1 ]] + [x2 , [x1 , x1 ]] ∈ L◦ , as required. 

Lemma 2.29. Let L be a Lie semialgebra over a semifield with a negation map R with base B =
{x1 , x2 }. Then there are α1 , α2 , β1 , β2 , γ1 , γ2 ∈ R such that
[x1 , x1 ] = α1 x1 + α2 x2 , [x2 , x2 ] = β1 x1 + β2 x2 , [x1 , x2 ] = (−) [x2 , x1 ] = γ1 x1 + γ2 x2
and α1 , α2 , β1 , β2 ∈ R◦ . Therefore, each such Lie semialgebra is obtained from Lemma 2.28.
Proof. The existence of α1 , α2 , β1 , β2 , γ1 , γ2 ∈ R follows from the fact that L is free, and by the
antisymmetry of [·, ·]. Since [·, ·] is alternating, combined with Lemma 1.70, α1 , α2 , β1 , β2 ∈ R◦ . 

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

25

2.3.3. 3-dimensional Lie Algebras with a negation map. We turn to studying free Lie semialgebras
with a base consisting of three elements. The purpose of this case is to define a Lie semialgebra that
will be parallel to sl (2, F ) in the classical theory.
We are now going to give a necessary and sufficient condition for a bilinear multiplication defined
on a module over a semifield with a negation map to be a negated Lie bracket. This is formulated
in Corollary 2.32 and in Lemma 2.33.
Throughout this part, R is a semifield with a negation map.
Lemma 2.30. Let L be a free R-module with base B = {x1 , x2 , x3 }. Define
∀i ≤ j : [xi , xj ] = (−) [xj , xi ] =

3
X

αi,j,ℓ xℓ

ℓ=1

where ∀i, j, ℓ : αi,j,ℓ ∈ R, and ∀i, ℓ : αi,i,ℓ ∈ R◦ . If i > j, we write αi,j,ℓ = (−) αj,i,ℓ . Assume also
that
3
XX
∀i, j, k, m :
αi,ℓ,m αj,k,ℓ ∈ R◦
where

X

i,j,k ℓ=1

is the cyclic sum over i, j, k. Extend [·, ·] to [·, ·] : L × L → L by bilinearity.

i,j,k

Then L equipped with [·, ·] is a Lie semialgebra over R.
Proof. Again, it is enough to check Jacobi’s identity. We need to ensure that
∀i, j, k : [xi , [xj , xk ]] + [xk , [xi , xj ]] + [xj , [xk , xi ]] ∈ L◦
Take some i, j, k. By calculating [xi , [xj , xk ]], we get:
!
#
"
3
3
3
3
3
X
X
X
X
X
αj,k,ℓ [xi , xℓ ] =
αj,k,ℓ
αi,ℓ,m xm =
[xi , [xj , xk ]] = xi ,
αj,k,ℓ xℓ =
ℓ=1

ℓ=1

By permuting the indices,

[xk , [xi , xj ]] =

3
X

m=1

[xj , [xk , xi ]] =

3
X

m=1

By summing all of the above, we get

3
X

ℓ=1
3
X

αi,j,ℓ αk,ℓ,m
αk,i,ℓ αj,ℓ,m

ℓ=1

[xi , [xj , xk ]] + [xk , [xi , xj ]] + [xj , [xk , xi ]] =

3
X

m=1

Thus, the condition
∀m :

3
XX

m=1

m=1

ℓ=1

!

!

3
X

αj,k,ℓ αi,ℓ,m

ℓ=1

xm
xm

3
XX

αj,k,ℓ αi,ℓ,m

cyc ℓ=1

!

xm

αi,ℓ,m αj,k,ℓ ∈ R◦

i,j,k ℓ=1

is equivalent to Jacobi’s identity (by Lemma 1.70), and the assertion follows.



The above condition involving the cyclic sum may seem rather strong, since it has to hold for
each choice of 1 ≤ i, j, k, m ≤ 3. However, it turns out that it is enough to check that it holds for
a specific choice of i, j, k in which they are all different – for example, (i, j, k) = (1, 2, 3). This is
formulated in the following lemma:
Lemma 2.31. Assume that ∀i, ℓ : αi,i,ℓ ∈ R◦ and ∀i, j, ℓ : αi,j,ℓ = (−) αj,i,ℓ . If
∀m :

3
X
ℓ=1

(α1,ℓ,m α2,3,ℓ + α3,ℓ,m α1,2,ℓ + α2,ℓ,m α3,1,ℓ ) ∈ R◦

!

xm

26

GUY BLACHAR

holds, then
3
XX

∀i, j, k, m :

αi,ℓ,m αj,k,ℓ ∈ R◦

i,j,k ℓ=1

Proof. We prove the lemma in two cases.
Case 1. We first assume that two of i, j, k are equal. Without loss of generality, we assume that
j = k, and thus ∀ℓ : αj,k,ℓ ∈ R◦ , and ∀ℓ : αi,j,ℓ = (−) αk,i,ℓ . We expand the cyclic sum:
3
XX

αi,ℓ,m αj,k,ℓ

=

i,j,k ℓ=1

3
X

(αi,ℓ,m αj,k,ℓ + αk,ℓ,m αi,j,ℓ + αj,ℓ,m αk,i,ℓ ) =

ℓ=1

=

=

3
X

ℓ=1
3
X

αi,ℓ,m αj,k,ℓ +

3
X

1◦R αi,ℓ,m αj,k,ℓ +

ℓ=1

=

1◦R

3
X

1◦R αk,ℓ,m αi,j,ℓ =

ℓ=1

3
X

αi,ℓ,m αj,k,ℓ +

ℓ=1

3
X

αk,ℓ,m αi,j,ℓ

ℓ=1

and thus the sum is quasi-zero.
Case 2.

