:‫اﻟﻤﺤﺎﺿﺮة اﻟﺜﺎﻟﺜﺔ‬
.‫ﻓﯿﻤﺎ ﯾﻠﻲ ﺳﻨﻌﺮف ﻣﺮﻛﺰ اﻟﺤﻠﻘﺔ ﺛﻢ ﻧﺜﺒﺖ أﻧﮭﺎ ﺣﻠﻘﺔ ﺟﺰﺋﯿﺔ ﻣﻦ اﻟﺤﻠﻘﺔ اﻷﺻﻠﯿﺔ‬
Definition 1.6 Let R be a ring, then cent R is the set of all elements of R
that commute with every element of R, that is
cent R = { a Î R| ar =ra for all r Î R}
.cent R= R
‫ ﺗﻜﻮن اﺑﺪاﻟﯿﺔ إذا وﻓﻘﻂ إذا‬R ‫ﻣﻦ اﻟﻮاﺿﺢ أن اﻟﺤﻠﻘﺔ‬
Theorem 1.4 For any ring R, cent R is a subring of R.
Proof: 0 Î cent R , since a0=0a =0 "a Î R, which implies cent R ¹ f .
Let a, b Î cent R, then (a- b)r = ar – br = ra – rb = r(a – b). Hence
a – b Î cent R. Also, (ab)r = a(br) =a(rb)= (ar)b=(ra)b=r(ab), which
implies ab Î cent R too. Therefore by Theorem 1.3 cent R is a subring.
‫ﻻﺣﻈﻨﺎ ﻓﻲ ﻣﺜ ﺎل ﺳ ﺎﺑﻖ اﻧ ﮫ ﻣ ﻦ اﻟﻤﻤﻜ ﻦ أن ﺗﻜ ﻮن ﺣﻠﻘ ﺔ ذات ﻋﻨ ﺼﺮ ﻣﺤﺎﯾ ﺪ وﻟ ﺪﯾﮭﺎ ﺣﻠﻘ ﺔ ﺟﺰﺋﯿ ﺔ‬
:‫ وﯾﻨﺒﻐﻲ أن ﻧﻌﻠﻢ أن اﻟﺤﺎﻻت اﻟﺘﺎﻟﯿﺔ ﻣﻤﻜﻨﺔ أﯾﻀﺎ‬،‫ﺑﺪون ﻋﻨﺼﺮ ﻣﺤﺎﯾﺪ‬
.‫ ﺣﻠﻘﺔ ﺑﺪون ﻣﺤﺎﯾﺪ وﺗﺤﻮي ﺣﻠﻘﺔ ﺟﺰﺋﯿﺔ ﺑﻤﺤﺎﯾﺪ‬-1
.‫ ﺣﻠﻘﺔ ﺑﻤﺤﺎﯾﺪ ﺗﺤﻮي ﺣﻠﻘﺔ ﺟﺰﺋﯿﺔ ﺑﻤﺤﺎﯾﺪ ﯾﺨﺘﻠﻒ ﻋﻦ ﻣﺤﺎﯾﺪ اﻟﺤﻠﻘﺔ اﻷﺻﻠﯿﺔ‬-2
‫ وﻓﯿﻤ ﺎ ﯾﻠ ﻲ‬.‫ﻓﻲ ﻛﻼ اﻟﺤﺎﻟﺘﯿﻦ أﻋﻼه ﯾﻜﻮن ﻣﺤﺎﯾﺪ اﻟﺤﻠﻘ ﺔ اﻟﺠﺰﺋﯿ ﺔ ﻗﺎﺳ ﻢ ﻟﻠ ﺼﻔﺮ ﻓ ﻲ اﻟﺤﻠﻘ ﺔ اﻷﺻ ﻠﯿﺔ‬
.‫ﺑﺮھﺎن ھﺬه اﻟﺤﻘﯿﻘﺔ‬
Let 1¢ ¹ 0 denote the identity element of the subring S; we assume further
that 1¢ does not act as an identity for the whole ring R. Accordingly, some
element aÎ R exists for which a1¢ ¹ a . It is clear that
(a 1¢ ) 1¢ = a(1¢ 1¢ )=a 1¢ .
or ( a 1¢ - a) 1¢ =0.
