## Friday, February 15, 2019

### 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:    Z } )

The congruence classes are:

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

  =  {0 + 5 j: j    Z}  =  { 5j: j   Z }  =  (5)

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

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

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

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

  =  {5  + 5j: : j    Z}   =   5 (j+ 1) : j    Z}   =   = 

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 SI =  Z 2     Thence or otherwise, find:

a (a)          b)        c) S/ I  =  Z 5

Solutions:

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

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

b)   =   {1, 3, -1, 5, -3, 7….}

c)If S /=  Z 5

S/I =  {, , , , }

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.