(αk,ℓ,m αi,j,ℓ + (−) αj,ℓ,m αi,j,ℓ ) =

ℓ=1

!

Now we may assume that i, j, k are different indices. Since 1 ≤ i, j, k ≤ 3, and by the
cyclic symmetry of i, j, k, there are two options: either (i, j, k) = (1, 2, 3), and the sum is
S(1,2,3) =

3
XX

αi,ℓ,m αj,k,ℓ =

3
XX

αi,ℓ,m αj,k,ℓ =

i,j,k ℓ=1

3
X

(α1,ℓ,m α2,3,ℓ + α3,ℓ,m α1,2,ℓ + α2,ℓ,m α3,1,ℓ )

3
X

(α1,ℓ,m α3,2,ℓ + α2,ℓ,m α1,3,ℓ + α3,ℓ,m α2,1,ℓ )

ℓ=1

or (i, j, k) = (1, 3, 2), and the sum is
S(1,3,2) =

ℓ=1

i,j,k ℓ=1

We notice that S(1,3,2) = (−) S(1,2,3) , since
S(1,3,2)

=

=

3
X

ℓ=1
3
X

(α1,ℓ,m α3,2,ℓ + α2,ℓ,m α1,3,ℓ + α3,ℓ,m α2,1,ℓ ) =
((−) α1,ℓ,m α2,3,ℓ + (−) α2,ℓ,m α3,1,ℓ + (−) α3,ℓ,m α1,2,ℓ ) =

ℓ=1

=

(−)

3
X

(α1,ℓ,m α2,3,ℓ + α2,ℓ,m α3,1,ℓ + α3,ℓ,m α1,2,ℓ ) = (−) S(1,2,3)

ℓ=1

But we assumed that S(1,2,3) ∈ R◦ , and thus S(1,3,2) ∈ R◦ , as required.

Corollary 2.32. Let L be a free R-module with base B = {x1 , x2 , x3 }. Define
∀i ≤ j : [xi , xj ] = (−) [xj , xi ] =

3
X

αi,j,ℓ xℓ

ℓ=1

where ∀i, j, ℓ : αi,j,ℓ ∈ R, and ∀i, ℓ : αi,i,ℓ ∈ R◦ . If i > j, we write αi,j,ℓ = (−) αj,i,ℓ . Assume that
also
3
X
∀m :
(α1,ℓ,m α2,3,ℓ + α3,ℓ,m α1,2,ℓ + α2,ℓ,m α3,1,ℓ ) ∈ R◦
ℓ=1

Extend [·, ·] to [·, ·] : L × L → L by bilinearity.

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

27

Then L equipped with [·, ·] is a Lie semialgebra over R.
Proof. By Lemma 2.31, the conditions of Lemma 2.30 hold, and thus the assertion follows.



Lemma 2.33. Let L be a Lie semialgebra over R with base B = {x1 , x2 , x3 }. Then there are αi,j,ℓ ∈
R such that
3
X
αi,j,ℓ xℓ
∀i ≤ j : [xi , xj ] = (−) [xj , xi ] =
ℓ=1

and ∀i, j, ℓ : αi,j,ℓ ∈ R, and ∀i, ℓ : αi,i,ℓ ∈ R◦ . If i > j, we denote αi,j,ℓ = (−) αj,i,ℓ . Moreover,
∀m :

3
X

(α1,ℓ,m α2,3,ℓ + α3,ℓ,m α1,2,ℓ + α2,ℓ,m α3,1,ℓ ) ∈ R◦

ℓ=1

Therefore, each such Lie semialgebra is obtained from Corollary 2.32.
Proof. The existence of αi,j,ℓ follows from the fact that L is free, and by the antisymmetry of [·, ·].
Since [·, ·] is also alternating, and by Lemma 1.70, ∀i, ℓ : αi,i,ℓ ∈ R◦ .
In addition, in the proof of Lemma 2.30, we showed that Jacobi’s identity is equivalent to
∀i, j, k, m :

3
XX

αj,k,ℓ αi,ℓ,m ∈ R◦

i,j,k ℓ=1

In particular, substituting (i, j, k) = (1, 2, 3),
∀m :

3
X

(α1,ℓ,m α2,3,ℓ + α3,ℓ,m α1,2,ℓ + α2,ℓ,m α3,1,ℓ ) ∈ R◦

ℓ=1

and thus we are finished.



We use the above way to construct a Lie semialgebra which has the same relations as sl (2, F ).
In section 2.4 we will see the naive way of constructing sl (n, R), which will be different from the
following Lie semialgebra:
Example 2.34. We take a free R-module L with base {e, f, h}, and define
[e, e] = [f, f ] = [h, h] =

0L

[e, f ] = (−) [f, e] =
[h, e] = (−) [e, h] =

h
2e

[h, f ] = (−) [f, h] =

2f

where 2e and 2f mean e + e and f + f . In the notations of Lemma 2.30, if x1 = e, x2 = f and
x3 = h, then
α1,2,3 = (−) α2,1,3 = 1R , α3,1,1 = (−) α1,3,1 = 2, α3,2,2 = (−) α2,3,2 = (−) 2
and all other coefficients are 0R . We want to prove that these coefficients satisfy the conditions of
Corollary 2.32, and thus L equipped with the bilinear extension of [·, ·] is a Lie semialgebra over R.
We need to ensure that
3
X
∀m :
(α1,ℓ,m α2,3,ℓ + α3,ℓ,m α1,2,ℓ + α2,ℓ,m α3,1,ℓ ) ∈ R◦
ℓ=1