Since neither a1¢ -a nor 1¢ is zero, the ring R has zero divisors, and in
particular, 1¢ is a zero divisor.
‫ﻓﯿﻤﺎ ﯾﻠﻲ ﻣﺜﺎل ﻋﻦ ﺣﻠﻘﺔ وﺣﻠﻘﺔ ﺟﺰﺋﯿﺔ ﻛﻼ ﻣﻨﮭﻤﺎ ذات ﻋﻨﺼﺮ ﻣﺤﺎﯾﺪ وﻟﻜﻦ ﻣﺤﺎﯾﺪ اﻟﺤﻠﻘﺔ ﯾﺨﺘﻠﻒ ﻋﻦ‬
.‫ﻣﺤﺎﯾﺪ اﻟﺤﻠﻘﺔ اﻟﺠﺰﺋﯿﺔ‬
Example 1.9
Consider the set R= Z ´ Z, consisting of ordered pairs of integers. One
converts R into a ring by defining addition and multiplication
componentwise:
(a, b) + (c, d) = (a + c, b +d); (a, b)(c, d) = (ac, bd)
A routine calculation will show that Z ´ {0} = {(a, 0)| a Î Z} forms a subring with
identity element (1, 0). While the identity of the entire ring R is (1, 1).
Also note that (1, 0)(0, 1)= (0, 0), that is, (1, 0) (the identity of the subring) is a zero
divisor of R.
.( ‫ﻓﯿﻤﺎ ﯾﻠﻲ ﻧﺘﻔﻖ ﻋﻠﻰ رﻣﺰ اﻟﻘﻮة ﻟﻌﻨﺎﺻﺮ اﻟﺤﻠﻘﺔ ) اﻟﻀﺮب اﻟﻤﺘﻜﺮر ﻟﻨﻔﺲ اﻟﻌﻨﺼﺮ‬
If R is an arbitrary ring and n a positive integer, then the nth power
a of an element a Î R is defined by the inductive conditions a 1 = a and
a n = a n -1a . From this the usual laws of exponents follow at once:
a n a m = a n + m , (a n ) m = a nm
( n, m Î Z + )
Observe that if two elements n, mÎ R happen to commute, then so do all
powers of a and b, whence (ab) n = a n b n for each positive integer n.
‫ ﯾﻤﻜﻦ اﻋﺘﻤﺎد اﻟﻘﻮى‬a ‫ﻛﺬﻟﻚ ﻓﻲ ﺣﺎﻟﺔ وﺟﻮد اﻟﻤﺤﺎﯾﺪ ﻓﻲ اﻟﺤﻠﻘﺔ ووﺟﻮد اﻟﻨﻈﯿﺮ اﻟﻀﺮﺑﻲ ﻟﻠﻌﻨﺼﺮ‬
:‫ أﯾﻀﺎ وﻛﻤﺎ ﯾﻠﻲ‬a ‫اﻟﺴﺎﻟﺒﺔ ﻟﻠﻌﻨﺼﺮ‬
0
-n
-1 n
a = (a ) , where n>0. With the definition a = 1 the symbol a n now has
a well-defined meaning foe every integer n – at least when the attention is
restricted to invertible elements.
‫ﺑﺎﻟﻨﺴﺒﺔ إﻟﻰ ﻋﻤﻠﯿﺔ اﻟﺠﻤﻊ ﯾﻤﻜﻦ أﯾﻀﺎ أن ﻧﻌﺒﺮ ﻋﻦ ﺗﻜﺮار اﻟﺠﻤﻊ ﻟﻌﻨﺼﺮ واﺣﺪ ﺑﺎﻟﻤﻀﺎﻋﻔﺎت‬
:‫اﻟﺼﺤﯿﺤﺔ وﻛﻤﺎ ﯾﻠﻲ‬
1a=a,
and na = (n-1) a + a, when n>1
0a= 0
and (-n) a= -(na).
