GEORGIAN MATHEMATICAL JOURNAL: Vol. 3, No. 1, 1996, 1-10

COMMUTATIVITY FOR A CERTAIN CLASS OF RINGS

HAMZA A. S. ABUJABAL

Abstract. We discuss the commutativity of certain rings with unity 1 and one-sided s-unital rings under each of the following conditions: x^{r}[x^{s}, y] = ±[x, y^{t}]x^{n}, x^{r}[x^{s}, y] = ±x^{n}[x, y^{t}], x^{r}[x^{s}, y] = ±[x, y^{t}]y^{m}, and x^{r}[x^{s}, y] = ±y^{m}[x, y^{t}], where r, n, and m are non-negative integers and t > 1, s are positive integers such that either s, t are relatively prime or s[x, y] = 0 implies [x, y] = 0. Further, we improve the result of [6, Theorem 3] and reprove several recent results.

Throughout the paper R will represent an associative ring (with or without unity 1). Let C(R) denote the commutator ideal of R, Z(R) the center of R, and H the heart of R. By (GF (q))_{2} we mean the ring of 2 × 2 ma-

trices over the Galois field GF (q) with q elements. Set e_{11} = (\begin{matrix}1 & 0 \\ 0 & 0 \end{matrix}),

e_{12} = (\begin{matrix}0 & 1 \\ 0 & 0 \end{matrix}), e_{21} = (\begin{matrix}0 & 0 \\ 1 & 0 \end{matrix}), and e_{22} = (\begin{matrix}0 & 0 \\ 0 & 1 \end{matrix}) in (GF (p))_{2} for a prime

p. Following [1], a ring R is said to be left (resp., right) s-unital, if x ∈ Rx (resp., x ∈ xR) for each element x in R. Further, R is called s-unital if x ∈ Rx ∩ xR (see [2] and [3]). The symbol [x, y] stands for the commutator xy − yx for any x, y ∈ R. In some particular cases several authors [1–3, 5] studied the commutativity of rings satisfying the following conditions:

(c_{1}) For every x, y ∈ R there holds x^{r}[x^{s}, y] = ±[x, y^{t}]x^{n} with integers

t > 1, s ≥ 1, n ≥ 0, r ≥ 0. (c_{2}) For every x, y ∈ R there holds x^{r}[x^{s}, y] = ±[x, y^{t}]y^{m} with integers

t > 1, s ≥ 1, m ≥ 0, r ≥ 0. (c_{3}) For every x, y ∈ R there holds x^{r}[x^{s}, y] = ±x^{n}[x, y^{t}] with integers

t > 1, s ≥ 1, n ≥ 0, r ≥ 0. (c_{4}) For every x, y ∈ R there holds x^{r}[x^{s}, y] = ±y^{m}[x, y^{t}] with integers

t > 1, s ≥ 1, m ≥ 0, r ≥ 0. To develop the commutativity of a ring R satisfying one of the above

conditions, we need some extra condition such as 1 Q(s): For any positive integer s s[x, y] = 0 implies [x, y] = 0 for all

1991 Mathematics Subject Classification. 16U80.

Key words and phrases. Commutativity of rings, ring with unity, s-unital rings.

1072-947X/96/0100-0001$09.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1996 Plenum Publishing Corporation

x, y ∈ R,

or

Q(s,t): s and t are relatively prime integers.

To prove our results we need a few preliminary lemmas. We begin with the following well-known result [6, p. 221].

Lemma 1. Let x, y ∈ R and [[x, y], x] = 0. Then [x^{k}, y] = kx^{k−1}[x, y] for any positive integer k.

Lemma 2 ([7]). Let R be a ring with unity 1, and let f : R → R be a function such that f(x + 1) = f(x) for every x ∈ R. If for some positive integer n we have x^{n}f(x) = f(x)x^{n} = 0 for all x ∈ R, then necessarily f(x) = 0.

Lemma 3 ([8]). Let f be a polynomial in the non-commuting indeterminates x_{1}, x_{2}, …, x_{n} with integer coefficients. Then the following statements are equivalent:

(i) C(R) is a nil ideal for any ring R satisfying f = 0.

(ii) (GF (p))_{2} fails to satisfy f = 0 for every prime p.

The main results of the paper are:

Theorem 1. Let R be a ring with 1 satisfying (c_{1}) together with either Q(s, t) or Q(s). Then R is commutative.

