On Four Dimensional Absolute Valued Algebras With nonzero
Omnipresent Idempotent
NOUREDDINE MOTYA 1, ABDELHADI MOUTASSIM 2
1Science Mathematics and Applications Laboratory(LaSMA).
Sidi Mohamed Ben Abdellah University, Faculty of Sciences Dhar El Mehraz, Fez.
2Regional Center for Education and Training, settat.
MOROCCO
Abstract: In this paper, we studies the absolute valued algebras of dimension four, containing nonzero om-
nipresent idempotent. And we construct algebraically some news classes of algebras.
Key-Words: Absolute valued algebras, omnipresent idempotent, central idempotent.
Received: December 18, 2022. Revised: June 4, 2023. Accepted: June 25, 2023. Published: July 25, 2023.
1 Introduction
An absolute valued algebra, is a nonzero real algebra,
that is equipped with a multiplicative norm (∥xy∥=
∥x∥∥y∥). These algebras have attracted the attention
of many mathematicians, [3], [7], [8], [9], [10], [11],
[12], [13], [14], [15]. In 1947 Albert, [1]. Proved that
the finite dimensional unital absolute valued algebras
are classified by R,C,H,O. And that every finite di-
mensional absolute valued algebra is isotopic to one
of the algebras R,C,H,O. And so has dimension 1,
2, 4 or 8, [1]. Note that, the norm ∥.∥of any finite-
dimensional absolute valued algebras, comes from an
inner product (./.), [2]. Urbanik and Wright proved
in 1960 that, all unital absolute valued algebras are
classified by R,C,H,O, [4]. It is easily to seen
that, the one-dimensional absolute valued algebras are
classified by R. And it is well-known that the two-
dimensional absolute valued algebras, are isomorphic
to, C, *C,C∗, or ∗
C, [5]. The four-dimensional ab-
solute valued algebras, have been described by M.I.
Ramírez Álvarez in 1997, [6]. The problem of classi-
fying all four (eight)-dimensional absolute valued al-
gebras seems still to be open.
Motivated by these facts, we became interested in
the study of four-dimensional absolute valued alge-
bras, with a nonzero omnipresent idempotent. which
generalizes the studies of M.L. El-Mellah, [3]. The
classification of these algebras containing only one
two-dimensional sub-algebra is still an open prob-
lem. We note that there are a four-dimensional ab-
solute valued algebras, with left unit not containing a
nonzero omnipresent idempotent, [6]. On the other
hand the four-dimensional absolute valued algebra
with a nonzero central idempotent, contains a subal-
gebras of dimension two. Which means that a cen-
tral idempotent is an omnipresent idempotent . The
reciprocal does not hold in general, and the counter-
example is given (remark 3.2). From the comments
below, it arises in a naturel way the following ques-
tion: what is the classification of four-dimensional
absolute valued algebras with a nonzero omnipresent
idempotent and containing two different sub-algebras
of dimension two?. This paper is devoted to shed
some lighe on this problem.
In section 2, we introduce the basic tools for the
study of four-dimensional absolute valued algebras,
with a nonzero omnipresent idempotent, and contain-
ing two different sub-algebras of dimension two.
Moreover, In section 3, we introduce news classes
of four-dimensional absolute valued algebras, with a
nonzero omnipresent idempotent, namely M1,M2,
M3,M4,∗
M1,∗
M2,∗
M3,∗
M4, *M1, *M2, *M3, *M4,
M∗
1,M∗
2,M∗
3and M∗
4.
In section 4, we classify algebraically, all four-
dimensional absolute valued algebras, containing at
least, two different subalgebras of dimension two.
In section 5, we summarize our study in the ta-
ble.6.
2 Notations and Preliminary Results
Throughout this paper, the word algebra refers to a
non-necessarily associative algebra, over the field of
real numbers R.
Definition 2.1 Let Abe an arbitrary algebra.
i) Ais called a normed algebra (resp, absolute val-
ued algebra) if it’s endowed with a space norm:
∥.∥such that ∥xy∥≤∥x∥∥y∥(resp, ∥xy∥=
∥x∥∥y∥), for all x, y ∈A.
