13 ‫اﻟﻤﺤﺎﺿﺮة‬
‫ﺳﻨﺒﺤﺚ ﻧﻈﺮﯾﺔ ﻗﺎﺑﻠﯿﺔ اﻟﻘﺴﻤﺔ ﻓﻲ اﻟﺴﺎﺣﺔ اﻟﺘﻜﺎﻣﻠﯿﺔ وھﻲ ﺗﻌﻤﯿﻢ ﻟﺼﻔﺎت ﻣﺸﺎﺑﮭﺔ ﻓﻲ ﺣﻠﻘﺔ اﻹﻋﺪاد‬
‫ وﻧﺒﺪأ ﺑﺘﻌﺮﯾﻒ ﻣﻌﻨﻰ أن ﯾﻘﺴﻢ ﻋﻨﺼﺮ ﻣﺎ ﻋﻨﺼﺮ آﺧﺮ ﻓﻲ اﻟﺤﻠﻘﺔ أو أن ﯾﻜﻮن ﻋﻨﺼﺮ‬.‫اﻟﺼﺤﯿﺤﺔ‬
.‫ﻣﻀﺎﻋﻒ ﻟﻌﻨﺼﺮ آﺧﺮ‬
Definition 1.27 If a  0 and b are elements of a ring R, then a is said to
divide b, in symbols a|b, provided that some c  R exists such that b= ac.
In case a does not divide b, we shall write a b.
Other language for the divisibility property a|b is that a is a factor of b,
that b is divisible by a, and that b is a multiple of a.
.‫اﻟﻤﻼﺣﻈﺎت اﻟﺘﺎﻟﯿﺔ ﺗﺴﺘﻨﺘﺞ ﻣﻦ اﻟﺘﻌﺮﯾﻒ ﻣﺒﺎﺷﺮة‬
Theorem 1.39 Let the elements a, b, c R. Then
a) a|0, 1|a, a|a;
b) a|1 if and only if a is invertible;
c) if a|b, then ac| bc;
d) if a|b and b|c, then a|c;
e) if c|a and c|b, then c|(ax + by) for every x, yR.
:‫ ﻗﺴﻤﺔ اﻟﻌﻨﺎﺻﺮ ﻓﻲ اﻟﺤﻠﻘﺔ ﺗﺮﺗﺒﻂ ﺑﻌﻼﻗﺔ اﻻﺣﺘﻮاء ﺑﯿﻦ اﻟﻤﺜﺎﻟﯿﺎت ﺑﻤﻌﻨﻰ‬a|b if and only if (b)  (a).
‫ ذو ﻣﻌﻜﻮس )ﺑﺎﻟﻨﺴﺒﺔ إﻟﻰ ﻋﻤﻠﯿﺔ اﻟﻀﺮب( ﻓﺎن أي ﻋﻨﺼﺮ ﻓﻲ اﻟﺤﻠﻘﺔ ﯾﻘﺒﻞ‬u ‫ ﻋﻨﺪﻣﺎ ﯾﻜﻮن اﻟﻌﻨﺼﺮ‬‫ وﺑﺎﻟﺨﺼﻮص ﻓﻲ اﻟﺤﻘﻞ ﻓﺎن أي ﻋﻨﺼﺮ ﯾﻘﺒﻞ اﻟﻘﺴﻤﺔ ﻋﻠﻰ أي ﻋﻨﺼﺮ آﺧﺮ ﻏﯿﺮ‬. u ‫اﻟﻘﺴﻤﺔ ﻋﻠﻰ‬
.‫اﻟﺼﻔﺮ‬
‫ ﻟﺘﺠﺎوز ﻣﺸﺎﻛﻞ‬.(‫ ﻟﯿﺲ ﻟﺪﯾﮫ أي ﻗﻮاﺳﻢ )ﻋﻮاﻣﻞ‬2 ‫ اﻟﻌﻨﺼﺮ‬،‫ ﺑﺎﻟﻤﻘﺎﺑﻞ ﻓﻲ ﺣﻠﻘﺔ اﻷﻋﺪاد اﻟﺰوﺟﯿﺔ‬.‫اﻟﻤﻼﺣﻈﺔ أﻋﻼه ﻧﺤﺘﺎج إﻟﻰ اﻟﺘﻌﺮﯾﻒ اﻟﺘﺎﻟﻲ‬
Definition 1.28 Two elements a, b R are said to be associated elements
or simply associated if a = bu, where u is an invertible element.
Example 1.33 In the case of the ring Z the only associates of an integer
n Z are ± n, since ± 1 are the only invertible elements of R.
