Let R a b ab R where a b c d a c b d and a bc d

Let R[] = {a + b : a,b R}, where (a + b) + (c + d) = (a + c) + (b + d)
and (a + b)(c + d) = ac + (ad + bc). These operations dene a commutative ring R[] called the ring of dual numbers. One writes a + 0 = a and 0 + b = b for all a,b R. The dual numbers are analogous to the complex numbers, but instead of i^2 = 1, one has ^2 = 0. One can think of as a nonzero “innitesimal” whose square is zero.

(a) Show that a + b is a zerodivisor of R[] if and only if a = 0.

(b) Show that (a + b)(ab) = a^2. Use this to show that a + b is a unit of R if and only if a does not equal 0.

(c) By part (b) one has R[] = {a + b R[] : a 6= 0}. Prove that the map g : R[] R×R given by   g(a + b) = (a, b/a) is a group isomorphism.

Solution

what is zero divisor .

zero divisor of an element is another element which makes that element zero.

so if a is zero

a+ be (c+ de) = ac + (ad + bc) e = (bc)e and this will be zero if and only if c is also zero

hence both the elements must be of type xe if their product is zero

hence if and only if condition is satisfied

b) (a +be) (a-be) = a.a + (ab-ab) e = a^2

if a is zero then a+be is zero divisor so it cannot be a unit.

if a is not zero we have 1/a defined

and (a+be) (1/a-be) = 1 .

hence (a+be) is a unit

c) R[e]* is agroup under multiplication as we saw that every element is a unit element in it.

g[(a+be)*(c+de)] = g(ac + (ad+bc)e) = (ac,(bc+ad)/ac) = (a, b/a) #(c,d/c) = g(a+be) # g(c + de)

this # is a group operation on R* XR

which is like if x1 is in R* , y1 is in R

and x2 is in R* , y2 is in R
(x1,y1) # (x2,y2) = (x1 y1 , x2+y2)

So its an homomorphism

Now we prove its one one and onto.

if g(a+be) = g(c+de)

then

(a,b/a) = (c,d/c)

a = c and b/a = d/c and hence b= d

so

a+be = c+ de

its one one

2 ) onto

take any element x in R* and y in R

and g(x +x y e) = (x, y)

hence it is onto

hence this function is onto