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ii) Ais called a division algebra if, for all nonzero
a∈A, the operators La(x) = ax and Ra(x) =
xa (for all x∈A) of left and right multiplica-
tion by aare bijectives. Note that every finite-
dimensional absolute valued algebra is a division
algebra.
iii) We mean by a nonzero omnipresent idempotent,
an idempotent which is contained in all two-
dimensional sub-algebras of A.
iv) A(x,y) denote the sub-algebra of A generated by
x, and y.
The most natural examples of absolute valued alge-
bras are R,C,H(the algebra of Hamilton quaternion),
and O(the algebra of Cayley numbers). The algebras
*C,C∗, and ∗
C(obtained by endowing the space C
with the products defined respectively by
x∗y= ¯xy, x ∗y=x¯y, and x ∗y= ¯x¯y(P)
Where x→¯xis the standard conjugation of C.
We shall also denote by *H,H∗, or ∗
Hthe real
algebras obtained by endowing the space Hwith the
products defined by (P) respectively, with x→¯xis
the standard conjugation of H. The reader is referred
to [11] for more informations of these classical
absolute valued algebras.
We need the following results.
Theorem 2.2 .[2]. The norm of any finite dimen-
sional absolute valued algebra come from an inner
product.
Lemma 2.3 .[9]. Every algebra in which x2= 0 only
if x= 0. Contains a nonzero idempotent.
Lemma 2.4 . Let Abe an absolute valued algebra of
dimension n≥2, containing a nonzero central idem-
potent e, and let B a 2-dimensional sub-algebra of A.
Then B contains a nonzero element orthogonal to e.
Proof.. Let a, b be an orthonormal basis of B. Then
there exists λ, β ∈Rand u, v ∈ {e}⊥such that a=
λe +u,b=βe +v. Now βa −λb =w∈B\ {0}
and w=βu −λv ∈ {e}⊥
Lemma 2.5 . Let Abe a four-dimensional absolute
valued algebra, containing a nonzero central idem-
potent f, then the following statements hold:
i) Acontains a 2-dimensional sub-algebra.
ii) x2=−∥x∥2f, for all x∈ {f}⊥.
iii) If e∈Ais another nonzero idempotent such that
e=f, then the subalgebra A(e, f )is isomorphic
to
∗
C.
Proof..
i) We can induce isometries from the commutative
linear isometries Lfand Rfon the orthogonal
space {e}⊥:= Eof dimension 3. So there ex-
ist common norm-one eigenvector u∈Efor
both Lfand Rfassociated to eigenvalues α, β ∈
{−1,1}. That is, u2=−f. Consequently
A(u, f)is a two-dimensional subalgebra of A.
ii) As Ahas an inner product space, we can assum
that ∥x∥= 1. We have
∥x2−f∥=∥x−f∥∥x+f∥= 2
That is (x2/f) = −1, then x2=−f.
iii) As e=f. We have
∥e−f∥=∥e2−f2∥=∥e−f∥∥e+f∥
That is ∥e+f∥= 1, this imply (e/f) = −1
2. So
e+f+ef = 0
Consequently, A(e, f)is isomorphic to ∗
C.
Lemma 2.6 . Let Abe a four-dimensional absolute
valued algebra containing 2-dimensional subalgebra
B.
1) If x∈B⊥, then x2∈B.
2) If fis a nonzero central idempotent of A, then f
is an omnipresent idempotent.
Proof.. By Rodríguez theorem, [5]. Bis isomorphic
to C, *C,C∗, or ∗
C, and by lemma 2.3, Bcontains a
nonzero idempotent e. We can set B=A(e, i), where
ei =±i,ie =±i, and i2=±e.
1) We have {e, i}is an orthonormal basis of B,
which can be extended to an orthonormal basis
F={e, i, j, k}of A. Since Ljis bijective, there
exist j1, and j2, such that
jj1=i and jj2=e
Let x=aj +bk ∈B⊥, we have
(j1/e) = (jj1/je) = (i/je) = ±(ie/je) = ±(i/j) = 0
And
(j1/i) = (jj1/ji) = (i/ji) = ±(ei/ji) = ±(e/j) = 0
Hence, j1=αj +βk, likewise j2=α′j+β′k.
