# Ideals and congruences in $L$-algebras and pre-$L$-algebras

Alberto Facchini  Dipartimento di Matematica “Tullio Levi-Civita”, Università
di Padova, 35121Padova, Italy<EMAIL_ADDRESS>,  Marino Gran  Institut
de Recherche en Mathématique et Physique, Université catholique de Louvain,
1348 Louvain-la-Neuve, Belgique<EMAIL_ADDRESS>and  Mara Pompili
University of Graz, Institute for Mathematics and Scientific Computing,
Heinrichstrasse 36, 8010 Graz, Austria<EMAIL_ADDRESS>
###### Abstract.

We link the recent theory of $L$-algebras to previous notions of Universal
Algebra and Categorical Algebra concerning subtractive varieties, commutators,
multiplicative lattices, and their spectra. We show that the category of
$L$-algebras is subtractive and normal in the sense of Zurab Janelidze, but
neither the category of $L$-algebras nor that of pre-$L$-algebras are Mal’tsev
categories, hence in particular they are not semi-abelian. Therefore
$L$-algebras are a rather peculiar example of an algebraic structure.

The first author wishes to express his gratitude to the Institut de Recherche
en Mathématique et Physique of the Université catholique de Louvain for the
hospitality received during his stay in Louvain-la-Neuve.

The second author’s work was supported by the Fonds de la Recherche
Scientifique - FNRS under Grant CDR no. J.0080.23

##  1\. Introduction

The aim of this paper is to link the recent fruitful theory of $L$-algebras
[15, 16, 17] to previous articles of Universal Algebra and Categorical Algebra
concerning subtractive varieties [18, 1, 11, 12], commutators and
multiplicative lattices, and their spectra [6].

$L$-algebras are related to right $\ell$-groups, projection lattices of von
Neumann algebras, quantum Yang–Baxter equation, MV-algebras, braidings, and
non-commutative logic. In this paper we prove that the category of
$L$-algebras is subtractive [18, 11] and normal (in the sense of [12]), but
neither the category of $L$-algebras nor that of pre-$L$-algebras are Mal’tsev
categories, hence in particular they are not semi-abelian. This shows that
$L$-algebras are a rather peculiar example of algebraic structure. In general,
in a pre-$L$-algebra $X$, there is a monotone Galois connection between the
lattice ${\mathcal{C}}(X)$ of congruences of $X$, and the lattice
${\mathcal{I}}(X)$ of ideals of $X$. We show that $L$-algebras do not form a
variety in the sense of Universal Algebra, but their category is a normal
subtractive quasivariety. The lattice ${\mathcal{I}}(X)$ is a distributive
lattice, as was proved in [17]. Thus the commutator of two ideals
(congruences) turns out to be the intersection of the two ideals
(congruences).

We are grateful to Professor Leandro Vendramin for several suggestions
concerning this paper.

##  2\. Basic notions

Recall that an $L$-algebra [15] is a set $X$ with a binary operation
$(x,y)\mapsto x\cdot y$ and a $0$-ary operation $1\in X$ such that

(2.1) $\displaystyle x\cdot x=x\cdot 1=1,\ 1\cdot x=x,$ (2.2)
$\displaystyle(x\cdot y)\cdot(x\cdot z)=(y\cdot x)\cdot(y\cdot z),\ \mbox{\rm
and}$ (2.3) $\displaystyle x\cdot y=y\cdot x=1\implies x=y$

for every $x,y,z\in X$.

Similarly, we define a pre-$L$-algebra assuming that only Properties (2.1) and
(2.2) hold. These pre-$L$-algebras are called unital cycloids in [15]. It is
easily seen that in a pre-$L$-algebra the element $1$ with Property (2.1) is
unique (this follows from $x\cdot x=1$.) It is called the logical unit of the
pre-$L$-algebra. Equation (2.2) holds in most generalizations of classical
logic, including intuitionistic, many-valued, and quantum logic.

