North Maharashtra University 2008 B.Sc Mathematics S.YBSc MTH – 212 (B) (Computational Algebra) - Question Paper

a) both field and integral domain
b) an integral domain but not a field
c) a field but not an integral domain
d) neither a field nor an integral domain
4) In (Z9 , +9 , ×9) , six is - - -
a) a zero divisor b) an invertible element
c) a zero element d) an identity element
5) Every Boolean ring is - - -
a) an integral domain b) a field
c) a commutative ring d) a division ring
6) If a is a unit in a ring R then a is - - -
a) a zero divisor b) an identity element
c) a zero element d) an invertible element
7) If R is a Boolean ring and a ? R then - - -
a) a + a = a b) a2 = 0 c) a2 = one d) a + a = 0
8) Value of ( seven )2 – seven in (Z8 , +8 , ×8) is - - -
a) six b) four c) two d) 0
3 : ques. of six marks
1a) describe i) a ring ii) an integral domain iii) a division ring.
b) Show that the set R = {0, 2, 4, 6} is a commutative ring under
addition and multiplication modulo 8.
2a) describe i) a commutative ring ii) a field iii) a skew field.
b) In 2Z, the set of even integers, we describe a + b = usual addition
of a and b and a ?? b =
2
ab . Show that (2Z , + , ??) is a ring.
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3 a) describe i) a ring with identity element ii) an unit element iii) a
Boolean ring.
b) Let (2Z , +) be an abelian group of even integers under usual
addition. Show that (2Z , + , ??) is a commutative ring with
identity 2, where a ?? b =
2
ab , for all a , b ? 2Z.
4) a) describe i) a zero divisor ii) an invertible element iii) a field.
b) Let (3Z , +) be an abelian group under usual addition where 3Z
= {3n ?n ? Z}. Show that (3Z , + , ??) is a commutative ring
with identity 3, where a ?? b =
3
ab , for all a , b ? 3Z.
5) a) Let (R, +, ??) be a ring and a, b, c ? R. Prove that
i) a ?? 0 = 0 ii) (a – b)c = ac – bc.
b) Show that (Z , ? , ??)is a ring, where a ? b = a + b – one and a
?? b = a + b – ab , for all a , b ? Z.
6) a) Let (R, +, ??) be a ring and a, b, c ? R. Prove that
i) a ?? (–b) = –(ab) ii) a (b – c)c = ab – ac.

Earning:   Approval pending.