Chapter – 3
SOME RESULTS ON
NON-ASSOCIATIVE
COMMUTATIVE RINGS
WITH CYCLIC IDENTITY
3.1 INTRODUCTION
The non-associative rings, namely, (1, 1) rings, derivation alternator rings and
generalized right alternative rings have been studied by Kleinfeld, Hentzel, Hogben,
Smith and C.Jaya Subba Reddy. They have proved the associativity and alternativity
of these rings by assuming some additional conditions and identities. In general, these
rings are not right alternative. In this chapter , without using the methods of Kleinfeld,
Hentzel and C.Jaya Subba Reddy we proved that not only above three rings but all
non-associative ring is right alternative rings by using
our condition:
Non-
associative ring satisfies commutative law and identity
x (yz) = y (zx) = z (xy) for all x, y, z in a ring R
... (3.1.1)
is known as cyclic identity same as in chapter II.
In section 3.2, we used the cyclic identity in a non-associative commutative ring.
Using this identity we proved the following theorems:
(i)
Let R be a non-associative commutative ring with cyclic identity. Then R is
a left alternative ring.
(ii)
Let R be a non-associative commutative ring with cyclic identity. Then R is
a right
(iii)
alternative ring.
Let R be a non-associative commutative ring with cyclic identity. Then R is
a flexible ring.
(iv)
Let R be an algebraic structure such that (R, +) is abelian group and (R, •) is
a commutative algebraic structure with cyclic identity. Then R is a
commutative ring.
(v)
Let R be a non-associative commutative ring with cyclic identity. Then
(y, x, n) (z, r, s) = 0.
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(vi)
Let R be a non-associative commutative ring with cyclic identity. Then
R = N.
(vii)
In a non-associative commutative ring with cyclic identity, then
(x, y, z)n = 0 for all x, y, z in R and for all natural numbers n.
(viii) In a non-associative commutative ring with cyclic identity, then
(xi, yj, zk) = 0 for all x, y, z in R and for all natural numbers i, j, k.
(ix)
Every non-associative commutative ring with cyclic identity is a derivation
alternator ring.
(x)
Every non-associative commutative ring with cyclic identity is a
(xi)
Every non-associative commutative ring with cyclic identity is a (1,1) ring.
(xii)
Every non-associative commutative ring with cyclic identity is a (-1, 1) ring.
,
ring.
(xiii) Every non-associative commutative ring with cyclic identity is a strongly
(-1, 1) ring.
(xiv) Every non-associative commutative ring with cyclic identity is a generalized
right alternative ring.
(xv)
Every non-associative commutative ring is a generalized alternative ring.
(xvi) Every non-associative commutative ring with cyclic identity is a Novikov
ring.
(xvii) Every non-associative commutative ring with cyclic identity is a strongly
Novikov ring.
In section 3.3, we have non-associative (or associative) ring with a cyclic
identity in a commutative. Using this identity we proved the following theorems:
(i)
A non-associative commutative ring with cyclic identity is an alternative
ring.
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(ii)
A lie ring is a non-associative commutative ring with cyclic identity, then
(xy) z = 0, ∀x, y, z∈R.
(iii) A non-associative commutative ring with cyclic identity, then R is a jordan
ring.
3.2. ON NON-ASSOCIATIVE COMMUTATIVE RING WITH CYCLIC
IDENTITY
Using this identity we proved the following theorems
Theorem 3.2.1: Let R be a non-associative commutative ring with cyclic identity.
Then R is a left alternative ring.
Let R be a non-associative commutative ring and
Proof:
let ∀ x, y ∈ R
Consider
(x, x, y)
= x (xy) – (xx) y.
… (3.2.1)
from equation (3.2.1),(x, x, y) = x( xy) – (xx ) y
Since R satisfies the cyclic identity (3.1.1), we have
= y (xx) – (xx) y
= (xx) y – (xx) y
= 0.
Thus (x, x, y) = 0 ∀ x, y ∈ R.
Hence R is a left alternative ring.