Example 1.34 Consider the domain Z(i) of Gausian integers, a
subdomain of the complex number field, whose elements form the set
Z(i) = { a + bi | a, b Z; i 2 = -1}.
Here the only invertible elements are ± 1 and ± i. So the class of
associates determined by any Gausian integer a + bi consists of exactly
four members :
a + bi, -a- bi, -b +ai, b- ai.
We have the following theorem:
Theorem 1.40 Let a, b be nonzero elements of an integral domain R.
Then the following statements are equivalent:
1. a and b are associates,
2. both a| b and b| a,
3. (a) = (b).
Proof: In the class.
‫اﻷﻋﺪاد‬
‫ﺻﻮل إﻟﻰ‬
‫ﻣﻦ اﺟﻞ ا ﻟ ﻮ‬
.‫ ﻧﺤﺘﺎج أوﻻ إﻟﻰ ﻣﻔﺎھﯿﻢ ﻣﺸﺎﺑﮭﮫ ﻟﻤﺎ ﻟﺪﯾﻨﺎ ﻓﻲ اﻷﻋﺪاد اﻟﺼﺤﯿﺤﺔ‬،‫اﻟﺼﺤﯿﺤﺔ إﻟﻰ ﻋﻮاﻣﻞ أوﻟﯿﺔ‬
Definition 1.29
1. A nonzero element p R is called a prime if and only if p is not
invertible and p| ab implies either p| a or else p| b.
2. A nonzero element q R is said to be irreducible (or
nonfactorizable) if and only if q is not invertible and in every
factorization q= bc with b, c R, either b or c is invertible.
Lemma In an integral domain R, any prime element p is irreducible.
Theorem 1.41 Let R be a principal ideal domain. A nonzero element
p R is irreducible if and only if it is prime.
Proof: In the class.
Theorem 1.42 Let R be a principal ideal domain. If { I n }, n Z  , is any
infinite sequence of ideals of R satisfying
I1  I 2  ...  I n  I n 1  ... ,
Then an integer m exists such that I n = I m for all n>m.
Proof: In the class.
A principal ideal of the ring R is said to be a maximal principal ideal
if it is maximal ( with respect to inclusion) in the set of proper principal
ideals of R.
Lemma Let R be an integral domain. For a nonzero element p R the
following hold:
a) p is an irreducible element of R if and only if (p) is a maximal
principal ideal;
b) p is a prime element of R if and only if the principal ideal (p)  R is
prime.
Proof: In the class.
.‫ ﺗﺘﻠﺨﺺ اﻟﺤﻘﺎﺋﻖ أﻋﻼه ﺑﻤﺎ ﯾﻠﻲ‬،‫ﻓﻲ اﻟﺴﺎﺣﺔ اﻟﺘﻜﺎﻣﻠﯿﺔ ذات اﻟﻤﺜﺎﻟﯿﺎت اﻟﺮﺋﯿﺴﯿﺔ‬
Theorem 1.43 Let R be a principal ideal domain. The nontrivial ideal (p)
is a maximal(prime) ideal of R if and only if p is an irreducible(prime)
element of R.
‫أن‬
.‫ﻣﻌﻜﻮس ﯾﺘﺤﻠﻞ إﻟﻰ ﻋﻮاﻣﻞ أوﻟﯿﺔ‬
Corollary Let a  0 be a noninvertible element of the principal ideal
domain R. Then there exists a prime p R such that p| a.
Proof . Since a is not invertible, the principal ideal (a)  R. Thus by
Theorem 1.28 a maximal ideal M of R exists such that (a)  M. But the
preceding result tells us that every maximal ideal is of the form M =(p),
where p is a prime element of R ( in this setting, there is no distinction
between prime and irreducible elements). Thus (a)  (p), which is to say
that p| a.
1.28
. 1.42‫ﺑﺎﻻﻋﺘﻤﺎد ﻋﻠﻰ ﻣﺒﺮھﻨﺔ‬
.‫اﻟﺴﺎﺣﺔ اﻟﺘﻜﺎﻣﻠﯿﺔ إﻟﻰ ﻋﻮاﻣﻞ أوﻟﯿﺔ‬
Definition 1.30 An integral domain R is a unique factorization domain
(U.F.D.) in case the following two conditions hold:
1. every a R that is neither zero nor invertible can be factored into a
finite product of irreducible elements;
2. if a= p 1 p 2 … p n = q 1 q 2 …q m are two factorizations of a into
irreducible elements, then n=m, and there is a permutation  of
the indices such that p i and q (i ) are associates (i= 1, 2, …, n).