:‫ﻛﻤﺎ ﯾﻤﻜﻦ ﺑﺴﮭﻮﻟﺔ ﺑﺮھﻨﺔ اﻟﻌﺒﺎرات اﻟﺘﺎﻟﯿﺔ‬
(n+ m) a = na + ma, (nm)a= n(ma), n(a +b) = na + nb,
n(ab) = (na) b = a(nb), and (na)(mb)= (nm)(ab).
‫ ﯾﻤﻜﻦ أن ﻻ ﯾﻜﻮن‬n ‫ ﺣﯿﺚ أن اﻟﻌﺪد‬، ‫ ﻻﺗﻤﺜﻞ ﺿﺮب ﻓﻲ اﻟﺤﻠﻘﺔ‬na ‫ﯾﻨﺒﻐﻲ أن ﻧﻔﮭﻢ أن اﻟﺼﯿﻐﺔ‬
.‫ وإﻧﻤﺎ ھﻮ رﻣﺰ)ﻣﺘﻔﻖ ﻋﻠﯿﮫ( ﻟﺠﻤﻊ ﻣﺎ‬،‫ﻋﻨﺼﺮ ﻓﻲ اﻟﺤﻠﻘﺔ‬
.‫وأﻻن ﻧﻌﺮف ﻣﻤﯿﺰ اﻟﺤﻠﻘﺔ‬
Definition 1.6 Let R be an arbitrary ring. If there exists a positive integer
n such that na= 0 for all aÎ R, then the smallest positive integer with this
property is called the characteristic of the ring . If no such positive integer
exists (that is, n=0 is the only integer for which na=0 for all a in R), then
R is said to be of characteristic zero. We shall write char R for the
characteristic of R.
.‫ اﻟﻨﺴﺒﯿﺔ واﻟﺤﻘﯿﻘﯿﺔ ھﻲ أﻣﺜﻠﺔ واﺿﺤﺔ ﻟﺤﻠﻘﺎت ﻣﻤﯿﺰھﺎ ﺻﻔﺮا‬، ‫ﺣﻠﻘﺎت اﻷﻋﺪاد اﻟﺼﺤﯿﺤﺔ‬
‫ ﺣﯿﺚ ان‬2 ‫( ھﻲ ﺣﻠﻘﺔ ﻣﻤﯿﺰھﺎ‬1.2 ‫ ) ﻣﺜﺎل‬P(X) ‫ﻣﻦ ﻧﺎﺣﯿﺔ أﺧﺮى اﻟﺤﻠﻘﺔ‬
2A = A ∆ A = f for every subset A of X.
‫ ﻟﻜﻦ ﻓﻲ اﻟﺤﻠﻘﺎت ذات‬،‫ﺑﺎﻟﺮﻏﻢ ﻣﻦ ان ﺗﻌﺮﯾﻒ اﻟﻤﻤﯿﺰ ﯾﺘﻄﻠﺐ ﺗﺤﻘﻖ ﺷﺮط ﻟﻜﻞ ﻋﻨﺎﺻﺮ اﻟﺤﻠﻘﺔ‬
.‫ ﻻﺣﻆ اﻟﻤﺒﺮھﻨﺔ اﻟﺘﺎﻟﯿﺔ‬.‫اﻟﻤﺤﺎﯾﺪ ﯾﻜﻔﻲ ﺗﺤﻘﻖ اﻟﺸﺮط ﻓﻲ اﻟﻤﺤﺎﯾﺪ ﻓﻘﻂ‬
n
Theorem 1.5 If R is a ring with identity 1, then R has characteristic n>0
if and only if n is the least positive integer for which n1=0.
Proof: In the class.
Corollary 1 In an integral domain R, all the nonzero elements have the
same additive order; this order is the characteristic of the domain when
char R> 0 and is infinite when char R = 0.