On any pre-$L$-algebra $X$ there is a natural preorder $\leq$, i.e., a
reflexive and transitive relation, defined by $x\leq y$ if $x\cdot y=1$. A
pre-$L$-algebra is an $L$-algebra if and only if this natural preorder is a
partial order, that is, if and only if it is also an antisymmetric relation.
Clearly, pre-$L$-algebras form a variety of algebras in the sense of Universal
Algebra. (We will see in Example 3 that this is not the case for
$L$-algebras.) In particular, a pre-$L$-algebra morphism is any mapping $f$
between two pre-$L$-algebras such that $f(1)=1$ and $f(x\cdot y)=f(x)\cdot
f(y)$ for all $x,y$. But notice that the first condition $f(1)=1$ follows from
the second, because

$f(1)=f(1\cdot 1)=f(1)\cdot f(1)=1$

by (2.1).

Let us now consider congruences on a pre-$L$-algebra $X$. We will see that
they correspond to suitably defined ideals of $X$. A subset $I$ of a
pre-$L$-algebra $X$ is an ideal of $X$ [15, Definition 1] if

(2.4) $\displaystyle 1\in I,$ (2.5) $\displaystyle x\in I\text{ and }x\cdot
y\in I\implies y\in I,$ (2.6) $\displaystyle x\in I\implies(x\cdot y)\cdot
y\in I,$ (2.7) $\displaystyle x\in I\implies y\cdot x\in I,$ (2.8)
$\displaystyle x\in I\implies y\cdot(x\cdot y)\in I$

for every $x,y\in X$. Clearly, $\\{1\\}$ and $X$ are ideals in any
pre-$L$-algebra $X$.

For instance, it is easily seen that if $f\colon X\to Y$ is a pre-$L$-algebra
morphism, then its kernel, that is, the inverse image $f^{-1}(1)$ of the
logical unit $1$ of $Y$, is an ideal of $X$.

Also, if $\sim$ is a congruence on a pre-$L$-algebra $X$, the equivalence
class $[1]_{\sim}$ of the logical unit $1$ of $X$ is an ideal of $X$.

Recall that if $(A,\leq)$ and $(B,\leq)$ are two partially ordered sets, a
monotone Galois connection between $A$ and $B$ consists of two order-
preserving mappings $f\colon A\to B$ and $g\colon B\to A$ such that $f(a)\leq
b$ if and only if $a\leq g(b)$ for every $a\in A$ and $b\in B$. A closure
operator on the partially ordered set $A$ is a mapping $c\colon A\to A$ such
that $x\leq c(y)$ if and only if $c(x)\leq c(y)$ for every $x,y\in A$.

######  Proposition 2.1.

Let $X$ be a pre-$L$-algebra, $\mathcal{C}(X)$ its lattice of congruences, and
$\mathcal{I}(X)$ the lattice of ideals of $X$. Define two mappings

$\varphi\colon\mathcal{C}(X)\to\mathcal{I}(X),\qquad\varphi\colon\sim{}\\!\in\mathcal{C}(X)\mapsto[1]_{\sim},$

that maps any congruence $\sim$ of $X$ to the equivalence class modulo $\sim$
of the logical unit $1$, and

$\psi\colon\mathcal{I}(X)\to\mathcal{C}(X),\qquad\varphi\colon
I\in\mathcal{I}(X)\mapsto{\sim}_{I},$

where ${\sim}_{I}$ is the congruence of $X$ defined, for every $x,y\in X$, by
$x\sim_{I}y$ if both $x\cdot y$ and $y\cdot x$ belong to $I$. Then:

  * (a)

$\varphi$ and $\psi$ are well-defined.

  * (b)

$\varphi\psi=\operatorname{id}_{\mathcal{I}(X)}$;

  * (c)

$\varphi$ and $\psi$ form a monotone Galois connection.

  * (d)

$\psi\varphi$ is a closure operator on $\mathcal{C}(X)$.

  * (e)

The image of $\psi$ is the set of all congruences $\sim$ of $X$ for wchich
$X/{\sim}$ is an $L$-algebra.

  * (f)

For every $\sim\,\in\mathcal{C}(X)$, $\psi\varphi(\sim)$ is the smallest
congruence $\equiv$ on $X$ that contains $\sim$ and is such that $X/{\equiv}$
is an $L$-algebra.