.‫ﻓﯿﻤﺎ ﯾﻠﻲ ﺳﻨﺮى أن ﻛﻞ ﺳﺎﺣﺔ ﺗﻜﺎﻣﻠﯿﺔ ذات ﻣﺜﺎﻟﯿﺎت رﺋﯿﺴﯿﺔ ﺗﻤﺘﻠﻚ اﻟﺼﻔﺔ أﻋﻼه‬
Theorem 1.44 If R is a principal ideal domain, then every element of R
that is neither zero nor invertible has a factorization into a finite product
of primes.
Proof. Let a R be a nonzero noninvertible element. Then  p 1 R, with
p 1 prime and p 1 |a ( by the last corollary). So, a= p 1 a 1 for some 0  a 1 R,
hence (a)  (a 1 ).
If (a) = (a 1 ), then a 1 = ra, (r R), hence a= p 1 a 1 = p 1 ra, or 1=p 1 r, a
contradiction. This means that (a)  (a 1 ).
Repeat the procedure, now starting with a 1 , to obtain an increasing
chain of principal ideals
(a)  (a 1 )  (a 2 )  …  (a n )  …, with
a n 1 =p n a n for some prime p n R. This process goes on as a n is not
invertible. But, by Theorem 1.42, the above chain terminates, that is, a n
must possess an inverse for some n, and
(a)  (a 1 )  (a 2 )  …  (a n )=R .
Therefore a= p 1 p 2 … p n 1 p n , where p n = p n a n , being an associate
of prime, is itself prime.
Corollary In a P.I.D. R, every nontrivial ideal is the product of a finite
number of prime(maximal) ideals.
‫ وﻓﻲ اﻟﻤﺒﺮھﻨﺔ اﻟﻘﺎدﻣﺔ ﺳﻨﺜﺒﺖ وﺣﺪاﻧﯿﺔ‬،(‫ﻓﻲ اﻟﻤﺒﺮھﻨﺔ اﻟﺴﺎﺑﻘﺔ أﺛﺒﺘﻨﺎ وﺟﻮد اﻟﺘﺠﺰﺋﺔ )اﻟﺘﺤﻠﯿﻞ‬
.‫اﻟﺘﺠﺰﺋﺔ‬
Theorem 1.45 Every P.I.D. R is a U.F.D.
Proof: In the class.
‫ ﻟ ﻜﻦ ﻣﻦ‬. P.I.D ‫ ﻟﯿﺲ‬U.F.D. ‫ ﻏﯿﺮ ﺻﺤﯿﺢ وﺳﻨﻌﻄﻲ )ﻻﺣﻘﺎ( ﻣﺜ ﺎل ﻋﻠﻰ‬1.45 ‫ﻋﻜﺲ اﻟﻤﺒﺮھﻦ‬
(‫اﻟﻤﻔﯿﺪ أن ﻧﻨﺘﺒﮫ إﻟﻰ اﻟﺤﻘﯿﻘﺔ اﻟﺘﺎﻟﯿﺔ )واﻟﺘﻲ ﺳﻨﻮﺿﺤﮭﺎ ﻓﻲ اﻟﺼﻒ‬
Remark. In a U.F.D. R, any irreducible element p R is necessarily
prime.
14 ‫اﻟﻤﺤﺎﺿﺮة‬
.‫ وﻧﺒﺪأ ﺑﺎﻟﺘﻌﺮﯾﻒ اﻟﺘﺎﻟﻲ‬،‫اﻟﺤﻠﻘﺎت ﯾﺴﻤﻰ اﻟﺴﺎﺣﺔ اﻻﻗﻠﯿﺪﯾﺔ‬
Definition 1.31 An integral domain R is said to be Euclidean if a
function  : R→Z   {0} (the Euclidean valuation) exists such that the
following conditions are satisfied:
1.  (a)= 0 if and only if a= 0;
2. for all a, b R,  (ab) =  (a)  (b);
3. given a, b R, with b  0, there exist elements q, r R ( the quotient and
the remainder) such that a= qb + r, where  (r)<  (b).
As simple examples of Euclidean domains, we may take
1. any field F, with valuation defined by  (a)= 1 for all nonzero a F
and  (0)= 0;
2. the ring Z of integers. With valuation defined by  (a)= |a| for all
a Z.
3. the Gaussian integers Z(i). with valuation defined by  (a + bi) =
a 2 + b 2 for all a + bi Z(i).