Corollary 2 An integral domain R has positive characteristic if and only
if na = 0 for some 0 ¹ a Î R and some integer nÎ Z + .
Theorem 1.6 The characteristic of an integral domain is either zero or a
prime number.
Proof: Assume that char R =n> 0 and n is not prime. Then n= n 1 n 2 , with
1< n 1 £ n 2 < n. It follows that
0 = n 1 = ( n 1 n 2 ) 1 = ( n 1 n 2 ) 1 2 = (n 1 1)( n 2 1).
By assumption , R is without zero divisors, so either n 1 = 0 or n 2 = 0.
Since both n 1 and n 2 are less than n, this is a contradiction. Therefore
char R must be prime.
Corollary If R is a finite integral domain, then char R = p, where p is a
prime.
Remark. Let R be a ring with identity and consider the set
Z1= { n1| nÎ Z}, then
1. Z1 forms a (commutative) ring with identity.
2. The order of the additive cyclic group (Z1, +) = char R.
3. If R is an integral domain, then Z1 is a subdomain of R . It is the
smallest subdomain of R.
4. If char R=p, where p is a prime, then Z1 is a field.
Proof:
1. n1 – m1 = (n – m) 1, (n1)(m1) = (n m)1 and (1) 1= 1.
2. Clear.
3. (n1)(m1) = 0 implies (n m)1 = 0 implies n m =0 implies n =0 or m = 0
implies n1= 0 or m1= 0.
If S is any subdomain of R, then 1S = 1R ( otherwise 1S is a zero divisor of
R), hence 1Î S, that is Z1 Í S.
4. Let char R=p, where p is a prime and R is an integral domain, then it is
enough to prove that each nonzero element of Z1 is invertible. Since Z1 is
an additive cyclic group of order p, it consists of p distinct elements,
namely, the p sums n1, where n = 0, 1, …, p – 1. Let n1 ¹ 0 ( 0 < n < p).
Since n and p are relatively prime, integers r and s exist for which
r p + s n =1. But then 1 = (r p + s n)1 = r(p1) + (s1)( n1). As p1 = 0,
we obtain the equation 1 = (s1)(n1), so that s1 serves as the multiplicative
inverse of n1 in Z1.
‫اﻟﻤﺤﺎﺿﺮة اﻟﺮاﺑﻌﺔ‬
Definition 1.7 Let R and R ¢ be two rings . By a (ring) homomorphism
from R into R ¢ is meant a function f : R→ R ¢ such that
f (a + b) = f ( a) + f ( b),
f (a ∙ b) = f ( a) ∙ f (b)
for every pair of elements a, bÎ R. A homomorphism that is also a one to
one mapping is called an isomorphism.
‫ﯾﻨﺒﻐﻲ اﻻﻧﺘﺒﺎه إﻟﻰ أن ﻋﻤﻠﯿﺘﺎ اﻟﺠﻤﻊ واﻟﻀﺮب اﻟﻈﺎھﺮة ﻋﻠﻰ اﻟﯿﺴﺎر ﻓﻲ اﻟﺘﻌﺮﯾﻒ أﻋﻼه ﺗﺨﺺ‬
. R ¢ ‫ واﻟﻈﺎھﺮة ﻋﻠﻰ اﻟﯿﻤﯿﻦ ﺗﺨﺺ اﻟﺤﻠﻘﺔ‬R ‫اﻟﺤﻠﻘﺔ‬
If f is homomorphism from R into R ¢ , then f ( R) is called the
homomorphic image of R. When R = R ¢ , we speak of f as a
homomorphism of R into itself or an endomorphism of R. Likewise, an
isomorphism of the ring R onto itself is termed an automorphism of R.
Example 1.10
Let R and R ¢ be arbitrary rings and f: R→ R ¢ be the function that sends
each element of R onto the zero element of R ¢ .
Example 1.11
Consider ring Z of integers and the ring Z n of integers modulo n. Define
f : Z ® Z n by taking f ( a) = [a]; that is, map each integer into the
congruence class containing it.
