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Problem 4
1. a) Prove that Z ´ Z e is a maximal ideal of the ring Z ´ Z.
b) Show that the ring R is a field if and only if {0} is a maximal ideal
of R.
solution: a) Let J be an ideal of Z ´ Z such that, Z ´ Z e Ì J Í Z ´ Z.
If (a,b)Î J\ Z ´ Z e , then b= 2n+1 for some integer n ,
but (a-1, 2n) Î Z ´ Z e Ì J, that is (1,1)=(a,2n+1)-(a-1,2n) Î J.
This implies J= Z ´ Z, hence Z ´ Z e is maximal.
3. Let f be a homomorphism from the ring R onto the ring S. Prove that
a) if M is a maximal ideal of R with M Ê ker f , then f(M) is maximal
ideal of S.
b) if N is a maximal ideal of S, then f -1 (N) is a maximal ideal of R.
solution: a) Since there is a one to one correspondence between the ideals
of S and the ideals of R containing kerf , and since f(M) correspond to M,
also, there is no ideal in R lying strictly between M and R, then there is
no ideal in S lying strictly between f(M) and S, that is f(M) is a maximal
ideal of S.
Another solution of (a). f(M) Ì J Í S implies
-1
M Í f (f(M)) Ì f -1 (J) Í f -1 (S)=R, implies M Ì f -1 (J) Í R
Now, by maximality of M , f -1 (J)=R, then J=S (since f is onto).
Therefore f(M) is a maximal ideal of S.
b) Let f -1 (N) Ì I Í R, then N= f(f -1 (N)) Ì f(I) Í f(R)=S (f is onto) ,
that is, N Ì f(I) Í S , by maximality of N f(I)=S, which implies I=R.
Therefore f -1 (N) is a maximal ideal of R.
4. If M and N are distinct maximal ideals of the ring R, establish the
equality MN = M Ç N.
Solution: MN Í M Ç N is always true (clear).
To prove the converse. Since M and N are distinct, let x Î N\M, then by
maximality of M, (M, x)= R and hence 1= m+ rx, for some mÎ M and
r Î R. Now for each a Î M Ç N, we have a = ma+ rxa. It is clear that the
right hand side is an element of MN( note that R is assumed commutative
with identity), hence a Î MN, that is, M Ç N Í MN.
5. Let M be a proper ideal of the ring R. Prove that M is a maximal ideal
if and only if, for each ideal I of R, either I Í M or else I +M =R.
Solution: ( Þ ) Assume that M is maximal, and I is any ideal.
If I Ë M, then M Ì I+ M Í R, and by definition of maximality I +M =R.
( Ü ) Assume that M is a proper ideal of the ring R, with the property that
for each ideal I of R, either I Í M or else I +M =R. If M Ì J Í R, then
J Ë M, hence J+ M= R, but J+ M= J( M Ì J). Therefore J= R, hence M is
maximal.
10. Given that f is a homomorphism from the ring R onto the ring S,
verify that a) S is a field if and only if ker f is a maximal ideal of R.
b) S is an integral domain if and only if kerf is a prime ideal of R.
Solution: f is onto implies S @ R/ker f
a) S is a field if and only if R/ker f is a field if and only if ker f is a
maximal ideal of R.
b) S is an integral domain if and only if R/ker f is an integral domain
if and only if ker f is a prime ideal of R.
12. a) Find all maximal and prime ideals of the ring Z 12 .
b) In the ring Z 36 obtain I for I = (4) and I = (6).
Solution: a) 2 Z 12 ={0,2,4,6,8,10}; 3 Z 12 ={0,3,6,9}
b) (4) =(2) = {0,2,4,…,34}, (6) =(6)= {0,6,…, 30}
14. Prove that if I is an ideal of the ring R and P is a prime ideal of I then
P is an ideal of the whole ring R.
Solution: It is enough to prove that r Î R and p Î P implies rp Î P.
Since p Î I (P Í I), and I is an ideal of R, then rp Î I. Also rpr Î I, so
(rpr)p Î P (since P is an ideal of I) , then we have (rp)(rp) Î P, but P is
prime in I, hence rp Î P or rp Î P, that is rp Î P.
15. Assume that P is a proper prime ideal of the ring R with the property
that the quotient ring R/P is finite. Show that P must be a maximal ideal
of R.
Solution: P is proper prime ideal of R implies R/P is an integral domain.
R/P is a finite integral domain implies R/P is a field, which implies that P
is a maximal ideal of R.
Problem 5.
4. Let n ¹ 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 ÎQ}.
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 = a+ b n in Q( n ), define a = a- b n (called the
conjugate of a ), and define N( a ) = a a (called the norm of a ) . Prove
that: 1. N( a )=0 if and only if a =0; 2. N( a b )=N( a )N( b ); 3. N(1)=1.
c) Let Z( n ) = { a+ b n | a, b ÎZ}. Show that Z( n ) is an integral
domain and the following two statements are hold:
1- N( a )= ±1 if and only if a is invertible in Z( n );
2- if N( a )=±p, where p is a prime number, then a is an
irreducible element of Z( n ).
Solution: a) It is enough to show that Q( n ) is a subring of C the field
of complex numbers and any non zero element of Q( n ) its
multiplication inverse is again in Q(n). (since the other conditions follows
from the fact that C is a field. Q(n) Í C is clear. 0=0 +0 n Î Q( n ) and
1=1+ 0 n Î Q( n ). (a+b n ) –(c+ d n )= a-c+(b-d) n Î Q(n) and
(a+b n )(c+ d n )=(ac+ bdn) + (ad+ bc) n Î Q( n ). Therefore Q( n )
is a subring (and hence a subdomain of C).
Now let a+b n ¹ 0, to verify the existence of the inverse in Q(n) we have
to solve the equation (a+ b n )(c+ d n )= 1for c and d. Then
ac+ bdn + (ad+ bc) n =1+ 0 n , which tends to two equations
ac+ bdn=1 ; ad+ bc=0 that have a unique solution since a 2 -b 2 n ¹ 0
(otherwise n= a 2 / b 2 , contradicts the assumption that n is square free).
b)1. N( a )=0 if and only if a 2 -b 2 n=0 if and only if a=b=0, that is a =0.
(since n ¹ a 2 / b 2 ).
2. N( a b )= a b a b = a a b b =N( a )N( b )
3. N(1)=1 is clear .
c) To show that Z( n ) is an integral domain is similar as in part (a).
1. From part a) a in Z( n ) is invertible if and only if , there exists b in
Z( n ) such that a b =1, then by by part (b) 2 and 3, N( a )N( b )=1, since
N( a ) and N( b ) are both integers, then the only solution of the last
equation is N( a )= ±1.
2. Suppose that N( a )=±p, where p is a prime number, the N( a )≠ 0, ±1,
that is, a is neither 0 nor invertible. If a = b g is a factorization of a in
Z( n ) , then N( b )N( g )=N( a )=±p from which it follows that one of
N( b ) or N( g ) must have the value ±1. By 1. b or g is invertible in Z( n )
n associate of a . Accordingly, a is an irreducible element of Z( n ).