Solutions To Abstract Algebra Problems

 (1) 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.

6)  Is Q [x] /x²  - 6 x  + 6 a field? Why or why not? 

Ans.   YES.  x²  - 6 x  + 6   is a maximal ideal of    Q [x] 

7)Let (G) and (H) be groups. Then a homomorphism of (G) into (H) is a map of the sets G and H which has the following property:  f(x o y) = f(x) o f(y)

 Let G = (0, 1, 2, 3) for the operation (+) which is addition in Z₄

Let H = (2, 4, 6, 8) for the operation (·) which is multiplication in Z₁₀

Prepare the respective tables for the isomorphism and give specific examples in terms of the function φ, i.e. show specific mappings.  (Where: φ(x) φ(y) = φ(xy) for example)

Soln.

The addition table for Z4 is:

+ /----0 ----1 ---2 ----3
-----------------------
0 --- 0------1 ---2 ----3

1 -----1 -----2 ---3---- 0

2 -----2 -----3--- 0 ----1

3----- 3 -----0 ---1---- 2
-----------------------------


The multiplication table in Z₁₀


-·--/----6 ----2 -----4 ----8

---------------------------
6 --- ---6------2 --- 4 ----8

2 ------2 ------4 ----8----6

4 ------4 ------8--- -6 ---2

8----- 8 -----6 ----2---- -4
-----------------------------

Since 6 · 6 = 6 then the identity element in  Z₁₀  is 6.. 

So φ(x) φ(x) = φ(x²) = (φ6)φ(6) =φ( I)

Then the mapping isomorphisms between the tables will be:


Φ: 0 -> 6 (identity to identity element)

Φ: 2 -> 4

Φ: 3 -> 8

Φ: 1 -> 2


So each of the numbers on the left side of the arrow is mapped to the corresponding number on the right side, and this is done using the two tables such that each element of Z₄  is mapped to each element of Z₁₀