Division ring or Skew field
(Monday, May 3, 2010)
Division ring or Skew field:
A ring with at least two elements is called a division ring or skew field, if it is a ring with unity element and is such that every non-zero element is invertible.
Thus, a skew-field is additionally a group, and the set of all non-zero elements of it forms a multiplicative non-commutative group.
Field:
A commutative division ring is called a field.
Theorem 1:
A commutative ring R is an integral domain, iff, for any non-zero element
a e R, a . u = a . v Þ u = v, u, v e R
Proof:
Let R be a commutative ring and
a . u = a . v Þ u = v where a (¹0) e R and u, v e R.
Then, for a¹0, a . b = a . 0 Þ b = 0 (taking u=b and v=0). Hence, R has no zero divisor and consequently R is an integral domain.
Conversely, let R be an integral domain, and let a(¹0) e R, u, v e R. Now, a . u = a . v Þ a . (u - v) = 0. Since, R is an integral domain and a ¹ 0, a cannot be a zero divisor.
\ u - v = 0, i. e., u = v
Hence, a . u = a . v Þ u = v. where a¹0.
Theorem 2:
A division ring does not contain any zero divisor.
Proof:
Let R be a division ring and let a, b R such that a . b = 0 and a ¹ 0. Since, in a division ring every non-zero element is invertible, a-1 exists and a-1 e R, where a . a-1 = a-1. a = 1.
\b = 1. b = (a-1 . a) . b = a-1(a.b) = a-1 . 0=0
\Thus a ¹ 0, a.b=0 Þ b=0.
Hence, R has no zero divisor.
Theorem 2:
A ring R is a division ring iff, the equations a . x = b and y . a = b have unique solutions in R where (a ¹ 0),
b e R.
Proof:
Let R be a division ring. As the non-zero elements or R are invertible, R is multiplicatively a group.
Therefore, the equations a . x = b and
y . a = b have unique solutions in R, provided b¹0, if y=0, since R has no zero divisors.
Conversely, let R be a ring and the equations a . x = b and y . a = b have unique solutions in, where a(¹0),b e R.
Then, if b=a, we have a . x = a and
y . a = a. Now, a.x2 =a.(x.x)=(a.x).x=a.x.
Hence, x2=x, i.e., x=1 (x¹0). Like wise, we can prove that y=1. Hence, R has the unity element 1. Further, a.x=1 and y.a=1 implies that a-1 e R, where a(¹0) e R. Hence, R is a division ring.
Field:
A set F (having at least two elements) equipped with two binary operations viz., addition and multiplication is said to form a field, if the following axioms are satisfied.
Properties due to addition (+):
F1:Closure law: For any two elements
a, b e F Þ a + b e R.
F2:Associative law:
For any three elements
a, b, c e F Þ a + (b + c) = (a + b) + c
F3: Existence of additive identity element:
The element 0 e F such that for all
a e F, a + 0 = 0 + a = a, 0 being the additive identity element of F
F4: Existence of additive inverse element:
The element a e F the element –a e F such that a +(-a ) = (-a) + a = 0, -a being the additive inverse element of a in F.
F5: Commutative law:
For all a, b e F, a+ b = b + a.
Properties due to multiplication (.):
F6:Closure law:
For any a, b, e F Þ a . b e F
F7: Associative law:
For any three elements
a, b, c e F Þ a . (b . c) = (a . b) . c
F8: Existence of multiplicative identity element:
The element 1 e F such that for all
a e F, a . 1 = 1 . a = a, 1 being the multiplicative identity element of F.
F9: Existence of multiplicative inverse element:
For all a e F, a ¹ 0 Þ a-1 e F such that
a.a-1 =a-1.a = 1. Thus, a-1 is called the inverse of a and vice versa.
F10: Commutative law:
For all a, b e F, a . b = b . a =
Properties due to addition and multiplication
F11: Distributive law:
For all a, b, c e F
a.(b+c) = a.c=b.c (Left cancellation)
(a+b).c = a.c+b.c (Right cancellation)
In other words, the algebraic structure áF, +, .ñ is called a field, if
(i) F is additive Abelian group
(ii) F' is a multiplicative Abelian group, where F' = F-{0}, and
(iii) Distributive law hold in F.
Recalling to the definition of an integral domain, we can easily see that an integral domain F is a field, if every non-zero element a e F have a multiplicative inverse a-1 such that
a.a-1=a-1.a=1. One can also say that a ring is a filed, if it forms an abelian group under both addition and multiplication with distributive laws satisfied in it.