###### Proof.

(a) This is proved in [15, Proposition 1]. Notice that here “$\varphi$ is a
well-defined mapping” means that $\varphi(\sim):=[1]_{\sim}$ is an ideal of
$I$ for every congruence $\sim$ on $X$. As far as “$\psi$ is well-defined” is
concerned, we mean that, for every ideal $I$ of $X$, the relation $\sim_{I}$
on $X$, defined, for every $x,y\in X$, by $x\sim_{I}y$ if $x\cdot y\in I$ and
$y\cdot x\in I$, is a congruence on $X$.

(b) is trivial.

(c) $\varphi$ and $\psi$ are clearly order-preserving. In order to show that
they form a monotone Galois connection, we must prove that if $\sim$ is any
congruence and $I$ is any ideal, then $[1]_{\sim}\subseteq I$ if and only if
$\sim{}\\!\subseteq{}\\!\sim_{I}$. Suppose $[1]_{\sim}\subseteq I$. Fix
$x,y\in X$ with $x\sim y$. Then $x\cdot y\sim y\cdot y=1\in I$. Similarly
$y\cdot x=1\in I$. Therefore $x\sim_{I}y$. The converse is easy.

(d) follows immediately from (c).

(e) Clearly, if $I$ is an ideal of $X$, then $X/{\sim_{I}}$ is an $L$-algebra.
Conversely, let $\sim$ be a congruence with $L/\sim$ an $L$-algebra, and $I$
be the ideal $\varphi(\sim)$. We must prove that $\sim{}\\!={}\\!\sim_{I}$,
i.e., that for every $x,y\in X$, $x\sim y$ if and only if $x\cdot y\in I$ and
$y\cdot x\in I$. Now $x\sim y$ implies $x\cdot y=y\cdot y=1$, so $x\cdot y\in
I$. Similarly, $y\cdot x\in I$. Conversely, $x\cdot y\in I$ and $y\cdot x\in
I$ are equivalent to $x\cdot y\sim 1$ and $y\cdot x\sim 1$. But $L/\sim$ is an
$L$-algebra, hence we have that $[x]_{\sim}=[y]_{\sim}$, that is, $x\sim y$,
as desired.

(f) From (d) and (e) we know that $\psi\varphi(\sim)$ is a congruence on $X$
that contains $\sim$ and that $X/{\psi\varphi(\sim)}$ is an $L$-algebra. If
$\equiv$ is any other congruence on $X$ that contains $\sim$ and is such that
$X/{\equiv}$ is an $L$-algebra, then $\equiv{}\\!\\!=\\!\\!{}\sim_{I}$ for
some ideal $I$ by (e), so that $\sim_{I}\supseteq\sim$. It follows that
$\varphi(\sim_{I})\supseteq\varphi(\sim)$, that is, $I\supseteq\varphi(\sim)$.
Therefore $\equiv{}={}\sim_{I}{}={}\psi(I)\supseteq\psi\varphi(\sim)$, as
desired.∎

As a trivial consequence of the previous proposition, we have that for any
pre-$L$-algebra $X$ there is a one-to-one correspondence between ideals of $X$
and congruences $\sim$ on $X$ for which $X/{\sim}$ is an $L$-algebra. Cf. [15,
Corollary 1].

Clearly, the category of all $L$-algebras is a full reflective subcategory of
the category of all pre-$L$-algebras. The left adjoint of the inclusion
associates with any pre-$L$-algebra $X$ the $L$-algebra $X/{\sim}$, where
$\sim$ is the congruence on $X$ defined, for every $x,y\in X$, by $x\sim y$ if
$x\cdot y=1$ and $y\cdot x=1$.

Also notice that for every pre-$L$-algebra morphism between two $L$-algebras,
the kernel pair always corresponds to an ideal of the domain. This occurs
because any pre-$L$-subalgebra of an $L$-algebra is an $L$-algebra.

##  3\. L-algebras do not form a variety

######  Example 3.1.

Let $X=\\{x,y,z,1\\}$ be the $L$-algebra given by Table 3.1 and let
$Y=\\{a,b,1\\}$ be the magma given by Table 3.2.

| $x$ | $y$ | $z$ | $1$
---|---|---|---|---
$x$ | $1$ | $y$ | $z$ | $1$
$y$ | $1$ | $1$ | $x$ | $1$
$z$ | $1$ | $x$ | $1$ | $1$
$1$ | $x$ | $y$ | $z$ | $1$
Table 3.1. An $L$-algebra. | $a$ | $b$ | $1$
---|---|---|---
$a$ | $1$ | $1$ | $1$
$b$ | $1$ | $1$ | $1$
$1$ | $a$ | $b$ | $1$
Table 3.2. A pre-$L$-algebra that is not an $L$-algebra.