Theorem 3.2.2: Let R be a non-associative commutative ring with cyclic identity.
Then R is a right alternative ring.
Proof: Let R be a non-associative commutative ring and
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let x, y ∈ R.
Since R satisfies the cyclic identity (3.1.1), we have
x (yz) = y (zx) = z (xy) ∀ x, y, z ∈ R.
Now consider
(x,y,y) = x (yy) – (xy ) y
= y (xy) – (xy) y
= (xy) y – (xy) y
= 0.
Thus (x, y, y) = 0 ∀ x, y ∈ R.
Hence R is a right alternative ring.
Corollary 3.2.1: Let R be a non-associative commutative ring with cyclic identity.
Then R is alternative ring.
Proof:
Already, we proved R is a left alternative ring and we proved R is a right
alternative ring
Therefore R is a alternative ring.
Theorem 3.2.3: Let R be a non-associative commutative ring with cyclic identity.
Then R is a flexible ring.
Proof: Let R be a non-associative commutative ring with cyclic identity and
let x, y ∈ R.
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Since R satisfies the cyclic identity (3.1.1), we have
x (yz) = y (zx) = z (xy) ∀ x, y, z ∈ R.
Now consider,
(x,y,x) = x (yx) – (xy) x
Since R satisfies the cyclic identity (3.1.1), we have
= x (xy) – (xy) x
= (xy) x – (xy) x
= 0.
Thus, (x, y, x) = 0 ∀ x, y ∈ R.
Hence R is a flexible ring.
Theorem 3.2.4: Let R be an algebraic structure such that (R, +) is abelian group and
(R, •) is a commutative algebraic structure with cyclic identity. Then R is a
commutative ring.
Proof:
Let x, y, z ∈ R
Now consider
(x, y, z) = x (yz) – (xy) z.
… (3.2.2)
Since R satisfies the cyclic identity (3.1.1), we have
= z (xy) – (xy) z
= (xy) z – (xy) z
= 0.
Thus (x, y, z) = 0 ∀ x, y, z ∈ R.
… (3.2.3)
i.e., x (yz) = (xy) z ∀ x, y, z ∈ R.
Hence R is a commutative ring.
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Theorem 3.2.5: If R be a non-associative commutative ring with cyclic identity, then
(y, x, n) (z, r, s) = 0.
Proof: From theorem (3.2.4), if R be a non-associative commutative ring with cyclic
identity
then (x, y, z) = 0 ∀ x, y, z ∈ R.
So in particular ((y, x, n) z, r, s) = 0.
… (3.2.4)
(y, x, n) (z, r, s) = 0.
Theorem 3.2.6: Let R be a non-associative commutative ring cyclic identity. Then
R = N.
Proof: we have to show that R = N
Trivially N ⊆ R.
… (3.2.5)
Next we show that R ⊆ N
Let n ∈ R
(y, n, z) = y (nz) - (yn) z ∀ y, z ∈ R.
= y (nz) – z (yn)
= y (nz) – y (nz)
= 0.
Hence (y, n, z) = 0 ∀ y, z ∈ R.
⇒ (R, n, R) = 0
⇒ n∈ N
⇒ R ⊆ N.
… (3.2.6)
From equation (3.2.5) and (3.2.6), we get R = N.
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Theorem 3.2.7: In a non-associative commutative ring with cyclic identity
(x, y, z)n = 0 for all x, y, z in R and for all natural numbers n.
Proof: consider (x, y, z)n = (x, y, z) . (x, y, z)…(x, y, z) (n-times).
… (3.2.7)
Now consider
(x, y, z) = x (yz) – (xy) z
Since R satisfies the cyclic identity (3.1.1), we have
= z (xy) – (xy) z
= (xy) z – (xy) z
= 0.
Thus (x, y, z) = 0 ∀ x, y, z ∈ R.
…(3.2.8)
Substituting equation (3.2.8) in (3.2.7), we get
(x, y, z)n = 0 for all x, y, z in R and for all natural numbers n.