As all non-zero elements in the ring of integers do not possess multiplicative inverse the ring of integers is not a field.
Sub-rings:
If the algebraic structure áR, +, .ñ be a ring and SÌR such that áS, +, .ñ is also a ring, then S is called a sub-ring of R.
Example 1: The set of integers is a sub-ring of the set of rational numbers.
Example 2: The ring of integers is a sub-ring of the field of real numbers.
Sub-fields:
If the algebraic structure áF, +, .ñ be a ring and SÌF such that áS, +, .ñ is a field, then S is called a sub-field of F.
Example 1: The field of rational numbers is a sub-field of the field of real numbers.
Example 1:
Let Q be the set of rational numbers, then show that áQ, +, .ñ is a ring. Is it an integral domain? Is it a field?
Solution: Under addition (+)
(i) Closure law: For any two rational number a, b e Q a + b e Q
(ii) Associative law: For any three rational numbers a, b, c e Q,
Þ a + (b + c)=(a + b) + c
(iii) Existence of additive identity element:
The rational number 0 is such that a+0=0+a=a for all aeQ. The number 0 is the additive identity elment in Q.
(iv) Existence of additive inverse element : For every rational number a, $ an element -a e Q, ' a+(-a) = (-a) + a =0. The number -a is called the additive inverse of a. Since 0+0 =0, therefore, 0 is the negative of itself.
(v) Commutative law:
For all a, b e Q, a+b=b+a e Q. Thus
áQ, +) is an Abelian group.
Under multiplication (.)
(vi) Closure law: For any two rational number a, b e Q Þa . b e Q
(vii) Associative law: For any three rational numbers a, b, c e Q,
Þ a . (b . c)=(a . b) . c
Under addition(+) and multiplication(.):
(viii) Distributive law: For any three rational number a, b, c e Q
Þa . (b + c) = a.b + a.c and
(a + b) . c = a.c + b.c
Thus áQ, +, .ñ is a ring.
(ix) Commutative law: For any two rational numbers a, b, e Q, Þ a . b = b . a
Hence, multiplication is commutative in Q.
\ áQ, +, .ñ is a commutative ring.
(x) Existence of multiplicative identity element: The rational number 1 such that
1.a=a.1=a for any a e Q. The element 1 is the multiplicative identity element in Q.
Thus áQ, +, .ñ is a commutative ring with a unity element.
(xi) Integral domain: For any two rational numbers a, b, e Q, Þ a . b = 0 Þ either a=0 or b=0. Hence, Q is a commutative ring without any zero divisor.
\ áQ, +, .ñ is also an integral domain.
(xii) Existence of multiplicative inverse element: If a be a non-zero rational number, then a-1 (or 1/a) is also a rational number
\ a.a-1 = a-1.a=1
\ a-1 is the multiplicative inverse of a (¹0).
By the virtue of the condition (i) to (x) and (xii) the algebraic structure áQ, +, .ñ is also a field.
Example 2:
Let C be the set of complex numbers, of the form a+ib, where a, b are real numbers,then show that áC, +, .ñ is a filed.
Solution: We use the notation of ordered pair of real numbers and unit a+ib=(a, b)
Under addition (+):
(i) Closure law:
C is closed under addition, since
(a, b) +(c, d) = (a+b, c+d) e C, where (a, b) e C and (c, d) e C. (Since, a, b, c, d e ÂÞ a+b, c+d e Â
(ii) Associative law: For a, b, c, d, e, f e Â, Þ a+c, b+d, a+c+e, b+d+f e Â
As
(a, b) +{(c, d) + (e, f)}=(a, b) +(c+e, d+f)
=(a+c+e, b+d+f) and {(a, b)+(c, d)}+(e, f) ={(a, c, d, f)}+(e, f)=(a+c+e, b+d+f)
\ Associative law under addition holds in C.
(iii) Existence of additive identity element:
For 0 e  and 0=0+i0=(0, 0), therefore, (0, 0) e C such that
(a, b) + (0, 0)=(a+0, b+0)=(a, b), where (a, b) e C for a, b e Â.
\ (0, 0) is the additive identity element in C
(iv) Existence of additive inverse element : If (a, b) e C, then (-a, -b) e C and
(a, b) + (-a, -b) = (a-a, b-b) = (0, 0).
So, (-a, -b) is the additive inverse of (a, b) in C.
(v) Commutative law:
For all (a, b) e C and (c, d) e C
Þ (a, b)+(c, d)
= (a+c, b+d)
= (c+a, b+d)
= (c, d) + (a, b)
\Addition is commutative in C
Note that áC, +, .ñ is an infinite Abelian group.