One checks that $Y$ is not an $L$-algebra, as $a\cdot b=b\cdot a=1$ but $a\neq
b$. The surjective map $f\colon X\to Y$, $f(x)=f(1)=1$, $f(y)=a$ and $f(z)=b$,
satisfies $f(u\cdot v)=f(u)\cdot f(v)$ for all $u,v\in X$ and $f(1)=1$. Since
the image of $f$ is not an $L$-algebra, it follows from Birkhoff’s theorem
(see for example [5, Theorem 3.1]) that the class of $L$-algebras is not a
variety.

##  4\. Subtractive and normal categories

Let us recall some definitions that are of interest in categorical algebra. A
variety of universal algebras $\mathbb{V}$ is subtractive [18] if its
algebraic theory contains a constant $0$ and a binary term $s(x,y)$ with the
properties that $s(x,x)=0$ and $s(x,0)=x$.

######  Lemma 4.1.

The variety $\mathsf{PreLAlg}$ of pre-$L$-algebras is a subtractive variety.

###### Proof.

It suffices to choose the term $s(x,y)=y\cdot x$ in the theory of
pre-$L$-algebras: this term is such that

$s(x,x)=1$

and

$s(x,1)=1\cdot x=x.$

∎

The definition of subtractive variety was extended to a categorical context by
Z. Janelidze in [11]. When a category $\mathcal{C}$ is _pointed_ , that is it
has a zero object $0$, the property of subtractivity can be defined as
follows. In a pointed category consider a reflexive relation
$(R,r_{1},r_{2},e)$ on an object $X$, where $r_{1}$, $r_{2}$ are the
projections and $e\colon X\rightarrow R$ is the morphism giving the
reflexivity: $r_{1}\circ e=1_{X}=r_{2}\circ e$. One says that the relation
$(R,r_{1},r_{2},e)$ is right (resp. left) punctual if there is a morphism
$t\colon X\rightarrow R$ (resp. $s\colon X\rightarrow R$) such that
$r_{2}\circ t=1_{X}$ and $r_{1}\circ t=0$ (resp. $r_{2}\circ s=0$ and
$r_{1}\circ s=1_{X}$).

######  Definition 4.2.

[11] A finitely complete pointed category is _subtractive_ if any right
punctual reflexive relation is left punctual.

As shown in [11], a pointed variety is subtractive in the sense of Definition
4.2 if and only if is subtractive in the sense of [18]. In particular the
variety $\mathsf{PreLAlg}$ is a subtractive category, and this implies the
following:

######  Lemma 4.3.

The category $\mathsf{LAlg}$ is subtractive.

###### Proof.

This follows from the fact that the inclusion functor
$U\colon\mathsf{LAlg}\rightarrow\mathsf{PreLAlg}$ is full, faithful, preserves
all limits (as any right adjoint), hence it preserves and reflects punctual
right (resp. left) reflexive relations, and the zero object. ∎

The next result we shall prove concerns $L$-algebras again. They turn out to
form a normal category in the following sense:

######  Definition 4.4.

[12] Let $\mathcal{C}$ be a finitely complete category with a zero object $0$.
Then $\mathcal{C}$ is a _normal_ category if

  1. (1)

any morphism $f\colon A\rightarrow B$ admits a factorization $f=m\cdot q$,
where $q$ is a normal epimorphism (=a cokernel) and $m$ is a monomorphism:

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{q}$$\textstyle{B}$$\textstyle{{C\,\,}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{m}$

  2. (2)

normal epimorphisms are stable under pullbacks, i.e., given any pullback

$\textstyle{{A\times_{C}B}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p_{2}}$$\scriptstyle{p_{1}}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{C}$

in $\mathcal{C}$, where $f$ is a normal epimorphism, then $p_{2}$ is a normal
epimorphism.

Equivalently, one can define a normal category as a regular category [2] with
a zero object with the property that any regular epimorphism is a cokernel.