Theorem 2. Let R be a ring with 1 satisfying (c_{2}) together with either Q(s, t) or Q(s). Then R is commutative.

Theorem 3. Let R be a ring with 1 satisfying (c_{3}) together with either Q(s, t) or Q(s). Then R is commutative.

Theorem 4. Let R be a ring with 1 satisfying (c_{4}) such that either Q(s, t) or Q(s) holds. Then R is commutative.

Remark 1. The well-known Grassmann algebra rules out the possibility of t = 1 in the above theorems. Moreover, if we drop the restriction that R has unity 1 in the above theorems, then the ring R may be poorly noncommutative. Indeed, the following example demonstrates this constraint: Let D_{k} be the ring of all k × k matrices over a division ring D, and let A_{k} = {(a_{ij} ∈ D_{k}|a_{ij} = 0, j ≥ i}. Then A_{k} is a non-commutative ring for any positive integer k > 2. But A_{3} satisfies (c_{1}), (c_{2}), (c_{3}), and (c_{4}) for all positive integers s, t and non-negative integers m, n, and r.

According to [9], let f be a polynomial in two non-commuting indeterminates with integral coefficients. Now write f in the form

f(x, y) = ∑_{r=1}^{d} ∑_{i=0}^{r} f_{ri}(x, y),

where f_{ri} denotes the sum of all terms of f with degree i in x and r − i in y. Let s_{ri} denote the sum of the coefficients of f_{ti}. Then we note that if

s_{ri} = 0 for all r and i,  (I)

then all commutative rings satisfy f = 0. The converse is also true as proved by Kezlan [9]. For this we take a transcendental extension field of rationals and use the fact that the polynomial

f(X^{d}, X^{d+1}) = ∑_{r=1}^{d} ∑_{i=0}^{r} s_{ri}X^{(d+1)r−i}

in one indeterminate X vanishes on it.

Thus if f is to be equivalent with the commutativity it must at least satisfy (I), and so we may write

f(x, y) = m[x, y] + ∑_{r=3}^{d} ∑_{i=1}^{r−1} f_{ri}(x, y)

for some integer m. Moreover, if m is divisible by a prime p, then the ring of strictly upper triangular 3 × 3 matrices over any field of characteristic p satisfies the identity, and so we assume that

m = ±1.  (II)

Let us consider the condition

f_{r1} = 0 for all r.  (III)

In [9] Kezlan proved the following

Theorem. If f satisfies (I), (II), and (III), then an arbitrary ring R is commutative if and only if it satisfies the identity f = 0.

Also, it should be remarked that (I) in the theorem could be replaced by

f_{r,r−1} = 0 for all r.  (IV)

An example was given in [9] to show that we must assume either (III) or (IV) or some other condition concerning the terms linear in x or in y. So (I) and (II) alone are not enough.

Further, in [9] Kezlan proved that for a polynomial f(x, y) the identity f(x, y) = 0 is equivalent with the commutativity for all rings if f(x, y) = ±[x, y] + ∑_{r=3}^{d} ∑_{i=1}^{r−1} f_{ri}(x, y), where f_{ri} denotes the sum of all terms of f(x, y) with degree i in x and r − i in y, where s_{si} denotes the sum of coefficients of f_{ri}. It was also shown that under certain restrictions on the terms linear in one variable or the other, the polynomial identity f(x, y) = 0 is indeed equivalent with the commutativity.

The following fact plays an important role in the proof of our results.

Lemma 4 ([9]). Let R be a ring satisfying the polynomial identity f(x, y) = ±[x, y], where each homogeneous component of f(x, y) has integer coefficients whose sum is zero and where f(x, y) has no linear terms either in x or in y. Then R is commutative.

Now we prove

Proposition 1. Let R be a ring with p[x, y] = 0 for all x, y in R, p a prime, and let s be a positive integer not divisible by p. Suppose that R satisfies a polynomial identity of the form f(x, y) = ±m[x, y], where m is any non-negative integer and f(x, y) satisfies the same condition as in Lemma 4. Then R is commutative.

Proof. Let qp + mn = 1. By hypothesis, we have

f(x, y) = ±m[x, y] for all x, y ∈ R.

Multiply the above identity by n and use p[x, y] = 0 to get nf(x, y) = ±[x, y]. Hence by Lemma 4 R is commutative.

Lemma 5. Let R be a ring with 1 satisfying (c_{1}) or (c_{2}) or (c_{3}) or (c_{4}) such that either Q(s, t) or Q(s) holds. Then C(R) is nil.

