More Abstract Algebra: Looking Again At Rings, Fields And Ideals

  Ring (def.)

By a ring (S, +, · ) one means an abstract set  S  which has two binary operations defined on it. The operation + is called addition, and the operation · is called multiplication. Moreover, the set (S, +, ·) is required to satisfy the following axioms:

Let a, b and c Î  S

A1):  a + (b + c) = (a + b) + c

A2)  ∃  an element  0 Î  S with the property that 0 + a = a

A3)  For each element a Î  S there exists an element

         (-a)  Î  S  such that  a + (-a) = 0

A4)   a + b = b + a

A5)  a · (b · c)     = (a · b)  · c

N.B.  Any field is a ring.

ii)                Commutative Ring

By a commutative ring (S, + · ) one means a ring (S, + · )  which satisfies, in addition to the ring axioms the axiom:

A6)   For all  a, b  Î  S  then a · b  = b · a

And there also exist the properties:

a)Closure: If a,b Î S, then the sum a+b and the product a·b are uniquely defined and belong to S.

b)Associative laws: For all a,b,c  Î S,

a+(b+c) = (a+b)+c and a·(b·c) = (a·b)·c.

c)Commutative laws: For all a,b  Î S,    a+b = b+a and a·b = b·a. 

d)Distributive laws: For all a,b,c Î  S,

a·(b+c) = a·b + a·c and (a+b)·c = a·c + b·c

e)Additive identity: The set R contains an additive identity element, denoted by 0, such that for all     a Î  S,

a+0 = a and 0+a = a.

f)Additive inverses: For each a Î  S, the equations:

a+x = 0 and x+a = 0

have a solution x  Î  S, called the additive inverse of a, and denoted by  (-a).

The commutative ring S  is called a commutative ring with identity if it contains an element 1, assumed to be different from 0, such that for all a Î  S,

a·1 = a and 1·a = a.

In this case, 1 is called a multiplicative identity element

iii)Commutative Ring with unity

By a commutative ring with unity one means a commutative ring (S, + · ) which satisfies the axiom:

A7)  There exists an element 1  Î  S  with the property:

1  · a  = a for all a  Î  S 

iv)Integral domain

By an integral domain  ring (S, + · )  one means a commutative ring with unity which satisfies the following axiom:

If a, b  Î  S    and a · b  = 0 then either a = 0 or b = 0.

v)Field

By a field (S, + · )  one means a commutative ring with unity which satisfies the additional axiom:

A7) For every non-zero element a  Î  S    there exists an element a ^{- 1}  Î  S  such that:    a · a ^{- 1} =   1.  The element    a ^{- 1}   is called the reciprocal or multiplicative inverse of a.

vi)                    Ordered Field

An ordered field is a field (S, + · )  which contains a subset P (called the positive cone in P) which has the following properties:

P is a proper subset of S.

O.1) P is closed under addition.

O.2)  P is closed under multiplication

O.3) If  a  Î  S   then one and only one of the following three conditions must hold:

a = 0,    a  Î  P, or    - a  Î  P

vii) Order relation on an ordered field

Let   (F , + · )  be an ordered field and let P represent the positive cone in F. We define a relation ‘<’  on F, called ‘less than’ by the formula:

<   =    {(a, b) :   a, b  Î  F, b – a Î  P }

Or, equivalently:

a  < b if an only if   b – a Î  P

We also define the relation ‘greater than’ on F by

>  as follows:

s > b if and only if b < a

Definition. The field K is said to be an extension field of the field  (F + · )   if  (F + · )    is a subset of K which is a field under the operations of K.

Definition. Let K  be an extension field of F and let u  Î K. If there exists a nonzero polynomial f(x)  Î  F[x] such that f(u)=0, then u is said to be algebraic over F. If there does not exist such a polynomial, then u is said to be

transcendental over K.

Proposition. Let K be an extension field of F , and let u Î K be algebraic over F. Then there exists a unique irreducible polynomial p(x)  Î  F[x] such that p(u)=0. It is characterized as the monic polynomial of minimal degree that has u as a root.

Furthermore, if f(x) is any polynomial in F[x] with f(u)=0, then p(x) | f(x).

Definition. Let F   be a field and let f(x) = a₀ + a₁ x + · · · + a_n xⁿ be a polynomial in F [x] of degree n> 0. An extension field K   of F   is called a splitting field for f(x) over F if there exist elements r₁, r₂, . . . , r_n   Î  F such that

(i)                 f(x) = a_n (x-r₁) (x-r₂) · · · (x-r_n), and

(ii)               K  = F (r₁,r₂,...,r_n).

In the above situation we usually say that f(x) splits over the field F. The elements r₁, r₂, . . . , r_n are roots of f(x), and so K  is obtained by adjoining to F  a complete set of roots of f(x).

Definition: Subfield of a field.  Let (F + · )   be a field. A subset  T  ⊂    F is  called a subfield if  T is closed under the operations of   +  and  · , and T is a field under those operations.    

Definition: Ideal:   Let (S,  + · )    be a commutative ring, by an ideal I in S one means the following:

i)I is a subring of S,

ii)If  a   Î  I ,  b   Î  S then a, b    Î  I

Definition: A homomorphism φ:  R -> R' from one ring to another is a map which is compatible with the laws of composition and which carries 1 to 1, i.e. a map such that: φ(a + b) =    φ(a ) + φ (b)  and  φ(a b) =   φ(a) φ (b)

for all a,b Î  R.

An isomorphism of rings is a bijective homomorphism.  If there is an isomorphism R  -> R', the two rings are said to be isomorphic.

Definition: Congruence modulo I.

Let S be a commutative ring. Let I be an ideal in S. We write  a  ≡  b modulo I for the two elements a, b    Î  S if

a   Î [ b ]. Hence,  b   Î [ a ].

Definition:  S is a commutative ring and I is an ideal. Then let S/ I denote  a quotient ring and the set of congruence classes modulo I. 

Perhaps the most critical aspect of ideals to note is that: “an ideal is to a ring as a normal subgroup is to a group”.  Also, in a similar manner to groups, one can have left and right ideals.

Let   a   Î  S,   a commutative ring. 

Then the set (a) defined by: (a) = {ax: ,  x Î  S }  is an ideal.

Now, let S be a commutative ring and I be an ideal in S. For each element a  Î  S let [a] be the subset of S defined by:

[a]  = I  + a = {a + j:  j   Î  S}

Further, let s be a subring of a ring S satisfying:

[a]  s  ⊆     s   and s[a]  ⊆   s  for all  a   Î  S is an ideal (or two-sided ideal of S). Also, a subring s of S satisfying [a]  s  ⊆     s    for all  a   Î  S is a left ideal of S. One satisfying  s [a] ⊆     s    for all  a   Î  S is a right ideal of S.

Example Problems:

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)  Is Q [x] /x²  - 5 x  + 6 a field? Why?   

Soln.   No.  x²  - 5 x  + 6   is not a maximum ideal of  Q [x]  since  x²  - 5 x  + 6  

= (x  -  3) (x -2) is not irreducible over Q

Problems for Math Mavens:

1)      Let Z be the integers. Then prove for each class:   

[a] mod n there exists an integer  r   Î  Z, for:         0 < r < n,   r   Î  [n]

2)     Let Z be the integers.  The ideal: 

           (5)  =   {5 j:  j Î  Z }

        Show all the congruence classes with respect to this ideal.    

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

3)      Let S be a commutative ring, and let I be an ideal in S. If A, B are subset of S, then use set notation to define: i) A ·  B,   A + 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)     [0]     b)   [1]     c) S/ I  =  Z ₅       

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

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

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)