Basic Elements Of Abstract Algebra Solutions (Part 3)

1)  Find all ideals I of the integer class Z 12    and in each case compute:

 

Z 12  / I  

 

I.e. find a known ring to which the quotient ring is isomorphic

 

Solutions:

I ₁ =  [0] ,   Z 12  / I ₁    ~    Z 12  

I ₂ =  {0, 2, 4, 6, 8, 10;  Z 12  / I ₂ }  ~    Z 2  

 

I ₃ =  {0, 3, 6, 9; 10,  Z 12  / I ₃ }  ~   Z ₃  

 

I ₄ =  {0, 4, 8;   Z 12  / I ₄ }  ~   Z ₄  

 

I ₅ =  {0, 6;   Z 12  / I ₅ }  ~   Z ₅  

 

I ₆ =  Z ₁₂ ,  Z ₁₂ / Z ₁₂ }  ~    0 

 

2) Find a subring of the ring Z + Z  which is not an ideal of Z + Z

 

Solution:

 

{n, n   |  n   Î Z}

3) Give all units in each of the following rings: 

 a) Z   b) Z + Z   c)  Z ₅    d) Q

Solutions:  

 a) 1, -1   

 b) (1,1), (1, -1), (-1, 1), (-1, -1)   

c) 1, 2, 3, 4  

d) all non zero q  Î Q  

4) Let S be a commutative ring with I an ideal a  Î S, with (a) = {ja : j Î S},  

Prove that S is closed under +   and ·   

Soln. 

We assume x Î (a),   y Î S,   

Then x = ja for some j Î S,   So:  y · x = y(ja)=  (yj) a Î a  And we can write the following formulation:  If x,y  Î (a) then x + y  Î (a),  

 Also: x = ja for some j Î S,  then x - y   Î (a)  and further, y = j' a for some j' Î S then x· y  Î (a)   

 Then for (+) closure: x + y = ja + j'a = (j + j') a  Î (a)   

 For (· ) closure:  x · y = ja  ·  j'a  =  (j - j') a Î (a)  

And for subtraction: x - y =   ja - j'a = (j - j') a Î (a)     

       5)  Take S as the set of integers, Z. Let the ideal I = (2) so that S / I =  Z ₂     Thence or otherwise, find:

a) [0]     b)   [1]     c) S/ I  =  Z ₅ 

    Solns.

      Take S =    Z.  I = (2)  so that S / I =  Z ₂    

        We have:

a) [0] =  {0, + 2, + 4, + 6…..}  =  I = (2)    

        

       b) [1]  =   {1, 3, -1, 5, -3, 7….}   

           c)If S / I  =  Z ₅         

             S/I =  {[0], [1], [2], [3], [4]}