Proof. Let x = e_{22} and y = e_{21} + e_{22} in (c_{1}) and (c_{2}). Then by Lemma 3, x and y fail to satisfy the polynomial identities (c_{3}) and (c_{4}) for any prime p. Similarly, x = e_{11} and y = e_{21} fail to satisfy (c_{3}) and (c_{4}). Thus C(R) is a nil ideal.

Proof of Theorem 1. According to Lemma 5, C(R) is nil. Now let R satisfy (c_{1}). Then by contradiction we assume that there exists a noncommutative ring with 1 satisfying (c_{1}). Another step is to pass to the subdirectly irreducible case, and with Q(s, t) this reduction can be obtained as in [9]. Therefore, without loss of generality, we assume that there is a ring R such that

(α) R is a non-commutative ring with 1, satisfies (c_{1}), and R is subdirectly irreducible with heart H = C(R) with H^{2} = (0).

Using the condition Q(s), we must slightly modify the arguments in [8] because this condition is not preserved under homomorphism. Then sssuming that we have a non-commutative ring A with unity satisfying (c_{1}) and that Q(s) holds, we have a, b ∈ A with s[a, b] ≠ 0. By Zorn’s lemma we get an ideal M which is maximal with respect to the exclusion of s[a, b]. Then the ring \overline{A} = A/M is not commutative, satisfies (c_{1}), and is subdirectly irreducible with the heart containing \overline{s}[\overline{a}, \overline{b}]. Hence the ring R = \overline{A} may not inherit Q(s). This shows that s does not annihilate all commutators of R. To summarize these paragraphs, if Q(s, t) holds, then we have a ring R satisfying (α). If Q(s) holds, then, in addition to (α), we have

(β) s does not annihilate all commutators of R.

Now we define a mapping F : R → R for fixed y, w ∈ R by

F (x) = ±[x, (y + w)^{t} − y^{t} − w^{t}] for all x ∈ R.  (1)

Replace x by x + 1 in (1) to get

F (x + 1) = ±[x, (y + w)^{t} − y^{t} − w^{t}] = F (x).

Multiplying (1) by x^{n} on the right, we get

F (x)x^{n} = ±[x, (y + w)^{t} − y^{t} − w^{t}]x^{n} =
= ±[x, (y + w)^{t}]x^{n} ∓ [x, y^{t}]x^{n} ∓ [x, w^{t}]x^{n} =
= x^{r}[x^{s}, y + w] − {±[x, y^{t}]x^{n}} − {±[x, w^{t}]x^{n}} =
= x^{r}[x^{s}, y] + x^{r}[x^{s}, w] − x^{r}[x^{s}, y] − x^{r}[x^{s}, w] = 0.

Using Lemma 2, we obtain F (x) = 0 for all x, y in R. Hence

± [x, (y + w)^{t} − y^{t} − w^{t}]x^{n} = 0,
± [x, (y + w)^{t}] ∓ [x, y^{t}] ∓ [x, w^{t}] = 0

and

±[x, (y + w)^{t}] = ±[x, y^{t}] ± [x, w^{t}] = 0.  (2)

Substituting w = 1 in (2), we get

±(t[x, y] + ∑_{k=2}^{t−1} \binom{t}{k}[x, y^{k}]) = 0.  (3)

By (3) we can write

t[x, y] + ∑_{k=2}^{t−1} \binom{t}{k}[x, y^{k}] = 0.  (3′)

Replacing y in turn by z_{1}y, z_{2}y, …, z_{t−1}y in (3′), we get

z_{1}t[x, y] + \binom{t}{2}z_{1}^{2}[x, y^{2}] + \binom{t}{3}z_{1}^{3}[x, y^{3}] + ⋯ + \binom{t}{t−1}z_{1}^{t−1}[x, y^{t−1}] = 0, z_{2}t[x, y] + \binom{t}{2}z_{2}^{2}[x, y^{2}] + \binom{t}{3}z_{2}^{3}[x, y^{3}] + ⋯ + \binom{t}{t−1}z_{2}^{t−1}[x, y^{t−1}] = 0,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

z_{t−1}t[x, y] + \binom{t}{2}z_{t−1}^{2}[x, y^{2}] + \binom{t}{3}z_{t−1}^{3}[x, y^{3}] + ⋯ + \binom{t}{t−1}z_{t−1}^{t−1}[x, y^{t−1}] = 0.