(1) If m = 1, αi,ℓ,1 = 0R unless (i, ℓ) ∈ {(1, 3) , (3, 1)}. We are left with two summands 6= 0R :
α1,3,1 α2,3,3 + α3,1,1 α1,2,1 = 0R

(2) If m = 2, αi,ℓ,2 = 0R unless (i, ℓ) ∈ {(2, 3) , (3, 2)}. We are left with two summands 6= 0R :
α2,3,2 α3,1,3 + α3,2,2 α1,2,2 = 0R
(3) If m = 3, αi,ℓ,3 = 0R unless (i, ℓ) ∈ {(1, 2) , (2, 1)}. We are left with two summands 6= 0R :
α1,2,3 α2,3,2 + α2,1,3 α3,1,1 = 1 · 2 (−) 1 · 2 = 2◦

28

GUY BLACHAR

Since the sum is quasi-zero for each choice of m, such L exists.
2.4. Symmetrized Versions of the Classical Lie Algebras. In this subsection we construct
negated versions of the classical Lie algebras: An , Bn , Cn and Dn . We assume again that our
underlying semiring R is a semifield with a negation map.
Definition 2.35. The general linear algebra, gl (n, R), is the Lie semialgebra of all matrices of
size n × n, where the Lie bracket is the negated commutator.
Definition 2.36. We define matrices ei,j ∈ gl (n, R) by
(
1R (i, j) = (k, ℓ)
(ei,j )k,ℓ =
0R (i, j) 6= (k, ℓ)
these matrices form a base of gl (n, R).
Remark 2.37. In gl (n, R), we have the formula
[ei,j , ek,ℓ ] = δj,k ei,ℓ + (−) δi,ℓ ek,j
where
δi,j =

(

1R
0R

i=j
i 6= j

Definition 2.38. An , or the negated special linear algebra, is
An = {A ∈ gl (n + 1, R) | s (tr (A)) = 0}
Lemma 2.39. An is a subalgebra of gl (n + 1, R).
Proof. Obviously, An is a submodule of gl (n + 1, R). Since tr (AB) = tr (BA), we have
s (tr ([A, B])) = s (tr (AB + (−) BA)) = 0
so An is a subalgebra of gl (n + 1, R).



Lemma 2.40. An has the following s-base:
B = {ei,j | i 6= j} ∪ {ei,i + (−) ej,j | 1 ≤ i < j ≤ n} ∪ {1◦R ei,i | 1 ≤ i ≤ n}
Proof. Assume that A = (ai,j ) ∈ An . Therefore, s (tr (A)) = 0. There are two options:
(1) If tr (A) is achieved from a dominant zero-layered element in the diagonal of A, without loss
of generality a1,1 , then
A=

n
X

ai,j ei,j +

n
X

ai,i (ei,i + (−) e1,1 )

i=1

i,j=1
i6=j

(2) Otherwise, tr (A) is achieved from (at least) two elements, not of layer zero, such that they
“cancel” each other. Assume without loss of generality that these elements are a1,1 , . . . , ak,k ;
that means that τ (a1,1 ) = · · · = t (ak,k ) = τ (tr (A)), s (a1,1 ) , . . . , s (ak,k ) 6= 0 and for
each ℓ > k, either τ (aℓ,ℓ ) < τ (tr (A)) or s (aℓ,ℓ ) = 0. Then
A=

n
X

ai,j ei,j +

i,j=1
i6=j

n
X

ai,i (ei,i + (−) e1,1 )

i=2

Therefore, B spans An . It is easy to see that no element of B is spanned by the others, and thus it
is an s-base.

Lemma 2.41. Considering A1 , we see that its s-base consists of the elements
e = e1,2 ,

f = e2,1 ,

h = e1,1 + (−) e2,2

they satisfy the relations
[e, f ] = h;

[e, h] = [−2]0e;

[f, h] = [2]0f

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

29

Proof. Considering the formula mentioned in Remark 2.37 we get:
[e, f ] =

[e1,2 , e2,1 ] = e1,1 + (−) e2,2 = h

[e, h] =

[e1,2 , e1,1 + (−) e2,2 ] = (−) e1,2 + (−) e1,2 = [−2]0e

[f, h] =

[e2,1 , e1,1 + (−) e2,2 ] = e2,1 + (−) (−) e2,1 = [2]0f


Let L be the Lie semialgebra constructed in Example 2.34. Note that A1 is similar to L; however,
A1 has no d,s-base, whereas L is free.
Unlike the definition of An , the Lie semialgebras Bn , Cn and Dn are all defined using involutions.
In this part, we follow [23].
Definition 2.42. Let R be a semifield with a negation map, and let A1 , A2 be nonassociative
semialgebras. An antihomomorphism is a function ϕ : A1 → A2 that satisfies:
(1) ∀x, y ∈ A1 : ϕ (x + y) = ϕ (x) + ϕ (y).
(2) ∀α ∈ R, ∀x ∈ A1 : ϕ (αx) = αϕ (x).
(3) ∀x, y ∈ A1 : ϕ (xy) = ϕ (y) ϕ (x).
Definition 2.43. Let R be a semifield with a negation map, and let A be a nonassociative semialgebra. An involution is an antihomomorphism ϕ : A → A for which
∀x ∈ A : ϕ (ϕ (x)) = x
In other words, ϕ is its own inverse. We also denote involutions as ∗ : A → A, x 7→ x∗ .
Example 2.44. Let R be a semifield with a negation map. Then the transpose on Mn×n (R) is an
involution.
Definition 2.45. Let R be a semifield with a negation map, and let A be a nonassociative semialgebra with an involution ∗. An element x ∈ A is called symmetric, if
x∗ = x
x is called skew-symmetric, if
x∗ = (−) x
Lemma 2.46. Let A be a nonassociative semialgebra over R. Denote by [·, ·] the negated commutator
on A, and let ∗ be an involution on A. Then the set of all skew-symmetric elements in A,
Ã = {x ∈ A | x∗ = (−) x}
is a Lie subalgebra of A.
Proof. ∗ is an involution, and in particular an antihomomorphism, so Ã is a submodule of A.
Assume x, y ∈ A. Then
∗

∗

[x, y] = (xy + (−) yx) = y ∗ x∗ + (−) x∗ y ∗ = [y ∗ , x∗ ]
If x, y ∈ Ã, then

[x, y]∗ = [y ∗ , x∗ ] = [(−) y, (−) x] = [y, x] = (−) [x, y]

so [x, y] ∈ Ã, as needed.