Observe then that $\mathsf{LAlg}$ is a quasivariety, since it is the class of
algebras determined by adding a finite number of implications to the theory of
$\mathsf{PreLAlg}$. Example 3 shows that these implications cannot be
“transformed” into an equivalent set of identities, since in that case
$\mathsf{LAlg}$ would then be a subvariety of $\mathsf{PreLAlg}$ and then it
would be stable in it under quotients, and this is not the case. As a
consequence $\mathsf{LAlg}$ is not a variety of algebras.

######  Proposition 4.5.

The category $\mathsf{LAlg}$ of $L$-algebras is a normal subtractive category.

###### Proof.

Let us check that in $\mathsf{LAlg}$ every surjective homomorphism is a
cokernel. The terms $t_{1}(x,y)=x\cdot y$ and $t_{2}(x,y)=y\cdot x$ have the
property that

$(\forall i\in\\{1,2\\}\,\,t_{i}(a,b)=1)\Leftrightarrow a=b.$

These terms will attest the $1$-regularity (in the sense of Beutler [3]) of
$\mathsf{LAlg}$ (also see [9, Theorem 2.2]).

Note that the terms $t_{1}(x,y)=x\cdot y$ and $t_{2}(x,y)=y\cdot x$ also
satisfy the identities $t_{i}(x,x)=1$, for $i\in\\{1,2\\}$. Let $f\colon
A\rightarrow B$ be a surjective homomorphism, $\kappa\colon K\rightarrow A$ be
the inclusion of its kernel $K$ in $A$ (that we shall omit to simplify the
notations), and $g\colon A\rightarrow C$ be any homomorphism such that
$(g\circ\kappa)(k)=g(k)=1$, for every $k\in K$. Now, for any $b\in B$ there is
an $a\in A$ such that $f(a)=b$. Let us prove that, by setting $\phi(b)=g(a)$,
we get a well-defined function. To this end, we will show that if $a$ and
$a^{\prime}$ are such that $f(a)=f(a^{\prime})$ one always has that
$g(a)=g(a^{\prime})$. First observe that

$f(t_{i}(a,a^{\prime}))=t_{i}(f(a),f(a^{\prime}))=t_{i}(f(a),f(a))=1,$

hence $t_{i}(a,a^{\prime})\in K$. This implies that

$t_{i}(g(a),g(a^{\prime}))=g(t_{i}(a,a^{\prime}))=1,$

so that $g(a)=g(a^{\prime})$, and $\phi$ is well-defined. It remains to prove
that $\phi$ is a homomorphism: this is easy, because $\phi\circ f=g$ is a
homomorphism by assumption and $f$ is a surjective homomorphism. The
uniqueness of the factorization $\phi$ is clear, and $f$ is then necessarily
the cokernel of its kernel $\kappa$.

To conclude that $\mathsf{LAlg}$ is a normal category, now it suffices to
observe that any quasivariety is a regular category (see [14], for instance)
since surjective homomorphisms are always stable under pullbacks in a
quasivariety. From Lemma 4.3 it follows that $\mathsf{LAlg}$ is indeed a
normal subtractive category. ∎

The subtractive variety $\mathsf{PreLAlg}$ of pre-$L$-algebras then contains
the normal subtractive quasivariety $\mathsf{LAlg}$ of $L$-algebras. One might
then wonder whether $\mathsf{LAlg}$ is also a Mal’tsev category [4]. The
answer to this question is negative, as we are now going explain. For this,
let us fix some notations. The kernel pair $(Eq(f),p_{1},p_{2})$ of a morphism
$f\colon A\to B$ is the (effective) equivalence relation obtained by the
pullback of $f$ along itself:

$\textstyle{Eq(f)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p_{2}}$$\scriptstyle{p_{1}}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{B.}$

In the case of a quasivariety of universal algebras, then the kernel pair of a
homomorphism $f\colon A\rightarrow B$ is the congruence

$Eq(f)=\\{(a,a^{\prime})\in A\times A\,\mid\,f(a)=f(a^{\prime})\\}.$

We’ll be interested in the following two elements $L$-algebra $X$ whose
multiplication is defined in the following Table:

| $0$ | $1$
---|---|---
$0$ | $1$ | $1$
$1$ | $0$ | $1$

Consider then the relation $R=\\{(0,1),(1,0),(1,1)\\}$ on $X$: this is easily
seen to be a subalgebra of the product $L$-algebra $X\times X$.