So we have
i=jj1=αj2+βjk
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and
e=jj2=α′j2+β′jk
As αβ′−βα′=±1, then j2∈B, and jk ∈B.
Similarly we show that k2∈Band kj ∈B, so
x2=a2j2+b2k2+ab(jk +kj)∈B
2) Let x be a nonzero element of A orthogonal to
f, so x2=−∥x∥2fthen f=−∥x∥−2x2∈B
which mean that f is an omnipresent idempotent.
Remark 2.7 . For any orthogonal two elements
x, y ∈e⊥, we have (xy/yx) = −(x2/y2).
Proof. A simple linearisation of the identity ∥x2∥=
∥x∥2give this result.
3 New class of four-dimensional
absolute valued algebras with a
nonzero omnipresent idempotent
In this paraghraph we construct some news classes
of four-dimensional absolute valued algebras with a
nonzero omnipresent idempotent.
3.1 Construction of M1,M2,M3, and M4
Let {e, i, j, k}be the orthonormal basis of the algebra
Hof quaternions with the usual multiplication table:
T able.1.H
e i j k
e e i j k
i i -e k -j
j j -k -e i
k k j -i -e
Let ϕ, ψ, Λthe linaer isometries of the euclidian
space Hwhose matrices with respect to the canonical
basis are given, respectively, by diag {1,1,1,−1},
diag {1,1,−1,−1}, diag{1,1,−1,1}. We define
news multiplications on the space H.
x∗1y=ϕ(x)ϕ(y)
x∗2y=ϕ(x)y
x∗3y=xϕ(y)
x∗4y=ψ(x)Λ(y)
we get new class of algebras with the multiplica-
tion tables defined respectively by:
T able.2.M1
e i j k
e e i j -k
i i -e k j
j j -k -e -i
k -k -j i -e
T able.3.M2
e i j k
e e i j k
i i -e k -j
j j -k -e i
k -k -j i e
T able.4.M3
e i j k
e e i j -k
i i -e k j
j j -k -e -i
k k j -i e
T able.5.M4
e i j k
e e i -j k
i i -e -k -j
j -j k -e -i
k -k -j -i e
Lemma 3.1 . The algebras M1,M2,M3, and M4are
absolute valued algebras with omnipresent idempo-
tent e.
Proof. All these algebras are trivially absolute val-
ued. We have also:
1. e is central idempotent for M1, so it’s an om-
nipresent.
2. e is left-unit for algebra M2so the only non zero
idempotent. It belongs to all subalgebras of M2,
[10]. So e is omnipresent.
3. e is a right-unit for algebra M3so it is om-
nipresent.
4. Let B be a two dimensional sub-algebra of
M4, then there exist an nonzero idempotent f
and t in B, such that (f/t) = 0 and t2=
±f. Using the basis {e, i, j, k}there exists
α1, β1, γ1, δ1, α2, β2, γ2, δ2∈Rsuch that f=
α1e+β1i+γ1j+δ1kand t=α2e+β2i+
γ2j+δ2kWe have
i2=j2=−e, k2=e, ie =ei =i
je =ej =−j, ke =−k, ek =k
And
ik =ki =−j, ij =−ji =−k, jk =kj =−i
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Since f2=f, then
α2
1−β2
1−γ2
1+δ2
1=α1(1)
2α1β1−2γ1δ1=β1(2)
−2α1γ1−2β1δ1=γ1(3)
δ1= 0 (4)
As ∥f∥= 1, then α2
1+β2
1+γ2
1+δ2
1= 1 also
α2
1−β2
1−γ2
1−δ2
1=α1(1), we get
2α2
1−α1−1 = 0
thus α1= 1 or α1=−1
2.