Example 1.12
In the ring map(X, R), define t a to be the function that assigns to each
f Î map(X, R) its value at a fixed element a Î X; in other words,
t a ( f)= f (a). Then is a homomorphism from map(X, R) into R, known as
the evaluation homomorphism at a.
.‫وأﻻن ﻧﺴﺘﻨﺘﺞ ﺑﻌﺾ اﻟﺨﻮاص اﻷوﻟﯿﺔ ﻟﻠﺘﺸﺎﻛﻞ ﻣﻦ ﺧﻼل اﻟﺘﻌﺮﯾﻒ‬
Theorem 1.7 If f is a homomorphism from the ring R into the ring R ¢ ,
the following hold:
1. f (0) = 0 and 2. f (- a ) = - f (a) for all aÎ R.
If, in addition, R and R ¢ are both rings with identity and f ( R)= R ¢ , then
3. f (1) = 1 and 4. f ( a -1 ) = f (a ) -1 for each invertible element aÎ R.
Proof: In the class.
Theorem 1.8 Let f be a homomorphism from the ring R into the ring
R ¢ . Then
a) for each subring S of R, f(S) is a subring of R ¢ ; and
b) for each subring S¢ of R ¢ , f -1 ( S¢ ) is a subring of R.
Proof: In the class.
In agreement with our previous use of the term, two rings R and R ¢
are said to be isomorphic if there exists an isomorphism from R onto R ¢ ;
we indicate this by writing R @ R ¢ .
Definition 1.8 A ring R is said to be imbedded in a ring R ¢ if there exists
some subring S¢ of R ¢ such that R @ S¢ .
In general, if a ring R is imbedded in a ring R ¢ , then R ¢ is referred
to as an extension of R, and we say that R can be extended to R ¢ .
Theorem 1.9 (Dorroh Extension Theorem) Any ring R can be
imbedded in a ring with identity.
Proof: Consider the Cartesian product R ´ Z, where
R ´ Z= {(r, n)| r Î R; nÎ Z}. If addition and multiplication are defined
by (a, n) + (b, m) = (a + b, n + m), (a, n)(b, m) = (ab + ma + nb, nm),
then it is a simple matter to verify that R ´ Z forms a ring. Note that this
system has a multiplicative identity, namely, (0, 1).
Next, consider the subset R ´ {0} of R ´ Z consisting of all pairs of the
form (a, 0), it is a subring and isomorphic to R under the mapping
f: R→R ´ {0} defined by f(a)= (a, 0). This process of extension therefore
imbeds R in R ´ Z, a ring with identity.
Problems 1
1. Consider the set Z of integers with the usual operations. If a new
addition and multiplication are defined by
a * b= a+b+1, a o b= a+ b+ ab, prove that the system (Z, * , o )
is a commutative ring with identity.
2. Establish the following statements:
a) The group Z * 5 of invertible elements of the ring Z 5 is
isomorphic to the cyclic group Z 4 .
b) The group Z * 9 of invertible elements of the ring Z 9 is cyclic
with generators 2 and 5.
3. In a ring R with identity, prove that each of the following is true:
a) The identity element for multiplication is unique.
b) If a Î R has a multiplicative inverse, then a -1 is unique.
c) If an element aÎ R is invertible, then so also is –a.
d) No zero divisor of R can possess a multiplicative inverse.
4. a) If the set X contains more than one element, show that every
nonempty proper subset of X is a zero divisor in the ring P(X).
c) For the ring map R # , verify that the function f(x) = x + |x| and
g(x) = x -|x| are both zero divisors.
5. An element a of a ring R is said to be idempotent if a 2 = a and
nilpotent if a n = 0 for some n Î Z + . Show that
a) a nonzero idempotent element cannot be nilpotent;
b) every nonzero nilpotent element is a zero divisor.
c) If R is an integral domain, the only nilpotent element is the
zero element;
d) If R is an integral domain, the identity element is the only
nonzero idempotent element.