Write $p_{1}\colon R\rightarrow X$ and $p_{2}\colon R\rightarrow X$ for the
first and the second projections, and $Eq(p_{1})$ and $Eq(p_{2})$ for the
congruences associated with these homomorphisms, namely

$Eq(p_{1})=\\{((0,1),(0,1)),((1,0),(1,0)),((1,1),(1,1)),((1,0),(1,1))\\}$

and

$Eq(p_{2})=\\{((0,1),(0,1)),((1,0),(1,0)),((1,1),(1,1)),((0,1),(1,1))\\}.$

Then, clearly,

$(1,0)Eq(p_{1})(1,1)Eq(p_{2})(0,1)$

showing that

$((1,0),(0,1))\in Eq(p_{2})\circ Eq(p_{1}).$

However,

$((1,0),(0,1))\not\in Eq(p_{1})\circ Eq(p_{2}),$

hence

$Eq(p_{1})\circ Eq(p_{2})\not=Eq(p_{2})\circ Eq(p_{1}).$

This shows that:

######  Proposition 4.6.

The categories $\mathsf{LAlg}$ and $\mathsf{PreLAlg}$ are not Mal’tsev
categories.

###### Proof.

The fact that $\mathsf{LAlg}$ is not a Mal’tsev category follows from the fact
that the two congruences $Eq(p_{1})$ and $Eq(p_{2})$ on the $L$-algebra $X$
above do not permute in the sense of composition of relations. Since the
category $\mathsf{LAlg}$ is stable in $\mathsf{PreLAlg}$ under subalgebras and
products in $\mathsf{LAlg}$, the same counter-example also shows that
$\mathsf{PreLAlg}$ is not a Mal’tsev category. ∎

In particular, the above proposition implies that $\mathsf{LAlg}$ and
$\mathsf{PreLAlg}$ are not semi-abelian categories.

######  Remark 4.7.

It is well-known that any subtractive variety $\mathbb{V}$ (with a constant
$1$) is “permutable at $1$”, which means that, for any pair of congruences $R$
and $S$ on any algebra $X$ in $\mathbb{V}$, the following implication holds:

$(x,1)\in S\circ R\quad\Leftrightarrow\quad(x,1)\in R\circ S$

In the case of the variety $\mathsf{PreLAlg}$ we have the term $s(x,y)=y\cdot
x$. Accordingly, when $(x,1)\in S\circ R$, from the existence of a $y$ such
that $xRyS1$, one deduces that

$x=s(x,1)Ss(x,y)Rs(y,y)=1,$

that is, $(x,1)\in R\circ S$. The variety $\mathsf{PreLAlg}$ is then
“permutable at $1$”, even though it is not congruence permutable.

##  5\. Commutators

In Proposition 2.1 we have considered, for a pre-$L$-algebra $X$, the lattice
of congruences $\mathcal{C}(X)$, and the lattice of ideals $\mathcal{I}(X)$.
These are complete lattices because any intersection of congruences (of
ideals) is a congruence (an ideal). Let us focus onto the case of ideals.
Notice that ideals of $X$ are pre-$L$-subalgebras of $X$ (essentially because,
for a congruence $\sim$, $x\sim 1$ and $y\sim 1$ imply $x\cdot y\sim 1$).
Since the lattice $\mathcal{I}(X)$ is complete, there is an obvious notion of
ideal of $X$ generated by a subset of $X$.

Let us consider the notion of commutator of two ideals of an $L$-algebra $X$.
In Group Theory, if we have two normal subgroups $M$ and $N$ of a group $G$,
the commutator $[M,N]$ is the smallest normal subgroup of $G$ for which group
multiplication $\mu\colon M\times N\to G/[M,N]$, $\mu(m,n)=mn[M,N]$, where
$mn[M,N]$ is the coset of $mn$ in the quotient $G/[M,N]$, is a group
homomorphism. This argument can be repeated for $L$-algebras, as follows.