(a) If α1= 1, therefore e=f∈B
(b) If α1=−1
2, then the equalities (2) and (4)
give β1=δ1= 0. So
f=−1
2e±√3
2j∈A(e, j)
On the other hand, we know that t2=±f
then
α2
2−β2
2−γ2
2+δ2
2=±α1(5)
2α2β2−2γ2δ2= 0 (6)
−2α2γ2−2β2δ2=±γ1(7)
δ2= 0 (8)
Since ∥t∥=∥f∥= 1, then α2
2+β2
2+
γ2
2+δ2
2= 1. The equalities (5)and (8) give
2α2
2−1 = ±α1, thus α2= 0 and γ2= 0.
Hence the equalities (6) and (8) imply that
β2=δ2= 0, that is, t=α2e+γ2j∈
A(e, j). Therefore B=A(e, j), so e∈B.
As a result eis a nonzero omnipresent idem-
potent of M4.
Remark 3.2 .eis a nonzero omnipresent idempotent
for the algebras M2,M3and M4which isn’t a central
idempotent.
3.2 Construction of the standard isotope of
M1,M2,M3and M4
Let Mdenote one of absolute valued algebras M1,
M2,M3or M4. We constructon the vectorial space
of Mby the news multiplications given respectively,
by x∗y= ¯x¯y,x∗y= ¯xy,x∗y=x¯y, where x→¯x
is the standard conjugation of M. The algebras ob-
tained called the standard isotopes of M, and denoted
respectively by ∗
M, *M,M∗.
Since the conjugation is an isometry, ∗
M, *M,M∗
are absolute valued algebras, As any two dimensional
sub-algebra of Mis invariant under conjugation, then
e is also an omnipresent idempotent of these news al-
gebras.
4 Main results
In this section, we assume that Ais a four dimensional
absolute valued algebra with omnipresent idempotent
eand having at least two different subalgebras B1and
B2of dimension two.
We have the following studies.
4.1 B1and B2are isomorphic to Cor
∗
C
Proposition 4.1 . If B1and B2are isomorphic to C.
Then Ais isomorphic to H,M1,M2or M3.
Proof. Let B1=A(e, i)and B2=A(e, j)be a two
subalgebras of Aisomorphic to C, then we have
i2=j2=−e, ie =ei =i and je =ej =j
We know also (e/i) = (e/j) = 0, so without loss of
generality we may assume that (i/j)=0. indeed, if
(i/j)= 0 then t=j−(i/j)i
∥j−(i/j)i∥is orthogonal to e. Since
te =et =tand ∥e∥=∥t∥= 1, we get t2=−e.
Which implies that A(e, t)is isomorphic to C.
Now in Athere exists an orthonormal subset {e, i, j}
which can be extended to an orthonormal basis
{e, i, j, k}for A. Since k∈ {e, i, j}⊥, then k2∈
A(e, i)∩A(e, j) = {e}(lemma 2.6.(1)). We get
k2=±e. But since
(ek/e) = (ek/e2) = (e/k) = 0
(ke/e) = (ke/e2) = (k/e) = 0
(ek/i) = (ek/ei) = (k/i) = 0
(ke/i) = (ke/ie) = (k/i) = 0
(ek/j) = (ek/ej) = (k/j) = 0
(ke/j) = (ke/je) = (k/j) = 0
we obtain ek =εk and ke =ζk, where |ε|=|ζ|= 1.
We conclude that A(e, k)is two-dimensional subalge-
bra of A, that is A(e, k)is isomorphic to C, *C,C∗or
∗
C. We distinguish the following cases:
1. If A(e, k)is isomorphic to C.
Then ewill be the unit element of Aand, there-
fore the multiplication of A is given by Table.1,
so Ais isomorphic to the quaternion H.
2. If A(e, k)is isomorphic to ∗
C.
So ke =ek =−kand k2=−e, since
(ij/e) = (ij/−i2) = −(i/j) = 0
(ij/i) = (ij/ie) = (i/j) = 0
and
(ij/j) = (ij/j) = (i/j) = 0
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Hence ij =kor ij =−k. In a similar manner,
we can show that
ik =j or ik =−j
and
jk =i or jk =−i
Assume that ij =k, in this case we have ik =j
and jk =−i. Indeed, if ik =−j, then
i(j+k) = k−j=−ek −ej =−e(k+j)
Which gives i=−e(Ahas no zero divisors),
contradiction. Also if jk =i, then
(i+j)k=j+i= (j+i)e
which implies k=e, absurd. Moreover, by re-
mark 2.7, we have
(ij/ji) = −(i2/j2) = −1
which means that
∥ij +ji∥2= 0
So ij =−ji, and by the same way we have
ik =−ki, and jk =−kj. Therefore, the mul-
tiplication of A is given by the Table.2, which
mean that A is isomorphic to M1.
