1. N. GANESAN proved,in [1], that if R is a commutative ring containing n + 1 zero divisors, where n is a positive integer, then the order of R is less than or equal to (n + 1) 2. In this paper, we will prove that if R is an arbitrary ring having n + 1 left (right) zero divisors, then R has at most (n + 1) 2 elements. We will also prove that if 1 e R, the order of R is even less than (n + 1) 2 unless n + 1 is a power of a prime p, and every minimal right ideal I has the property 12 = 0. Actually in case R contains a minimal right ideal I which is not nilpotent, the 1 2 order of R is less than or equal to -~-n + n + 1. Finally we prove that if the set of all left (right) zero divisors of R has the infinite cardinality, say ~t then the cardinality of R is precisely ~.2. An element x of an arbitrary ring R is called a left (right) zero divisor if there is a e R, a 4: 0, such that ax = O(xa = 0). Theorem I. Any ring having only n + 1 left (right) zero divisors, where n is a positive integer, is necessarily finite and does not coniain more than (n + 1) 2 elements.Proof. If a ~ R, let R a be the right annulator of a, i. e. the set of all elements r such that ar = 0. Since R contains n + 1 left zero divisors, there is a e R, a ~ 0, such that the order of R a is at most n + 1. Let y 4 0 be an element of Ra. Then the order of yR is at most n + 1 since yR is a subset of Ra. The order of yR is the order of R/Ry since yR ~ R/Ry. Thus the order of R is the product of the order ofyR and the order of Ry, which is less than or equal to (n + 1) 2. Theorem II. Let R be a ring with 1. If R contains n + 1 left (right) zero divisors then the order of R is even less then (n + 1) 2 unless n + 1 is a power of a prime p, and every minimal right (left) ideal I has the property I 2 = O.Proof. Let I be a minimal right ideal of R such that 12 + 0. Then there is an idempotent 0 ~: e e I such that Rec~eR = 0. The order of R e is less than or equal to n + 1 since every element of Re is a left zero divisor. Similarly every element of eR is also a left zero divisor since 1 -e :~ 0 annihilates eR. Let ml be the order ofR e and m2 be the order of oR. Then ml • m 2 is the order of R and m 1 + m 2 ~< n + 2. Since 2m 1 • m 2 _< m 2 + m 2, 4ml • m 2 ~ (ml + m2) 2, i. e. m t • m2 < = 2 -J < \~-2--/ =-4-+ n + 1. Hence the order of R is less than (n + 1) 2. Now, suppose that if I is a minimal right ideal of R then 12 = 0. If the
In [3, p. 149], J. Lambek gives a proof of a theorem, essentially due to Grothendieck and Dieudonne, that if R is a commutative ring with 1 then R is isomorphic to the ring of global sections of a sheaf over the prime ideal space of R where a stalk of the sheaf is of the form R/0P, for each prime ideal P, and . In this note we will show, this type of representation of a noncommutative ring is possible if the ring contains no nonzero nilpotent elements.
If R is a ring and I is a right ideal of R then I is called faithful if R - I is a faithful right R-module, i.e. if { r ∊ R: Rr⊆ I} = (0). I is called irreducible [ 1 ] provided that if J1 and J2 are right ideals such that J1 ∩ J2 = I, then J1 or J2 = I. Let N(I){ r ∊ R: rI⊆ I} and [ I: a ] = { r ∊ R: ar⊆ I} for a ∊ R. We write (a)r for [(0): a ].
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