The above identities can be written in the matrix form:

A_{(t−1)×(t−1)}W_{(t−1)×1} =

= (\begin{matrix}z_{1} & z_{1}^{2} & z_{1}^{3} & ⋯ & z_{1}^{t−1} \\ z_{2} & z_{2}^{2} & z_{2}^{3} & ⋯ & z_{2}^{t−1} \\ z_{3} & z_{3}^{2} & z_{3}^{3} & ⋯ & z_{3}^{t−1} \\ ⋮ & ⋮ & ⋮ & ... & ⋮ \\ z_{t−1} & z_{t−1}^{2} & z_{t−1}^{3} & ⋯ & z_{t−1}^{t−1} \end{matrix}) ( t[x, y] ; \binom{t}{2}[x, y^{2}] ; \binom{t}{3}[x, y^{3}] ; ⋮ ; \binom{t}{t−1}[x, y^{t−1}]) = 0.

Multiplying the above by adj(A), we have det(A). In particular, det(A)t[x, y] = 0 for all x, y ∈ R.

Let z_{i}^{j} be the (i, j)th entry in the matrix A. Since factoring z_{i} out of the ith row of A gives a Vandermonde matrix, ⟨R, +⟩ is torsion free. Therefore [x, y] = 0 for all x, y ∈ R, and R is commutative. (In short, if ⟨R, +⟩ is torsion free, then the standard Vandermonde determinant argument shows that the homogeneous components must vanish on R (see [9]). So [x, y] = 0, since R is torsion free. Thus R is commutative.)

As in [9], the subdirect irreducibility gives a unique prime p such that R has elements of additive order p, from which it follows that pH = (0) (as a special case p annihilates all commutators).

Let p not divide t. Then (3) yields a polynomial identity of the type in Proposition 1 and hence R is commutative. Thus p must divide t. Therefore p cannot divide s, which is obvious if Q(s, t) holds, and is also true if Q(s) holds, since s does not annihilate all commutators as p does. Hence in either of the cases Q(s, t) and Q(s) we get

p divides t but does not divide s.  (4)

By interchanging the roles of x and y in (2) we obtain

±[y, (x + w)^{t}] = ±[y, x^{t}] ± [y, w^{t}].  (5)

Replace x by x + 1 in (c_{1}) to get the identities

(x + 1)^{r}[(x + 1)^{s}, y] = ±[x, y^{t}](x + 1)^{n}

and

∑_{j=0}^{r} ∑_{k=1}^{s} \binom{r}{j}\binom{s}{k}x^{j}[x^{k}, y] = ± ∑_{j=1}^{n} \binom{n}{j}[x, y^{t}]x^{s}.  (6)

Now we prove that H ⊆ Z(R). Let x ∈ H and y ∈ R. Thus using H^{2} = (0), from (6) we have

s[x, y] = ±[x, y^{t}] for all x ∈ H, y ∈ R.  (7)

Substitute w = y into (5) to get

±[y, xy^{t−1} + yxy^{t−2} + ⋯ + y^{t−1}x] = 0

and so

±[x, y^{t}] = 0 for all x ∈ H, y ∈ R.  (8)

From (7), (8) and the fact that p does not divide s we have [x, y] = 0 for x ∈ H, y ∈ R. Hence

H ⊆ Z(R).  (9)

Thus all commutators are central. By Lemma 5 we get

[x, y^{t}] = ty^{t−1}[x, y] = 0 for all x, y ∈ R.  (10)

This condition shows that t divisible by p annihilates all commutators. Hence (6) can be rewritten as

s[x, y] + ∑_{k=2}^{s} \binom{s}{k}[x^{k}, y] = ± ∑_{j=1}^{r} ∑_{k=1}^{s} \binom{r}{j}\binom{s}{k}x^{j}[x^{k}, y]

which is the form of Proposition 1. Hence R is commutative. Proof of Theorem 2. Let R satisfy (c_{2}). As above, it is easy to observe that the reduction to a subdirectly irreducible ring R satisfying (α) with Q(s, t) holds and so do both conditions (α) and (β) with Q(s). Replacing x + 1 by x and y + 1 by y in c_{2} gives the following identities:

s[x, y] = ±G(x, y),
t[x, y] = ±H(x, y),

where G and H satisfy the conditions of Lemma 4. As in the proof of Theorem 1, we have a unique prime p such that pH = (0). Thus by Proposition 1, p must divide both t and s, which is impossible if Q(s, t) holds. Otherwise, since s is divisible by p, it must annihilate all commutators. Thus (β) gives a contradiction if Q(s) holds. Hence R is commutative.