Now we can define Bn , Cn and Dn by defining their involutions.
Definition 2.47. Let R be a semifield with a negation map. We define an involution ∗ on gl (k, R)
by the transpose (A∗ = At ). As stated in Lemma 2.46, we get a Lie subalgebra consisting all of the
skew-symmetric elements.
(1) When k = 2n + 1 is odd, this Lie semialgebra is called Bn , the negated odd-dimensional
orthogonal algebra.
(2) When k = 2n is even, it is called Dn , the negated even-dimensional orthogonal algebra.

30

GUY BLACHAR

Remark 2.48. Bn has the following s-base:
{ei,j + (−) ej,i | 1 ≤ i < j ≤ 2n + 1} ∪ {1◦R ei,i | 1 ≤ i ≤ 2n + 1}
whereas Dn has the following s-base:
{ei,j + (−) ej,i | 1 ≤ i < j ≤ 2n} ∪ {1◦R ei,i | 1 ≤ i ≤ 2n}
Definition 2.49. Let R be a semifield with a negation map. We define an involution ∗ on gl (2n, R)
by

∗ 

A B
Dt
(−) B t
=
C D
(−) C t
At
By taking the skew-symmetric elements, we get the Lie subalgebra Cn , the negated symplectic
algebra.
Lemma 2.50. Cn has the following base:
{ei,j + (−) en+j,n+i | 1 ≤ i, j ≤ n} ∪ {ei,n+j + ej,n+i | 1 ≤ i, j ≤ n} ∪ {en+i,j + en+j,i | 1 ≤ i, j ≤ n}


A1 A2
Proof. Denote the above set by B , and assume
∈ Cn . Then
A3 A4
∗ 




At4
(−) At2
A1 A2
A1 A2
=
=
(−)
A3 A4
A3 A4
(−) At3
At1

we get the following conditions:

At4 = (−) A1 ;

At2 = A2 ;

At3 = A3

If (Ak )i,j = ak,i,j , then


n
n
n
X
X
X
A1 A2
a3,i,j (en+i,j + en+j,i )
a2,i,j (ei,n+j + ej,n+i )+
a1,i,j (ei,j + (−) en+j,n+i )+
=
A3 A4
i,j=1

i,j=1

i,j=1

and this combination is unique, implying B is a base of Cn .



3. Solvable and Nilpotent Symmetrized Lie Algebras
3.1. Basic Definitions.
3.1.1. Solvability.
Definition 3.1. Let R be a semifield with a negation map, and let L be a Lie semialgebra with a
negation map over R. The
derived series
of L is the series defined by L(0) = L, L(1) = L′ = [L, L],
 (n−1)

(n)
(n−1)
and in general: L = L
,L
.

Definition 3.2. Let R be a semifield with a negation map, and let L be a Lie semialgebra with a
negation map over R. We say that L is solvable, if
∃n ∈ N : L(n) ⊆ L◦
Example 3.3. All abelian Lie semialgebras with a negation map are solvable.
Lemma 3.4. Any Lie semialgebra with a negation map L, for which L′ ⊆ L◦ , is abelian.
Proof. The condition is equivalent to ∀x, y ∈ L : [x, y] ∈ L◦ , which shows that L is abelian.



Lemma 3.5. Let R be a semifield with a negation map, and let L be a Lie semialgebra with a
negation map over R. If L is solvable, then so are all of its subalgebras and homomorphic images.
Proof. If K is a subalgebra of L, then it is easy to see by induction that ∀n ∈ N : K (n) ⊆ L(n) .
Also, if ϕ : L → L1 is an R-epimorphism, then



∀n ∈ N : (L1 )(n) = ϕ L(n)

Therefore, if L is solvable, there is some n ∈ N such that L(n) = L◦ ; by the above formula,
(n)

(L1 ) ⊆ L1 ◦ .

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

31

Lemma 3.6. Let R be a semifield with a negation map, and let L be a Lie semialgebra with a
negation map over R. L is solvable if and only if there exists a chain of subalgebras Lk ⊳ Lk−1 ⊳
· · · ⊳ L0 = L, such that Lk ⊆ L◦ and L′i ⊆ Li+1 .
Proof. It is easy to check that in such a chain, L(i) ⊆ Li ; so, if such a chain exists, L(k) ⊆ L◦ ,
implying L is solvable.
If L is solvable with L(k) ⊆ L◦ , than the chain L(k) ⊳ · · · ⊳ L(0) = L works.



Lemma 3.7. If I, J ⊳ L are two ideals of L, and if I is solvable, then I ∩ J is also solvable.
Proof. I ∩ J ⊳ I, so this is a direct corollary of Lemma 3.5.



