Question

Let a belong to a ring R with unity and suppose that a^n=0 for some positive integer n. (Such an element is called nilpotent). Prove that 1-a has a multiplicative inverse in R.

Answer 1

consider \[(1-a)(1+a+a^2+...+a^{n-1})\]

Answer 2

not sure if you can read the latex or not, let me know

Answer 3

can you read what i wrote?

Answer 4

(1-a)(1+a+a^2+a^3+...+a^(n-1))=1-a^n

Answer 5

and since a^n = 0, then you get 1

Answer 6

so is 1 the multiplicative inverse? It was that simple? huh, I thought it would be super hard!

Answer 7

no no not 1

Answer 8

lets go slow

Answer 9

you are trying to show 1 - a has a multiplicative inverse, if a^n = 0

Answer 10

i.e. you are trying to solve (1-a)x = 1 for x

Answer 11

then
(1-a)(1+a+a^2+a^3+...+a^(n-1))=1-a^n = 1-0 = 1

Answer 12

so the inverse of 1 - a is 1+a+a^2+a^3+...+a^(n-1)

Answer 13

OOOHhh ok!

Answer 14

good

Answer 15

Thank you! That's all I have to do then? Just explain that 1+a+a^2.....+a^n-1 is the inverse?