Abstract Algebra 19.15 (15) Define a ring D in a manner similar to how one defin
ID: 3184244 • Letter: A
Question
Abstract Algebra 19.15 (15) Define a ring D in a manner similar to how one defines the ring of complex numbers, as follows: D = {a+be : a, b E R), where E2-0 (instead of =-1). Elements of D are called dual numbers. The plane of all dual numbers is an "alternative complex plane" (a) Show that D is a commutative ring under addition and multiplication given by (a+be) + (e+de) = (a + c) + (D+d)- and (a + be) (c + de) = ac + (ad + bc)e, respectively. (b) Show that D is not an integral domain. (c) Show that a + be is a unit of D if and only if a 0.Explanation / Answer
a) Let us check all the properties of a commutative ring one by one-
i) D must be an abelian group under the given addition:
. Let (a+b) , (c+d) and (e+f) be three elements of D then
(a+b) + [(c+d) + (e+f)] = (a+b) + [(c+e) + (d+f)] = (a+c+e) + (b+d+f) = [(a+c) + (b+d)] + (e+f) =
[(a+b) + (c+d)] + (e+f)
Hence given addition is associative.
. Also we can check that (0+0) is additive identity and for (a+b), (-a-b) is additive identity.
also this addition is commutative. Hence D is abelian group under addition as defined on D
ii) Similarly we can check easily that D is a monoid under multiplication and Multiplication is distributive with respect to addition.
Hence D is a ring with the given operations.
Now (a+b).(c+d) = ac + (ad+bc) = ca + (bc+ad) = ca + (cb+da) = (c+d).(a+b)
Thus D is commutative ring.
b) Now D does not have multiplicative identity for all elements so D is not integral domain.
c) Suppose (a+b) is unit of D then-
(a+b).(c+d) = ac + (ad+bc) = 1 hence ac = 1, ad+bc = 0 it gives 'a' should not be zero otherwise (a+b) will not be unit. similarly we can show otherway round.