Lemma 3.8. If I, J ⊳ L are two solvable ideals of L, then I + J is also solvable.
Proof. We shall prove by induction that
(k)

∀k ∈ N ∪ {0} : (I + J)

⊆ I (k) + J (k) + I ∩ J

If k = 0 the assertion is clear. Assume it is true for k, i.e.
(k)

(I + J)

⊆ I (k) + J (k) + I ∩ J

We get

i h
i
h
(k)
(k)
⊆ I (k) + J (k) + I ∩ J, I (k) + J (k) + I ∩ J =
(I + J) , (I + J)
h
i h
i h
i h
i
= I (k) , I (k) + I (k) , J (k) + I ∩ J + J (k) , J (k) + J (k) , I (k) + I ∩ J +
h
i
+ I ∩ J, I (k) + J (k) + I ∩ J ⊆ I (k+1) + J (k+1) + I ∩ J

 
 

Note that I (k) , J (k) + I ∩ J , J (k) , I (k) + I ∩ J , I ∩ J, I (k) + J (k) + I ∩ J ⊆ I ∩ J.
(I + J)

(k+1)

=

I and J are solvable, so ∃m, n ∈ N : I (m) , J (n) ⊆ L◦ . Taking k = max {m, n}, we get
(k)

(I + J)
(k)

We also know that (I ∩ J)

⊆ L◦ + I ∩ J

⊆ L◦ , and thus,

(I + J)

(2k)

⊆ (L◦ )

(k)

+ (I ∩ J)

(k)

+ L◦ ∩ I ∩ J ⊆ L◦

implying I + J is solvable.



3.1.2. Nilpotency.
Definition 3.9. Let R be a semifield with a negation map, and let L be a Lie semialgebra with a
negation map over R. The descending central series, or the lower
 central series, is the series
defined by L0 = L, L′ = L1 = [L, L], and in general: Ln = L, Ln−1 .
Definition 3.10. Let R be a semifield with a negation map, and let L be a Lie semialgebra with a
negation map over R. L is called nilpotent, if
∃n ∈ N : Ln ⊆ L◦
Example 3.11. All abelian Lie semialgebras with a negation map are nilpotent.
Lemma 3.12. ∀n ∈ N ∪ {0} : L(n) ⊆ Ln .
Proof. By induction: for n = 0, it is clear, because L(0) = L = L0 .
Assume the statement is true for n ∈ N ∪ {0}, and let x ∈ L(n+1) . Then there exist y1 , y2 ∈ L(n)
such that
x = [y1 , y2 ]
So, y1 ∈ L, and by the induction hypothesis, y2 ∈ Ln . Thus,
x = [y1 , y2 ] ∈ [L, Ln ] = Ln+1
and we get L(n) ⊆ Ln .



32

GUY BLACHAR

Corollary 3.13. Any nilpotent Lie semialgebra with a negation map is solvable.
Lemma 3.14. Let R be a semifield with a negation map, and let L be a nilpotent Lie semialgebra
with a negation map over R.
(1) All of the subalgebras and homomorphic images of L are nilpotent.
(2) If L 6= L◦ , then L◦ 6= Z (L) (in fact, L◦ ( Z (L)).
Proof.
(1) If K is a subalgebra of L, then it is easy to see by induction that ∀n ∈ N : K n ⊆ Ln . Also,
n
if ϕ : L → L′ is an R-epimorphism, then ∀n ∈ N : (L′ ) = ϕ (Ln ).
(2) Assume n is the maximal index for which Ln * L◦ . Then [L, Ln ] ⊆ L◦ , so Ln ⊆ Z (L).
∃x0 ∈ Ln \L◦ , so x0 ∈ Z (L) \L◦ , as required.

Lemma 3.15. If L′ is nilpotent, then L is solvable.
Proof. We shall prove that

n

∀n ∈ N ∪ {0} : L(n+1) ⊆ (L′ )
We use induction on n, the case n = 0 being trivial. We get
h
i h
i h
i
n−1
n−1
n−1
n
L(n+1) = L(n) , L(n) ⊆ (L′ )
, (L′ )
⊆ L′ , (L′ )
= (L′ )

Thus, if L′ is nilpotent, for some n we have (L′ )n ⊆ (L′ )◦ ⊆ L◦ , implying L(n+1) ⊆ L◦ , so L is
solvable.

Similarly to Lemma 3.7 and Lemma 3.8, we have the following lemmas:
Lemma 3.16. If I, J ⊳ L are two ideals of L, and if I is nilpotent, then I ∩ J is also nilpotent.
Lemma 3.17. If I, J ⊳ L are two nilpotent ideals of L, then I + J is also nilpotent.
3.1.3. Solvability and Nilpotency Modulo Ideals. In the semiring, the quotient algebra structure fails,
and the replacement is congruences. However, we would define equivalent definitions to solvability
and nilpotency of quotient algebras.
Definition 3.18. Let R be a semifield with a negation map, let L be a Lie semialgebra with a
negation map over R, and let I ⊳ L. We say that L is solvable modulo I, if
∃n ∈ N : L(n) ⊆ I
Remark 3.19. Solvability modulo L◦ is equivalent to solvability.
Lemma 3.20. If L is solvable modulo I, and if I is solvable (as a Lie semialgebra with a negation
map), then L is also solvable.
Proof. L is solvable modulo I, so there exists n ∈ N for which L(n) ⊆ I. But I is also solvable, so
there exists m ∈ N for which I (m) ⊆ I ◦ ⊆ L◦ . To conclude,
(m)

⊆ I (m) ⊆ L◦
L(m+n) = L(n)
implying L is solvable.



Definition 3.21. Let R be a semifield with a negation map, let L be a Lie semialgebra with a
negation map over R, and let I ⊳ L. We say that L is nilpotent modulo I, if
∃n ∈ N : Ln ⊆ I
Remark 3.22. Nilpotency modulo L◦ is equivalent to nilpotency.
Definition 3.23. Let R be a semifield with a negation map, let L be a Lie semialgebra with a
negation map over R, and let I ⊳ L. We say that L is abelian modulo I, if [L, L] ⊆ I.
Remark 3.24. Lemma 3.6 can be rephrased as follows: L is solvable if and only if there exists a
chain of subalgebras Lk ⊳ Lk−1 ⊳ · · · ⊳ L0 = L such that each Li+1 is abelian modulo Li and Lk
is abelian.