Theorem 3.2.8: In a non-associative commutative ring with cyclic identity, then
(xi, yj, zk) = 0 for all x, y, z in R and for all natural numbers i, j, k.
Proof: From theorem (3.2.4), we have
(x, y, z) = 0 for all x, y, z in R
Since x, y, z are in a non-associative commutative ring R,
so xi, yj, zk are in R.
... (3.2.9)
From (3.2.3) and (3.2.9), we have
(xi, yj, zk) = 0 for all x, y, z in R and for all natural numbers i, j, k.
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Theorem 3.2.9: Every non-associative commutative ring with cyclic identity is a
derivation alternator ring.
Proof: Let x, y, z in R.
Trivially (x, x, x) = 0 for all x in R.
... (3.2.10)
From theorem (3.2.4), we have (x, y, z) = 0 for all x, y, z in R.
From equation (3.2.3), (yz, x, x) = 0, (z, x, x) = 0, (y, x, x) = 0
Thus, (yz, x, x) = y (z, x, x) + (y, x, x) z.
... (3.2.11)
Once again from equation (3.2.3), (x, x, yz) = 0, (x, x, z) = 0, (x, x, y) = 0
Thus, (x, x, yz) = y (x, x, z) + (x, x, y) z.
… (3.2.12)
From equations (3.2.3), (3.2.11) and (3.2.12), we have
R becomes a derivation alternator ring.
Hence every non-associative commutative ring with cyclic identity is a derivation
alternator ring.
Theorem 3.2.10: Every non-associative commutative ring with cyclic identity is a
( γ , δ ) ring.
Proof: From theorem (3.2.4), we have (x, y, z) = 0 for all x, y, z in R.
From equation (3.2.3), (z, x, y) = 0, (x, z, y) = 0, (y, z, x) = 0
Thus, (z, x, y) + γ (x, z, y) + δ (y, z, x) = 0.
… (3.2.13)
Once again from equation (3.2.3), (x, y, x) = 0, (y, z, x) = 0, (z, x, y) = 0
Thus, (x, y, z) + (y, z, x) + (z, x, y) = 0.
... (3.2.14)
From equations (3.2.13) and (3.2.14), we have R is a ( γ , δ ) ring.
Hence every non-associative commutative ring with cyclic identity is a ( γ , δ ) ring.
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Theorem 3.2.11: Every non-associative commutative ring with cyclic identity is a
(1, 1) ring.
Proof: From theorem (3.2.4), we have (x, y, z) = 0 for all x, y, z in R.
From equation (3.2.3), we have (x, z, y) = 0.
... (3.2.15)
Thus from equations (3.2.3) and (3.2.15), we have
(x, y, z) = (x, z, y).
… (3.2.16)
Adding equations (3.2.3), (3.2.15), we have
(x, y, z) + (x, z, y) = 0.
… (3.2.17)
From equations (3.2.16) and (3.2.17), we have every non-associative
commutative ring with cyclic identity is a (1, 1) Ring
Theorem 3.2.12: Every non-associative commutative ring with cyclic identity is a
(-1, 1) ring.
Proof: From theorem (3.2.4), we have (x, y, z) = 0 for all x, y, z in R.
From equation (3.2.3), we have
(y, z, x) = 0, (z, x, y) = 0.
… (3.2.18)
Adding equations (3.2.3), (3.2.18), we get
(x, y, z) + (y, z, x) + (z, x, y) = 0.
… (3.2.19)
Again from equation (3.2.3), we have
(x, z, y) = 0.
... (3.2.20)
Adding equations (3.2.3) and (3.2.20), we get
(x, y, z) + (x, z, y) = 0.
… (3.2.21)
From equations (3.2.19) and (3.2.21), we have every non-associative
commutative ring with cyclic identity is a (-1, 1) ring
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Theorem 3.2.13: Every non-associative commutative ring with cyclic identity is
strongly (-1, 1) ring.
Proof: From theorem (3.2.4), we have (x, y, z) = 0 for all x, y, z in R.
From equation (3.2.3), we have (x, y, y) = 0.