.‫ ﺗﻈﮭﺮ ﻓﻲ اﻟﻤﺄﺧﻮذة اﻟﺘﺎﻟﯿﺔ‬،‫ﻋﺪة ﺧﻮاص أوﻟﯿﺔ ﻟﻠﺴﺎﺣﺎت اﻻﻗﻠﯿﺪﯾﺔ‬
Lemma. Let R be a Euclidean domain with valuation  . Then
1.  (1)= 1;
2. an element 0  a R is invertible if and only if  (a)= 1;
3. if two elements a, b R are associates, then  (a)=  (b).
Proof: In the class.
،
r‫ و‬q ،‫ ﻓﻲ ﺧﻮارزﻣﯿﺔ اﻟﻘﺴﻤﺔ‬،‫ﺑﺸﻜﻞ ﻋﺎم‬
.‫اﻻﻗﻠﯿﺪﯾﺔ ﯾﻤﻜﻦ اﻟﺒﺮھﻨﺔ ﻋﻠﻰ وﺣﺪاﻧﯿﺘﮭﻤﺎ‬
Theorem 1.46 The quotient and remainder in condition (3) of the
definition of a Euclidean domain R are unique if and only if
 (a+ b)  max{  (a),  (b)} for all a, b R.
Proof: In the class.
:‫ﻣﻼﺣﻈﺎت‬
‫ ﻣﺜﺎل ﻋﻠﻰ ﺳﺎﺣﺔ اﻗﻠﯿﺪﯾﺔ ﻻ ﺗﺤﻘﻖ‬،(‫ ﺳﺎﺣﺔ اﻷﻋﺪاد اﻟﺼﺤﯿﺤﺔ )ﻣﻊ اﻟﻘﯿﻤﺔ اﻟﻤﻄﻠﻘﺔ داﻟﺔ اﻟﻘﯿﻤﺔ‬.1
.1.46 ‫اﻟﺸﺮط ﻓﻲ اﻟﻤﺒﺮھﻨﺔ‬
(
)
.2
،‫اﻟﻘﺴﻤﺔ‬
‫ وﻧﺴﺘﻨﺞ ذﻟﻚ ﻣﻦ ﺧﻼل إﺛﺒﺎت أن ﻛﻞ ﺳﺎﺣﺔ‬، ‫ﻋﻠﻰ ﻏﺮار ﻣﺎ ﻣﻮﺟﻮد ﻓﻲ اﻷﻋﺪاد اﻟﺼﺤﯿﺤﺔ‬
.
.‫ﻣﺜﺎل‬
Theorem 1.47 Every Euclidean domain is a P.I.D.
Proof: In the class.
Corollary. Every Euclidean domain is a U.F.D.
Problems 5.
1. If u is a nonzero element of an integral domain R, prove that the
following statements are equivalent:
a) u and 1 are associates;
b) u is an invertible element of R;
c) the principal ideal (u)= R.
2. Assuming that R is a Euclidean domain with valuation  , prove the
statements below:
a) For nonzero a, b R, if a| b and  (a)=  (b), then a and b are
associates. [ Hint: Show that b|a.]
b) For nonzero a, b R,  (ab) >  (a) if and only if b is not an invertible
element.[ Hint: Use the " division algorithm" to write a= q(ab) + r.]
c) If n is any integer such that  (1) + n  0, then the function
 : R-{0}→ Z defined by  (a) =  (a) + n is also a Euclidean valuation
on R.
3. In the domain Z(i), let a= -4 + I and b= 5 + 3i. Show that :
a) the Euclidean valuation  defined by  (r + si) = r 2 + s 2 does not
satisfy  (a + b)  max{  (a),  (b)};
b) the element a has the two representations
a= (-1 + i)b + (4- i) = (-1)b + (1 + 4i)
and hence uniqueness fails in the "division algorithm" for Z(i).
4. Let n  1 be a square-free integer ( an integer that is not divisible by
square of any positive integer >1), let Q be the set of rational numbers,
and let Q(n) = { a+ b n | a, b Q}.
a)Prove that for each square-free integer n, Q(n) is a field, it is a
subfield of the field of complex numbers C.
b)For each  = a+ b n in Q(n), define  = a- b n (called the
conjugate of  ), and define N(  ) =   (called the norm of  ) . Prove
that: 1. N(  )=0 if and only if  =0; 2. N(   )=N(  )N(  ); 3. N(1)=1.
c) Let Z( n ) = { a+ b n | a, b Z}. Show that Z( n ) is an integral
domain and the following two statements are hold:
1- N(  )= ±1 if and only if  is invertible in Z( n );
2- if N(  )=±p, where p is a prime number, then  is an
irreducible element of Z( n ).