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

33

3.1.4. The Symmetrized Radical and Semisimplicity.
Definition 3.25. Let R be a semifield with a negation map, let L be a Lie semialgebra with a
negation map over R. The (negated) radical of L is
X
RadL =
I
I⊳L is solvable

That is, the negated radical is the sum of all solvable ideals of L.

Remark 3.26. In the classical theory, the radical of a finitely generated Lie algebra is its unique
maximal solvable ideal. However, in our theory, there is no guarantee that there would be one
maximal solvable ideal, since being finitely generated as a module with a negation map usually does
not mean being Noetherian (and thus, there may be infinite ascending chains of solvable ideals).
Definition 3.27. Let R be a semifield with a negation map, let L be a Lie semialgebra over R. We
say that L is locally solvable, if any finitely generated subalgebra of L is solvable.
Lemma 3.28. RadL is locally solvable.
Proof. Let L1 ⊆ RadL be a finitely generated subalgebra of L. Write
L1 = Span {x1 , . . . , xn }
By the definition of RadL, for each i there are solvable ideals Ii,1 , . . . , Ii,mi such that
xi ∈

mi
X

Ii,j

j=1

Therefore,
L1 ⊆

mi
n X
X

Ii,j

i=1 j=1

Since the sum is a finite sum of solvable ideals, by Lemma 3.8, it is also solvable. Therefore, by
Lemma 3.5, also L1 is solvable, as required.

Definition 3.29. Let R be a semifield with a negation map, let L be a Lie semialgebra over R.
If RadL = L◦ , then L is called semisimple.
Lemma 3.30. L is semisimple if and only if any abelian ideal of L is contained in L◦ .
Proof. Let L be semisimple, and let I be an abelian ideal of L. Then I is solvable, implying
I ⊆ RadL = L◦ .
Now suppose that any abelian ideal of L is contained in L◦ . If L is not semisimple, it has a
solvable ideal I * L◦ . Take a maximal n such that I (n) * L◦ ; then I (n) is an abelian ideal, a
contradiction.

3.2. Lifts of Lie Semialgebras with a Negation Map. As a continuation of subsection 1.4, we
give a definition of a lift of a Lie semialgebra with a negation map.


b ϕ
Definition 3.31. Let R be a semiring with a negation map with a lift R,
b , and let L be a Lie
b L,
b and a map ψb : L
b → L, such that the
semialgebra over R. A lift of L is a Lie algebra over R,
following properties hold:


b ψb is a lift of L as modules.
(1) L,
h
i
b : ψb ([x1 , x2 ])  ψb (x1 ) , ψb (x2 ) .
(2) ∀x1 , x2 ∈ L
We also have two parallel lemmas to Lemma 1.36:
 
b ⊆L
b be a subalgebra. Then ψb K
b is a subalgebra of L.
Lemma 3.32. Let K

34

GUY BLACHAR

 
b is a submodule of L. Therefore, we need to check that it is closed
Proof. By Lemma 1.36, ψb K
with respect to the negated Lie bracket.
 
b . Take x
b such that x  ψb (b
Let x, y ∈ ψb K
b, yb ∈ K
x) and y  ψb (b
y). Then
h
i
[x, y]  ψb (b
x) , ψb (b
y)  ψb ([b
x, yb])

b we are done.
Since [b
x, yb] ∈ K,



 
b be an ideal. Then ψb Ib is an ideal of L.
Lemma 3.33. Let Ib ⊳ L

 
Proof. By Lemma 3.32, ψb Ib is a subalgebra of L.

 
b and y ∈ L. Take x
b such that y  ψb (b
Let x ∈ ψb K
b ∈ Ib such that x  ψb (b
x), and take yb ∈ L
y).
Then
h
i
[x, y]  ψb (b
x) , ψb (b
y)  ψb ([b
x, yb])
b we are done.
Since [b
x, yb] ∈ I,



Although in general there is no obvious way to find a lift of a given Lie semialgebra (even if there
is a lift as modules), there is some consolation:


b ϕ
Theorem 3.34. Let R be a semiring with a negation map with a lift R,
b . Consider the function


b → gl (n, R)
ψb : gl n, R
 
 
b , ψb is a lift of gl (n, R) as Lie
which applies ϕ
b on each entry of a given matrix. Then gl n, R
semialgebras.


bB
b ∈ gl n, R
b ,
Proof. Since this is already a lift of modules, we will prove that given A,
   


b ψb B
b  ψb A
bB
b
ψb A

Indeed,

    
b ψb B
b
ψb A

i,j

=

n
X

k=1

n


X
b
ϕ
b (b
ai,k ) ϕ
b bbk,j  ϕ
b
ai,kbbk,j
k=1

!

 

bB
b
= ψb A

i,j

Now the assertion follows, since
h    i
   
   




b , ψb B
b
b ψb B
b + (−) ψb B
b ψb A
b  ψb A
bB
b + (−) ψb B
bA
b 
ψb A
= ψb A
i

h

b B
b
bB
b + (−) B
bA
b = ψb A,
 ψb A




b ϕ
3.2.1. Solvability, Nilpotency and Semisimplicity of the Lift. Throughout this subsection, R,
b


b ψb
will be a lift of a semifield with a negation map R, L will be a Lie semialgebra over R, and L,
will be a lift of L.
b1, L
b 2 be subalgebras of L.
b Then
Lemma 3.35. Let L
    
i
h
b 1 , ψb L
b 2 ⊆ ψb L
b 1, L
b2
ψb L

MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

35

i
 
 
h
b 1 and
b 1 and y ∈ ψb L
b 2 . We will prove that [x, y] ∈ ψb L
b 1, L
b 2 . Take x
Proof. Let x ∈ ψb L
b∈L
b 2 such that x  ψb (b
yb ∈ L
x) and y  ψb (b
y). Then
h
i
[x, y]  ψb (b
x) , ψb (b
y)  ψb ([b
x, yb])
i
h
b1, L
b 2 , we are finished.