… (3.2.22)
We know that (x, y) = xy – yx.
… (3.2.23)
and
((x, y), z) = (xy – yx ) z = (xy – xy ) z = 0.
… (3.2.24)
Thus from (3.2.22) and (3.2.24), we have every non-associative commutative ring is
strongly (-1, 1) ring.
Theorem 3.2.16: Every non-associative commutative ring with cyclic identity is a
generalized right alternative ring.
Proof: From theorem (3.2.4), we have (x, y, z) = 0 for all x, y, z in R.
From equation (3.2.3), we have (wx, y, z) = 0.
and
(w, x,(y, z)) = 0.
… (3.2.25)
…(3.2.26)
Adding equations (3.2.25) and (3.2.26), we get
(wx, y, z) + (w, x, (y, z)) = 0.
… (3.2.27)
From equation (3.2.3), we have
(w, y, z) x = 0.
and w (x, y, z) = 0.
… (3.2.28)
…(3.2.29)
Adding equations (3.2.28) and (3.2.29), we get
(w, y, z) x + w (x, y, z) = 0.
… (3.2.30)
From equations (3.2.27) and (3.2.30), we get
(wx, y, z) + (w, x, (y, z)) = (w, y, z) x + w (x, y, z).
Thus every non-associative commutative ring is a generalized right alternative ring.
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Theorem 3.2.15: Every non-associative commutative ring with cyclic identity is a
generalized alternative ring.
Proof: From theorem (3.2.4), we have (x, y, z) = 0 for all x, y, z in R.
From equation (3.2.3), we have (w, x, yz) = 0.
… (3.2.31)
and
…(3.2.32)
((w, x), y, z)) = 0.
Adding equations (3.2.31) and (3.2.32), we get
(w, x, yz) + ((w, x), y, z) = 0.
… (3.2.33)
From equation (3.2.3), we have
y (w, x, z) = 0.
…(3.2.34)
and (w, x, y) z = 0.
…(3.2.35)
Adding equations (3.2.34) and (3.2.35), we get
y (w, x, z) + (w, x, y) z = 0.
… (3.2.36)
From equations (3.2.33) and (3.2.36), we get
(w, x, yz) + ((w, x), y, z) = y (w, x, z) + (w, x, y) z.
Thus, every non-associative commutative ring is a generalized alternative ring.
Theorem 3.2.16: Every non-associative commutative ring with cyclic identity is a
Novikov ring.
Proof: From theorem (3.2.4), we have (x, y, z) = 0 for all x, y, z in R.
But (x, y, z) = x (yz) – (xy) z
= x (yz) – z (xy)
Since R satisfies the cyclic identity (3.1.1), we have
= x (yz) – x (yz)
= x (yz) – y (xz).
… (3.2.37)
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From equations (3.2.3) and (3.2.37), we have
x (yz) = y (xz).
…(3.2.38)
Again (x, y, z) = x (yz) – (xy) z
= x (zy) – (xz) y
= (x, z, y).
… (3.2.39)
From (3.2.38) and (3.2.39), we have every non-associative commutative ring
with cyclic identity is a Novikov ring.
Theorem 3.2.17: Every non-associative commutative ring with cyclic identity is a
strongly Novikov ring.
Proof: From theorem (3.2.4), we have (x, y, z) = 0 for all x, y, z in R.
But (x, y, z) = x (yz) – (xy) z
= x (yz) – z (xy)
Since R satisfies the cyclic identity (3.1.1), we have
= x (yz) – y (zx)
= x (yz) – y (xz).
… (3.2.40)
From (3.2.3) and (3.2.40), we have
x (yz) = y (xz).
…(3.2.41)
Theorem 3.2.18: A non-associative commutative ring with cyclic identity Satisfies
the Flexible law.
Proof: It is sufficient to show that, the associator, (x, y, x) = 0,
Now consider, (x, y, x) = (xy) x – x (yx)
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Since R satisfies the cyclic identity (3.1.1), we have
= (xy) x – y (xx)
= (xy) x – x (xx)
= (xy) x – (xy)x
= 0.