Under multiplication (.)
(vi) Closure law:
If a, b, c, d e  then (a, b) e C and (c, d) e C
Now, {(a, b).(c, d)} = (ac-bd, bc+ad)}, where ac-bd, bc+ad e Â
\(a, b).(c, d) e C is closed under multiplication.
(vii) Associative law: For any (a, c),
(b, c), (e, f) e C, we have
[ (a, b).(c, d)].(e, f)=(ac-bd, bc+ad) (e, f)
=[(ac-bd)e-(bc+ad)f, (bc+ad)e+(ac-bd)f]
= [a(ce-df)-b(de+cf), b(ce-df)+(de+cf)]
=(a, b){(c,d).(e,f)}
\Multiplication is associative in C
(viii) Existence of multiplicative identity element:
(1, 0) e C and is such that
(a, b).(1, 0) = (a, b) = (1, 0).(a, b) for
(a, b) eC
\ (1, ) is the multiplicative identity element in C
(ix) Existence of multiplicative inverse element:
Let (0,0) ¹(a, b) e C. Then there exists (c, d) e C such that
(a, b).(c, d) = (1, 0) = (c, d).(a, b), where
c=(a/(a2 +b2)) and d=(-b/(a2 + b2))
Further as (a, b) ¹ (0, 0), at least one of a and b is non-zero so that a2 + b2 ¹ 0.
So, c and d are define and at least one of c and d is non-zero.
\(c, d) i.e., ((a/(a2+b2), -b/(a2+b2)) is the inverse of (a, b) ¹ (0, 0).
(x) For (a, b), (c, d) e C, we have
(a, b).(c, d) = (ac-bd, bc+ad)
= (ca-db, cb+da)
= (c, d).(a, b)
\ Multiplication is commutative in C.
Under addition(+) and multiplication(.):
(viii) Distributive law: If (a, b), (c, d), (e, f) e C then
(a, b).{(c + d + (e + f)} = (a, b).(c + e, d + f)
= {a(c + e) - b(d + f), b(c + e) + a(d + f)}
={(ac-bd)+(ae-bf), (bc + ad) + (be + af)}
= (ac-bd, bc+ad) + (ae-bf, be+af)
= (a, b)/(c, d0 + (a, b).(e, f)
\ Left cancellation law holds in C.
Similarly, the right distributive law also holds in C, and is the same as the left distributive law, since multiplication in commutative in C.
Hence, áC, +, .ñ is a field.
Example:
Prove that the set S={0, 1, 2, 3} forms a ring under addition and multiplication modulo 4, but not a field.
Solution
We construct the tables below for the composition of addition modulo 4 and multiplication modulo 4.
From the table we observe that,
(i) The set S is closed under the operations of addition and multiplication modulo 4.
(ii) The operations are also associative and commutative
(iii) 0 is the additive identity and 1 is the multiplicative identity.
(iv) Each element of S has an additive inverse and
(v) The distributive laws holds in S.
Hence, áS, +, .ñ is a commutative ring under addition and multiplication modulo 4.
Further, the ring contains unity elements, but not every non-zero element has an inverse,
since we observe that 2.2=0. So, the ring has zero divisor. Thus, S is a commutative ring but not a division ring, and so it is not a field.
Theorem:
A field is necessarily an integral domain.
Proof:
Let F be a field. We are to prove that F has no zero divisor, i.e., for a, b eF, a.b=0 Þa=0 or b=0.
If a¹0, when a-1 exists and we have
a-1 .(a.b)=(a-1.a).b=1.b=b.
Since, a.b=0, therefore b=a-1.0=0.
Similarly, we can show that a=0, if b¹0.
Hence, the result follows
Note: Converse of this theorem is not true. For example, the ring of all integers is an integral domain but not a field.
Theorem:
A finite integral domain is a field.
Proof:
Let F be a finite integral domain with n distinct elements. F being an integral domain, multiplication is commutative in F and it admits of no zero-divisors.
We shall now show that every non-zero element of F possess a multiplicative inverse.
Let a(¹0) e F. Then, F being an integral domain, a.x=a.y Þx=y for x, y e F.
But, x ¹y, since F contains n distinct elements.
\ a.x ¹a.y, a¹0.
\ a.x ¹a.y, a¹0.
Hence, there exists an element x(¹0) eF such that a.x=1, for every a(¹0) eF.
Hence, the result follows.
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