3. A(e, k)is isomorphic to *C.
We have ek =k,ke =−kand k2=e. Using
remark 2.7, we have (ik/ki) = −(i2/k2) = 1
which means that
∥ik −ki∥2= 0
So ik =ki, similarly, we get
jk =kj and ij =−ji
By simple calculations, we show that
ij =k or ij =−k
,
ik =j or ik =−j
and
jk =i or jk =−i
Assume that ij =k, in this case we have ik =
−jand jk =i. Indeed, if ik =j, then
i(j+k) = k+j=ek +ej =e(k+j)
Which gives i=e(Ahas no zero divisors), con-
tradiction. Also if jk =−i, then
(i+j)k=−j−i=−je −ie =−(j+i)e
which implies k=−e, absurd. So the multi-
plication of Ais given by the Table.3, and A is
isomorphic to M2.
4. A(e, k)is isomorphic to C∗,
We have ek =−k,ke =kand k2=e. By
remark 2.7, we get
ik =ki, jk =kj and ij =−ji
And by simple calculations, we show that
ij =k or ij =−k
,
ik =j or ik =−j
and
jk =i or jk =−i
Assume that ij =k, in this case we have ik =j
and jk =−i. Indeed, if ik =−j, then
i(j+k) = k−j=−ek −ej =−e(k+j)
This implies that i=−e(Ahas no zero divi-
sors), contradiction. Also if jk =i, then
(i+j)k=j+i=je +ie = (j+i)e
which implies k=e, absurd. Then the product
of Ais given by Table.4, So A isomorphic to M3.
Proposition 4.2 . If B1and B2are isomorphic to
∗
C.
Then Ais isomorphic to
∗
H,
∗
M1,
∗
M2or
∗
M3.
Proof.. we define a new multiplication on Aby
x∗y= ¯x¯y, we obtain an algebra ∗
Awhich contains two
different subalgebras isomorphic to C. Therefore, ap-
plying proposition 4.1, ∗
Ais isomorphic to H,M1,M2
or M3. Consequently, Ais isomorphic to ∗
H,∗
M1,∗
M2
or ∗
M3.
4.2 B1isomorphic to Cand B2isomorphic
to
∗
C
We assume that B1=A(e, i)isomorphic to Cand
B2=A(e, j)isomorphic to ∗
C, we have
∥i+j∥2=∥e∥2∥i+j∥2=∥ei +ej∥2=∥i−j∥2
That is
2 + 2(i/j) = 2 −2(i/j)
Hence (i/j) = 0.
Proposition 4.3 . If B1is isomorphic to Cand B2is
isomorphic to
∗
C. Then Ais isomorphic to M1,M4,
∗
M1, or
∗
M4.
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Proof.. We can form an orthonormal basis {e, i, j, k}
of A. Since k∈ {e, i, j}⊥, then k2∈A(e, i)∩
A(e, j) = {e}(lemma 2.6.(1)). We get k2=±e.
But since
(ek/e) = (ek/e2) = (e/k) = 0
(ke/e) = (ke/e2) = (k/e) = 0
(ek/i) = (ek/ei) = (k/i) = 0
(ke/i) = (ke/ie) = (k/i) = 0
(ek/j) = (ek/−ej) = (k/j) = 0
(ke/j) = (ke/−je) = (k/j) = 0
We obtain ek =εk and ke =ζk, where |ε|=|ζ|= 1.
We conclude that A(e, k)is two-dimensional subalge-
bra of A, that is A(e, k)is isomorphic to C, *C,C∗or
∗
C. We have the following cases:
1. If A(e, k)is isomorphic to C, or ∗
C.