Since Theorems 3 and 4 can be proved in the same way, we omit their proofs.

A recent commutativity study [10] deals with rings satisfying related conditions of the form

[xy − p(yx), x] = 0

or

[xy − p(yx), y] = 0.

This becomes possible by interchanging the roles of x and y as

[yx − p(xy), y] = 0

or

[yx − p(xy), x] = 0.

The following result is proved in [10].

Theorem 5. Let R be a ring with 1 such that for each x, y ∈ R, there exist p(t), q(t) ∈ t^{2}Z[t] for which [xy −p(yx), x] = 0 and [xy −q(yx), y] = 0. Then R is commutative.

Now we generalize Theorem 5.

Theorem 6. Let R be a ring with unity 1 such that for each x, y ∈ R there exists p(x) ∈ x^{2}Z[x] for which either [yx − p(yx), x] = 0 or [yx − p(yx), y] = 0. Then R is commutative.

Proof. Suppose that p(x) = a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+⋯+a_{n}x^{n}, where a_{1}, a_{2}, ..., a_{n} are integers. By hypothesis, we can write for all x, y ∈ R

\begin{matrix}[yx, x] = [p(yx), x], \\ [x, y]x = [p(yx), x]. \end{matrix}  (1′)

Putting x + 1 for x in (1′) gives

[x, y](x + 1) = [p(y(x + 1)), x],
[x, y]x + [x, y] = [a_{2}(y(x + 1))^{2} + a_{3}(y(x + 1))^{3} + ⋯ + a_{n}(y(x + 1))^{n}, x],
[x, y]x + [x, y] = [a_{2}y(x + 1)y + a_{3}y(x + 1)y(x + 1)y + ⋯ +
+ a_{n}y(x + 1)y(x + 1) ⋯ y(x + 1)y, x](x + 1) =

= [a_{2}(yxy + y^{2}) + a_{3}(yxyxy + yxy^{2} + y^{2}xy + y^{3}) + ⋯ +

+ a_{n}(yxyx ⋯ y + (x + 1), x](x + 1),
[x, y]x + [x, y] = [a_{2}yxy + a_{3}yxyxy + ⋯ + a_{n}yxyxyx ⋯ yx, x] +
+ [a_{2}y^{2} + a_{3}y^{3} + ⋯ + a_{n}y^{n}, x] + H(x, y),

where each homogeneous component of G has the sum of coefficients which is equal to zero. Thus H has no terms linear in y, and each term of H has a degree greater than 1 in x. Hence

[x, y]x + [x, y] = [p(yx), x] + [p(y), x] + H(x, y).  (2′)

Using (1′) and (2′), we obtain

[x, y] = [x, p(y)] + G(x, y).

Thus R is commutative by Kezlan’s theorem [11] or Lemma 4. Similarly, with the help of Lemma 4, R is commutative if R satisfies [yx − p(xy), y] = 0.

Similarly to the proof of Theorem 6, we can reprove the next theorem using Lemma 4.

Theorem 7 ([12]). Let R be a ring with unity 1 satisfying

[xy − p(yx), y] = 0 for all x, y ∈ R,

where p(x) ∈ x^{2}Z[x]. Then R is commutative.

Now let P be a ring property. If P is inherited by every subring and every homomorphic image, then P is called an h-property. More weakly, if P is inherited by every finitely generated subring and every natural homomorphic image modulo the annihilator of a central element, then P is called an H-property.

A ring property P such that a ring R has the property P if and only if all its finitely generated subrings have P is called an F -property.

Proposition 2 ([13, Proposition 1]). Let P be an H-property, and P′ be an F -property. If every ring R with unity 1 having the property P has the property P′, then every s-unital ring having P has P′.

Finally, Theorems 1–4, 6, and 7 are automatically generalized from a unital ring to s-unital ones due to Proposition 2. Indeed, we have

Theorem 8. Let R be a left (resp., right) s-unital ring satisfying (c_{1}) (resp., (c_{2})). Then R is commutative.

Theorem 9. Let R be a left s-unital (resp., right) ring satisfying (c_{3}) (resp., (c_{4})). Then R is commutative.

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(Received 19.04.1994; revised version 02.11.1994)

Author’s address: Department of Mathematics, King Abdul Aziz University P.O.Box 31464 Jeddah 21497, Saudi Arabia