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Integral Domains and Fields
A. The Set M(2Z) and Integral Domains
For a set, like M(2Z), of 2x2 matrices with even integer entries to be an integral domain it
must be a commutative ring with unity that has no zero divisors.
1. Counterexample
With this knowledge, one must examine the set to see if the ring is commutative, which is
that a ring has the commutative property of multiplication. Consider the following matrices:
A= [ 68 62 ] and [ 44 62 ]
Matrices A and B ϵ M(2Z), where matrix addition and multiplication are well defined.
A ∙ B= [ 68 62 ] ∙ [ 44 62 ]
Using matrix multiplication with the set of even integers:
A ∙ B=
[ ( 8 ∙ 4 )+( 2∙ 4 ) ( 8 ∙ 6 )+( 2∙ 2)
( 6 ∙ 4 )+( 6 ∙ 4 ) ( 6 ∙ 6 )+( 6 ∙ 2) ]
A ∙ B=
[ ( 32)+( 8 ) ( 48 )+( 4 )
( 24 )+( 24 ) ( 36 )+(12) ]
A ∙ B= [ 4048 5248 ]
Now, one will reverse the order of Matrices A & B and multiply again using matrix
multiplication:
B ∙ A= [ 44 62 ] ∙ [ 68 62 ]
B ∙ A=
[ ( 4 ∙ 8 )+( 6 ∙ 6 ) ( 4 ∙ 2)+( 6 ∙ 6 )
( 4 ∙ 8 )+( 2∙ 6 ) ( 4 ∙ 2)+( 2∙ 6 ) ]
, 2
B ∙ A=
[ ( 32)+( 36 ) ( 8 )+( 36 )
( 32)+(12) ( 8 )+(12) ]
B ∙ A= [ 6844 4420 ]
The Matrices A ∙ B and B ∙ A are not the same product. Therefore, the set M(2Z) does not
have the commutative property of multiplication and is not a commutative ring. If the set M(2Z)
is not a commutative ring, it cannot be an integral domain.
B. The Set Z m and Integral Domain
Given the set of integers mod m denoted Zm, the elements of Zm are denoted [x]m,
where x is an integer from 0 to m – 1. Each element [x]m is an equivalence class of integers that
has the same integer remainder as x when divided by m.
In this set, let m = 12. This defines the set as follows:
Z12 = {[0]12, [1]12, [2]12, [3]12, [4]12, [5]12, [6]12, [7]12, [8]12, [9]12, [10]12, [11]12}.
1. Counterexample
To be an integral domain, the set of integers mod m must be a commutative ring with
unity that has no zero divisors as stated in the previous section. With the zero-divisor property in
mind, a ring will have zero divisors if two nonzero elements multiply to a zero element ( [ 0 ]12 ).
Consider the elements of Z12, specifically [2]12 and [6]12.
Multiply these two elements using modulo multiplication of the set of integers in the ring
Z12:
[ 2 ]12 ∙ [ 6 ]12= [ 0 ]12
[ 2∙ 6 ]12= [ 0 ]12
[ 12 ]12= [ 0 ]12
[ 12 ]12 is equal to the principal representative [ 0 ]12.
Integral Domains and Fields
A. The Set M(2Z) and Integral Domains
For a set, like M(2Z), of 2x2 matrices with even integer entries to be an integral domain it
must be a commutative ring with unity that has no zero divisors.
1. Counterexample
With this knowledge, one must examine the set to see if the ring is commutative, which is
that a ring has the commutative property of multiplication. Consider the following matrices:
A= [ 68 62 ] and [ 44 62 ]
Matrices A and B ϵ M(2Z), where matrix addition and multiplication are well defined.
A ∙ B= [ 68 62 ] ∙ [ 44 62 ]
Using matrix multiplication with the set of even integers:
A ∙ B=
[ ( 8 ∙ 4 )+( 2∙ 4 ) ( 8 ∙ 6 )+( 2∙ 2)
( 6 ∙ 4 )+( 6 ∙ 4 ) ( 6 ∙ 6 )+( 6 ∙ 2) ]
A ∙ B=
[ ( 32)+( 8 ) ( 48 )+( 4 )
( 24 )+( 24 ) ( 36 )+(12) ]
A ∙ B= [ 4048 5248 ]
Now, one will reverse the order of Matrices A & B and multiply again using matrix
multiplication:
B ∙ A= [ 44 62 ] ∙ [ 68 62 ]
B ∙ A=
[ ( 4 ∙ 8 )+( 6 ∙ 6 ) ( 4 ∙ 2)+( 6 ∙ 6 )
( 4 ∙ 8 )+( 2∙ 6 ) ( 4 ∙ 2)+( 2∙ 6 ) ]
, 2
B ∙ A=
[ ( 32)+( 36 ) ( 8 )+( 36 )
( 32)+(12) ( 8 )+(12) ]
B ∙ A= [ 6844 4420 ]
The Matrices A ∙ B and B ∙ A are not the same product. Therefore, the set M(2Z) does not
have the commutative property of multiplication and is not a commutative ring. If the set M(2Z)
is not a commutative ring, it cannot be an integral domain.
B. The Set Z m and Integral Domain
Given the set of integers mod m denoted Zm, the elements of Zm are denoted [x]m,
where x is an integer from 0 to m – 1. Each element [x]m is an equivalence class of integers that
has the same integer remainder as x when divided by m.
In this set, let m = 12. This defines the set as follows:
Z12 = {[0]12, [1]12, [2]12, [3]12, [4]12, [5]12, [6]12, [7]12, [8]12, [9]12, [10]12, [11]12}.
1. Counterexample
To be an integral domain, the set of integers mod m must be a commutative ring with
unity that has no zero divisors as stated in the previous section. With the zero-divisor property in
mind, a ring will have zero divisors if two nonzero elements multiply to a zero element ( [ 0 ]12 ).
Consider the elements of Z12, specifically [2]12 and [6]12.
Multiply these two elements using modulo multiplication of the set of integers in the ring
Z12:
[ 2 ]12 ∙ [ 6 ]12= [ 0 ]12
[ 2∙ 6 ]12= [ 0 ]12
[ 12 ]12= [ 0 ]12
[ 12 ]12 is equal to the principal representative [ 0 ]12.