Since [b
x, yb] ∈ L
Corollary 3.36.

and

h i
bn
∀n ∈ N ∪ {0} : Ln ⊆ ψb L

i
h
b (n)
∀n ∈ N ∪ {0} : L(n) ⊆ ψb L

Proof. We will prove the first assertion; the second is proven similarly.
We use induction on n, the case n = 0 being trivial. If the assertion holds for n ∈ N ∪ {0}, then,
by Lemma 3.35 and the induction hypothesis,
    
i
h
b , ψb L
b n ⊆ ψb L,
b L
b n = ψb (Ln+1 )
Ln+1 = [L, Ln ] ⊆ ψb L

as required.



b is nilpotent (resp. solvable), then L is nilpotent (resp. solvable).
Corollary 3.37. If L

Proof. We will prove the assertion for nilpotency; the assertion for solvability is proved similarly.
b is nilpotent, there is n such that L
b n = 0. By Corollary 3.36,
Since L
 
b n = ψb (0) = L◦
Ln ⊆ ψb L
and thus L is nilpotent.

Corollary 3.38.




 
b
⊆ Rad (L)
ψb Rad L

 

 
b ⊳ L,
b and thus, by Lemma 3.33, ψb Rad L
b
Proof. Rad L
⊳ L. In addition, by Corollary 3.37,

 
b
ψb Rad L
is solvable. Therefore, the assertion follows.


b is semisimple.
Corollary 3.39. If L is semisimple, then L

Proof. L is semisimple, hence Rad (L) ⊆ L◦ . By Corollary 3.38,

 
b
ψb Rad L
⊆ Rad (L) ⊆ L◦
 
b = 0, as required.
Hence Rad L



3.3. Cartan’s Criterion for Lie Algebras Over ELT Algebras. For this subsection, we work
only with ELT algebras. Specifically, we are interested in the following type of ELT algebra:

Definition 3.40. Let R = R (F , L ) be an ELT algebra. We say that R is a divisible ELT field,
if L is a field, and F is a divisible group.
In our version of Cartan’s criterion, we will use the essential trace defined in [6]. Let us recall its
definition:

36

GUY BLACHAR


Definition 3.41. Let R be a divisible ELT field, and let A ∈ Mn R . Let
n
X

αi λn−i
pA (λ) = det λI + [−1]0A = λn +
i=1

be its ELT characteristic polynomial. We define


τ (αk )
τ (αℓ )
≥
L (A) =
ℓ ≥ 1 | ∀k ≥ n :
ℓ
k
µ (A) = min L (A)
In words, αµ λn−µ is the first monomial after λn which is not inessential in pA (λ).
The essential trace of A is
(
etr (A) =

Recall:

[0]

tr (A) ,

τ (αµ )
µ

[−1]0tr (A)

,

is essential in pA (λ)
Otherwise

Corollary ([6, Corollary 2.22]). If A is ELT nilpotent, then s (etr (A)) = 0.
Returning to Lie semialgebras, we specialize our symmetrized Lie semialgebra to be free; therefore,
given a homomorphism ϕ : L → L, its trace and essential trace are well-defined.
Definition 3.42. Let R be a divisible ELT field, and let L be a free symmetrized Lie semialgebra
over R. The Killing form of L is
κ (x, y) = tr (adx ◦ ady )
The essential Killing form of L is
κes (x, y) = etr (adx ◦ ady )
Definition 3.43. Let R be a divisible ELT field, let L be a free symmetrized Lie semialgebra
over R, and let κ be the Killing form of L. We say that κ is non-degenerate, if
Radκ = {x ∈ L|∀y ∈ L : s (κ (x, y)) = 0} ⊆ L◦
We say that κes is non-degenerate, if
SRadκ = {x ∈ L|∀y ∈ L : s (κes (x, y)) = 0} ⊆ L◦
Remark 3.44. Since Radκ ⊆ SRadκ (by [6, Lemma 2.11]), if κes is non-degenerate, then κ is nondegenerate.
Theorem 3.45 (Cartan’s criterion). If L has a non-degenerate essential Killing form, then L is
semisimple.
Proof. By Lemma 3.30, it is enough to prove that any abelian ideal I of L is of layer zero.
Let I be an abelian ideal, and let x ∈ I. Take an arbitrary y ∈ L. Since adx ◦ ady maps L to
I, (adx ◦ ady )2 maps L to [I, I], which is of layer zero. So adx ◦ ady is ELT nilpotent. Thus, by [6,
Corollary 2.22], s (etr (adx ◦ ady )) = 0.
This argument implies x ∈ SRadκ; so I ⊆ SRadκ. But κ is strongly non-degenerate, and thus
I ⊆ L◦ , as required. So L is semisimple.