Hence associator (x, y, x) = 0 ∀ x, y ∈ R.
The given non-associative commutative ring with cyclic identity satisfies the Flexible
law.
Theorem 3.2.19: Every non-associative commutative ring with cyclic identity is
satisfies associative law.
Proof: Let x, y, z in R
Now (x, y, z) = x (yz) – (xy) z.
… (3.2.42)
= x (yz) – z (xy)
Since R satisfies the cyclic identity (3.1.1), we have
= x (yz) – x (yz)
= 0.
… (3.2.43)
From equation (3.2.42) and (3.2.43), we get x (yz) = (xy) z.
Hence non-associative commutative ring with cyclic identity satisfies the
associative law.
Theorem 3.2.20: Every non-associative commutative ring with cyclic identity is a
weakly Novikov ring.
Proof: From theorem (3.2.4), we have (x, y, z) = 0 for all x, y, z in.
Thus from equation (3.2.3), we have
(w, x, yz) = 0.
… (3.2.44)
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and (w, x, z ) = 0.
…(3.2.45)
So, from equations (3.2.44) and (3.2.45), we have
(w, x, yz) = y (w, x, z)
Thus every non-associative commutative ring with cyclic identity is a weakly
Novikov ring.
3.3. ON NON-ASSOCIATIVE (OR ASSOCIATIVE) COMMUTATIVE RING
WITH CYCLIC IDENTITY
Using this identity we proved the following theorems:
Theorem 3.3.1: A non-associative commutative ring with cyclic identity is an
alternative ring.
Proof: We have to show that a non-associative commutative ring with cyclic identity
is an alternative ring. It is sufficient to show that,
(xx) y = x (xy) & y (xx) = (yx) x.
Consider,
(xx) y = (xy) x
Since R satisfies the cyclic identity (3.1.1), we have
= x (xy)
(xx) y = x (xy).
…(3.3.1)
Similarly, Now Consider,
y (xx) = x (xy)
Since R satisfies the cyclic identity (3.1.1), we have
= (xy) x
= (yx) x
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y (xx) = (yx) x.
… (3.3.2)
From equations (3.3.1) & (3.3.2), we get
(xx) y = x (xy) & y (xx) = (yx) x.
A non-associative commutative ring with cyclic identity is an alternative
ring.
Theorem 3.3.2 : A lie ring is a non-associative commutative ring with cyclic
identity, then (xy) z = 0 ∀x, y, z∈R.
Proof: Given that R is a Lie ring, we have
(i) xy = - (yx) ∀x, y∈R,
& (ii) (xy) z + (yz) x + (zx) y = 0, ∀x, y, z∈R.
… (3.3.3)
Also, Lie ring satisfies the cyclic identity.
i.e., (xy) z = (yz) x = (zx) y.
… (3.3.4)
Substituting equation (3.3.4) in equation (3.3.3), then,
(xy) z + (xy) z + (xy) z = 0
⇒ 3(xy) z = 0
⇒
(xy) z = 0.
Hence A lie ring is a non-associative commutative ring with cyclic identity is
(xy) z = 0 ∀x, y, z∈R.
Theorem 3.3.3: A non-associative commutative ring with cyclic identity, then R is a
Jordan ring.
Proof: Given that R is a non-associative commutative ring with cyclic identity.
We have show that R satisfies the following laws.
(i)
Commutative law : xy = yx, ∀ x, y∈R.
(ii)
Jordan Identity: (xy)2 = x (yx2), ∀ x, y∈R.
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Since R is commutative ring with cyclic identity, so, it is sufficient to show
that R Satisfies the Jordan Identity.
Consider, (xy) x2 = (yx2) x
Since R satisfies the cyclic identity (3.1.1), we have
= x (yx2)
(xy) x2 = x (yx2).
Thus R Satisfies Jordan Identity
R Satisfies (i) Commutative law &
(ii) Jordan Identity.
R is a Jordan Ring
Thus, every non-associative commutative ring with cyclic identity is a Jordan Ring.
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