Question #226101
Suppose that a and b belong to a commutative ring R with unity. If a is a unit of R and b^2 = 0, show that a + b is a unit of R.
1
Expert's answer
2021-08-16T17:03:38-0400

We look at this problem from two perspective.


Case 1 (b=0)

If b=0 then a+b = a. And since a is a unit of R, we can conclude that a+b=a is a unit of R.


Case 2 (b0)b \neq 0)

Since a is a unit, let c be the inverse of a.

In order to show that a+b is a unit, it suffices to show that 1+bc is a unit.

Given that b2=0b^2=0 , we have that b3=0b^3 = 0 because b3=b2b=0(b)=0b^3=b^2b=0(b)=0

Also, (bc)3=0(bc)^3 = 0

But

1=1+(bc)3=(1+bc)(1bc+(bc)2)1= 1+ (bc)^3 = (1+bc)(1-bc+(bc)^2)

That is 1+bc is a unit. Hence, a+b is a unit.


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Comments

Rohit Mete
18.08.21, 03:24

Thank you

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