Then Ahas two different subalgebras isomor-
phic to Cor isomorphic to ∗
C. Hence Ais iso-
morphic to M1,∗
M1, (Proposition 4.1, and Propo-
sition 4.2.).
2. If A(e, k)is isomorphic to *C.
We have ek =k,ke =−kand k2=e.
According to remark 2.7, we have (ik/ki) =
−(i2/k2) = (e/e) = 1 which means that
∥ik −ki∥2= 0
So ik =ki, and similarly, we get
jk =kj and ij =−ji
We can also show that
ij =k or ij =−k
ik =j or ik =−j
and
jk =i or jk =−i
If ij =−k, then ik =−jand jk =−i. Indeed,
if ik =j, then
i(j+k) = −k+j=−ek −ej =−e(k+j)
So i=−e(Absurde).
Also if jk =i, then
(i+j)k=−j+i=je +ie = (j+i)e
which implies k=e, absurd. So the multiplica-
tion of Ais given by Table.5, and A is isomor-
phic to M4.
3. If A(e, k)is isomorphic to C∗,
On Awe can define a new algebra ∗
Aby the mul-
tiplication x∗y= ¯x¯y. Then ∗
Acontains three
different subalgebras isomorphic to ∗
C,Cand *C
respectively. Hence the last result imply that ∗
A
is isomorphic to M4. So Ais isomorphic to ∗
M4.
4.3 B1and B2are isomorphic to *Cor C∗
We have the following results
Proposition 4.4 . If B1and B2are isomorphic to *C.
Then Ais isomorphic to *H, *M1, *M2or *M3.
Proof.. We define a new multiplication on Aby
x∗y= ¯xy, we obtain an algebra *Awhich contains
two different subalgebras isomorphic to C. There-
fore, applying proposition 4.1, *Ais isomorphic to
H,M1,M2or M3. Consequently, Ais isomorphic to
*H, *M1, *M2or *M3.
Proposition 4.5 . If B1and B2are isomorphic to C∗.
Then Ais isomorphic to H∗,M1∗,M2∗or M3∗.
Proof.. We change the product of A by x∗y=x¯y, we
get the algebra noted A∗which contains two different
subalgebras isomorphic to C. So by Proposition 4.1.
A∗is isomorphic to H,M1,M2or M3. Which mean
that Ais isomorphic to H∗,M1∗,M2∗, or M3∗.
4.4 B1isomorphic to Cand B2isomorphic
to *C
We can pose B1=A(e, i), and B2=A(e, j)isomor-
phic to *C, we have (i/j) = 0.
Proposition 4.6 . If B1isomorphic to Cand B2iso-
morphic to *C. Then Ais isomorphic to M2,M4,
*M2, or *M4.
Proof.. We can construct an orthonormal basis
{e, i, j, k}of A. and we have by the same argument in
the precedent case, A(e, k)is two-dimensional subal-
gebra of A.
1. If A(e, k)is isomorphic to C, or *C.
Then Ahas two different subalgebras isomor-
phic to Cor isomorphic to *C. Hence Ais iso-
morphic to M2, or *M2(Propositions 4.1, and
Proposition 4.4).
2. If A(e, k)is isomorphic to ∗
C.
By Proposition 4.3. Ais isomorphic to M4
3. If A(e, k)is isomorphic to C∗.
We considere the product x∗y= ¯xy, we ob-
tain an algebra *Awhich contains three different
subalgebras isomorphic to ∗
C,Cand *Crespec-
tively. Therefore, applying the last result, *Ais
isomorphic to M4. Consequently, Ais isomor-
phic to *M4.
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4.5 B1isomorphic to Cand B2isomorphic
to C∗
We pose B1=A(e, i), and B2=A(e, j). we always
have (i/j) = 0. the identity
∥i+j∥2=∥e∥2∥i+j∥2=∥ei +ej∥2=∥i−j∥2
Give
2 + 2(i/j) = 2 −2(i/j)
So (i/j) = 0.
Proposition 4.7 . If B1isomorphic to Cand B2iso-
morphic to C∗. Then Ais isomorphic to M3,M3∗,
*M4or
∗
M4.
