Solutions to Revisiting Rings, Fields, Ideals

 (1)      Proof:   By the Euclidean (division) algorithm there exist numbers q, r ∈  Z

Such that 0 < r < n,   and a is of the form:

a = q n + r

By definition of   [a ]  mod n  =   [a]

 r  ∈   [a] ,   And:   r = a + (-q) n

  (2) Let Z be the integers.  The ideal:

 

           (5)  =   {5 j:  j ∈  Z }

Show all the congruence classes with respect to this ideal.    

(Hint:  [a] =  {a + j: j ∈ [(5)}  = {a +  5j: j ∈   Z } )

The congruence classes are:

[a]   =  {a + j: j  ∈  (5)}  =  {a + 5j: j ∈  Z }

[0]  =  {0 + 5 j: j  ∈  Z}  =  { 5j: j ∈  Z }  =  (5)

[1] =  {1  + 5 j: j  ∈  Z}   =  {1, 6, -4, 11, -9}

[2]  =  {2  + 5 j: j  ∈  Z}   =  {2, 7, -3, 12, -8}

[3]  =  {3  + 5 j: j  ∈  Z}   =  {3, 8, -2, 13, -7}

[4]  =  {4  + 5 j: j  ∈  Z}   =  {4, 9, -1, 14, -6}

[5]  =  {5  + 5j: : j  ∈  Z}   =   5 (j+ 1) : j  ∈  Z}   =  [5] = [0]

3) Using set notation we may define:

A ·  B  =  {x ·  y: x ∈ A,  y ∈ B}

A +  B  =  =  {x +  y:  x ∈ A,  y ∈ B}

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

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

Solutions:  

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

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]}

5 ) Show every ring S has two ideals: S itself and {0}.

Every ring S always has at least two subrings, namely S and the zero ring,  S _0.

Further, if S is also a  field then the only ideals in S are S and {0}.  Further:

 if 1   ∈ I  then I = S. (Let I = S be an ideal,  I = {0} is also an ideal. )

 (In a field F, the only ideals are 0 and F)

Basically, an ideal of a ring S is an additive subgroup a  of  S with the property that:

a   ∈ S    and a  ∈  a    imply  ra  ∈  a  .  Clearly then the set containing the single element 0 and the set containing the whole ring S are ideals.