MODULES AND LIE SEMIALGEBRAS OVER SEMIRINGS WITH A NEGATION MAP

37

4. PBW Theorem
In this subsection we will construct the universal enveloping algebra of an arbitrary Lie semialgebra with a negation map. Then, we will give a counterexample to a naive version of PBW theorem.
The tensor product is a well-known process in general categories (for example, [4], [11], [19] and
[26]). Regarding tensor product of modules over semirings, there are two (nonisomorphic) notions of
tensor product of modules over semirings in the literature, both denoted by ⊗R . For our purposes, we
use the tensor product discussed in [20], [21] and [4]. The other notion of tensor product is discussed
in [26] and [10]. Their construction satisfies a different universal property, and the outcome is a
cancellative monoid.
4.1. Tensor Power and Tensor Algebra of Modules over a Semirings.
Definition 4.1. Let R be a commutative semiring, and let M be an R-module. The k-th tensor
product of M is defined as
M ⊗k = M ⊗ · · · ⊗ M
{z
}
|
k times

If k = 0, we define M ⊗0 = R.

We note that this definition is well-defined, because the tensor product is associative.
Definition 4.2. Let R be a commutative semiring, and let M be an R-module. The tensor algebra
of M , denoted T (M ), is
∞
M
T (M ) =
M ⊗k
k=0

Remark 4.3. The natural isomorphism

M ⊗k ⊗ M ⊗ℓ ∼
= M ⊗(k+ℓ)
yields a multiplication
M ⊗k × M ⊗ℓ → M ⊗(k+ℓ)
given by (a, b) 7→ a ⊗ b. Endowed with this multiplication, T (M ) is an R-semialgebra.
Lemma 4.4. Let A be an R-semialgebra, and assume that ϕ : M → A is a homomorphism of
R-modules. Then there exists a unique homomorphism of R-semialgebras ϕ̂ : T (M ) → A, such that
ϕ̂|M = ϕ.
Proof. Uniqueness forces that
ϕ̂ (x1 ⊗ · · · ⊗ xk ) = ϕ̂ (x1 ) ⊗ · · · ⊗ ϕ̂ (xk ) = ϕ (x1 ) ⊗ · · · ⊗ ϕ (xk )
This defines a homomorphism of R-semialgebras.



4.2. The Universal Enveloping Algebra of a Lie Algebra with a Negation Map. We now
return to our motivation, where we work with a Lie semialgebra L over a semifield with a negation
map R.
Definition 4.5. Let R be a semifield with a negation map, and let L be a Lie semialgebra over R.
We define a congruence ∼ on T (L) as the congruence generated by
∀x, y ∈ L : [x, y] ∼ x ⊗ y + (−) y ⊗ x
The universal enveloping algebra of L is
U (L) = T (L) /∼
If ρ : T (L) → U (L) is the canonical map, we have the PBW homomorphism of L, which we
denote ϕL : L → U (L), defined by ϕL = ρ|L .
However, the PBW homomorphism of L need not be injective, as the next subsection demonstrates.

38

GUY BLACHAR

4.3. Counterexample to the Naive PBW Theorem. We now present a Lie semialgebra with
a negation map, which cannot be embedded into a Lie semialgebra with a negation map obtained
from an associative algebra together with the negated commutator operator. In particular, ϕL will
not be injective.
Example 4.6. The example will use the ELT structre. For convenience, we assume that our underlying ELT field is R = R (C, R). Recall that in the ELT theory, the relation  is denoted .
We take a free R-module L with base B = {x1 , x2 }, and define:
[x1 , x1 ] =

[0]1x
1

+ [0]1x2

[x2 , x2 ] =

[0]1x
1

+ [0]1x2

[x1 , x2 ] =
[x2 , x1 ] =

x1 + x2
(−) [x1 , x2 ]

By Lemma 2.28, L equipped with [·, ·] is a Lie semialgebra with a negation map over R.
Consider
y1 = [x1 , [x1 , x2 ]] + [x1 , [x2 , x1 ]]
and
y2 = [x2 , [x1 , x1 ]] .
From Jacobi’s identity, we know that y1 + y2 ∈ L◦ . Let us expand them, using the definition of
the Lie bracket:


 
y1 = [x1 , [x1 , x2 ]] + [x1 , [x2 , x1 ]] = x1 , [0]0 [x1 , x2 ] = x1 , [0]0x1 + [0]0x2 =
= [0]0 [x1 , x1 ] + [0]0 [x1 , x2 ] = [0]1x1 + [0]1x2 + [0]0x1 + [0]0x2 = [0]1x1 + [0]1x2

whereas



y2 = [x2 , [x1 , x1 ]] = x2 , [0]1x1 + [0]1x2 = [0]1 [x2 , x1 ] + [0]1 [x2 , x2 ] =


= [0]1 [−1]0x1 + [−1]0x2 + [0]1 [0]1x1 + [0]1x2 = [0]2x1 + [0]2x2
Therefore, y2  y1 , but y1 6= y2 .

Now, assume that there is an ELT associative algebra A, which is a Lie semialgebra with a negation
map together with the ELT commutator, such that there exists an R-monomorphism ϕ : L → A
of Lie semialgebra with a negation maps. Denote a1 = ϕ (x1 ), a2 = ϕ (x2 ). We will show that
ϕ (y1 ) = ϕ (y2 ), which will show that ϕ is not injective.
We note that
b1 = ϕ (y1 ) = [a1 , [a1 , a2 ]] + [a1 , [a2 , a1 ]]
and
b2 = ϕ (y2 ) = [a2 , [a1 , a1 ]]
Since any associative algebra with the symmetrized commutator satisfies the strong Jacobi’s identity
(Example 2.12), b1  b2 . However, since y2  y1 , also b1 = ϕ (y2 )  ϕ (y1 ) = b2 .
By Corollary 1.6, since R◦ is idempotent,  is a partial order on L. In particular, ϕ (y1 ) = b1 =
b2 = ϕ (y2 ). But ϕ is a monomorphism, implying y1 = y2 , a contradiction.
References
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Department of Mathematics, Bar-Ilan University, Ramat-Gan 52900, Israel.
E-mail address: blachag@biu.ac.il