Let $X$ be an $L$-algebra and $I,J$ be two ideals of $X$. Define their
commutator $[I,J]$ as the smallest ideal of $X$ for which the multiplication
$\cdot$ in $X$, i.e., the mapping $\mu\colon I\times J\to X/[I,J]$,
$\mu(i,j)=[i\cdot j]_{\sim_{[I,J]}}$, is an $L$-algebra morphism. Notice that
the ideal $[I,J]$ is always contained in $I\cap J$. This follows from the
remark that the mapping $\mu\colon I\times J\to X/I\cap J$ is clearly an
$L$-algebra morphism. One actually has the following

######  Proposition 5.1.

For every pair $I,J$ of ideals of an $L$-algebra $X$, one has

$[I,J]=I\cap J.$

###### Proof.

We only need to prove that $I\cap J\subseteq[I,J]$. For this it will suffice
to show that, for any $x\in I\cap J$, its equivalence class $[x]_{\sim_{I\cap
J}}$, that will be simply written $[x]$, is the neutral element in the
quotient $X/[I,J]$:

$[x]=[1].$

By assumption, for any $i\in I$, $j\in J$, one has the equality

$([x]\cdot[x])\cdot([i]\cdot[j])=([x]\cdot[i])\cdot([x]\cdot[j]).$

By choosing $i=1$ and $j=x$ we get

$([x]\cdot[x])\cdot([1]\cdot[x])=([x]\cdot[1])\cdot([x]\cdot[x]),$

from which it follows that $[x]=[1],$ as desired. ∎

In particular, this result implies that the only abelian algebras in the
quasivariety of $L$-algebras are the trivial ones:

######  Corollary 5.2.

Let $X$ be an abelian $L$-algebra. Then $|X|=1$.

###### Proof.

By Proposition 5.1 the condition $[X,X]=\\{1\\}$ gives $X\cap X=X=\\{1\\}.$ ∎

######  Remark 5.3.

_The fact that the commutator $[I,J]$ of two ideals is simply their
intersection $I\cap J$ is not surprising, since the category $\mathsf{LAlg}$
is congruence distributive, as it follows from Proposition 2.1 and the fact
that the lattice of ideals on each $L$-algebra is distributive [17]. It would
then be interesting to revisit the results on commutators of congruences in
terms of pseudogroupoids in varieties [10] in the more general context of
quasivarieties (also see [13])._

We now consider the multiplicative lattice $(\mathcal{I}(X),\cap)$ in the
sense of [6], which has been implicitly considered in [17]. Now, primes ideals
studied in [17] agree with the notion of prime elements of [6]. Notice that an
ideal $P$ of an $L$-algebra $X$ is prime if and only if $P$ is a
$\wedge$-irreducible element of the lattice $\mathcal{I}(X)$.

Recall that an ideal $I$ of $X$ is semiprime if, for every ideal $J$,
$[J,J]\subseteq I$ implies $J\subseteq I$. Hence Proposition 5.1 trivially
implies that:

######  Corollary 5.4.

In an $L$-algebra, every ideal is semiprime.

Thus, in an $L$-algebra, every ideal is an intersection of prime ideals [6].
Now, in any multiplicative lattice, the lattice of all semiprime elements is
isomorphic to the lattice of all open subsets of the Zariski spectrum. The
Zariski spectrum is always a sober space [6, 17]. Hence we find that:

######  Proposition 5.5.

[17] For an $L$-algebra $X$, the lattice $\mathcal{I}(X)$ of all ideals of $X$
is isomorphic to the lattice of all open subsets of the sober topological
space $\operatorname{Spec}(X)$. In particular, the lattice $\mathcal{I}(X)$ is
a complete distributive lattice.

In view of Proposition 5.1, several classical notions of Algebra trivialize
for $L$-algebras. For instance, solvable $L$-algebras, nilpotent $L$-algebras,
$L$-algebras with empty Zariski spectrum are only those with one element, the
centralizer of any nontrivial ideal is the trivial ideal, the center of any
$L$-algebra is the trivial ideal, and the central series and the derived
series are always stationary.

Notice that maximal ideals are prime, because the zero element is
$\wedge$-irreducible in the lattice of two elements. It would be interesting
to describe simple $L$-algebras.

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