Proof.. We have (i/j) = 0 so {e, i, j}is an or-
thonormal familly which can be extend to an or-
thonormal basis {e, i, j, k}of A. Since A(e, k)is two-
dimensional subalgebra of A. We have the following
cases:
1. If A(e, k)is isomorphic to Cor C∗.
Then Ahas two different subalgebras isomor-
phic to Cor isomorphic to C∗. Hence Ais
isomorphic to M3,or M3∗(Proposition 4.1, and
Proposition 4.5).
2. If A(e, k)is isomorphic to ∗
C.
By Proposition 4.3. Ais isomorphic to ∗
M4.
3. If A(e, k)is isomorphic to *C, then Ais isomor-
phic to *M4(Proposition 4.6).
Remark 4.8 .
1. If Ahas two subalgebras B1=A(e, i), isomor-
phic to
∗
C, and B2=A(e, j)isomorphic to *C.
We can define a new algebra
∗
A, with product
x∗y= ¯x¯y. So
∗
Acontains two different subalge-
bras isomorphic to Cand C∗respectively. So
∗
A
is isomorphic to M3,M3∗, *M4, or
∗
M4(Propo-
sition 4.7). Consequently Ais isomorphic to M4,
,
∗
M3, *M3, or M4∗.
2. if Ahas two subalgebras B1=A(e, i)iso-
morphic to
∗
C, and B2=A(e, j)isomorphic
to C∗. We define a new multiplication on Aby
x∗y= ¯x¯y, and we obtain an algebra
∗
A, which
contains two different subalgebras isomorphic
to C, and *Crespectively. By Proposition 4.6,
∗
Ais isomorphic M2,M4, *M2, or *M4. Hence
Ais isomorphic to
∗
M2,
∗
M4,M2∗or M4∗.
4.6 B1isomorphic to *Cand B2isomorphic
to C∗
Let’s B1=A(e, i), and B2=A(e, j). We have
∥i+j∥2=∥i+j∥2∥e∥2=∥ie +je∥2=∥−i+j∥2
So
2 + 2(i/j) = 2 −2(i/j)
Hence (i/j) = 0.
Proposition 4.9 If B1isomorphic to *Cand B2iso-
morphic to C∗. Then Ais isomorphic to *M1, , *M4,
M1∗,or M4∗
Proof. We construct an orthonormal basis {e, i, j, k}
of A.:
1. If A(e, k)is isomorphic to C∗, or *C.
Then Ahas two different subalgebras isomor-
phic to *Cor isomorphic to C∗. Hence Ais iso-
morphic to *M1,or M1∗(Propositions. 4.4, and
Proposition. 4.5).
2. If A(e, k)is isomorphic to C, the result is a con-
sequence of the Proposition. 4.6, thus Ais iso-
morphic to *M4.
3. If A(e, k)is isomorphic to ∗
C, the result is a con-
sequence of the remark. 4.8. Hence Ais isomor-
phic to M4∗.
Remark 4.10 If ij =−k, we subistitute −k=twe
obtain ij =t, so we use the basis {e, i, j, t}, we again
get the same classifications.
5 Conclusion
In this section, we have the following main result.
Theorem 5.1 Let Abe a four dimensional absolute
valued algebra with a nonzero omnipresent idempo-
tent e, and having two different subalgebras B1and
B2of dimension two. The following table specifies
the isomorphisms classes.
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T able.6. All classifications
B1B2A
C C H,M1,M2,M3
∗
C
∗
C
∗
H,
∗
M1,
∗
M2,
∗
M3
C
∗
C M1,M4,
∗
M1,
∗
M4
*C*C*H, *M1, *M2, *M3
C∗C∗H∗,M1∗,M2∗,M3∗
C*C M2,M4, *M2, *M4
C C∗M3,M3∗, *M4,
∗
M4
∗
C*C M4,
∗
M3, *M3,M4∗
∗
C C∗∗
M2,
∗
M4,M2∗,M4∗
*C C∗*M1, , *M4,M1∗,or M4∗
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