# Commutative Ring

• Nov 19th 2008, 12:50 PM
pkr
Commutative Ring
• Nov 19th 2008, 06:41 PM
ThePerfectHacker
We will first prove that if $b^2 = 0$ then $1+b$ is a unit. This is simple, $(1+b)(1-b) = 1 - b^2 = 1$. Thus, $1+b,1-b$ are inverses of eachother. Say that $a$ is a unit. Since $b^2 = 0$ it means $a^{-2}b^2 = 0$ and so $(a^{-1}b)^2 = 0$ since the ring is commutative. Using the above result it means $1 \pm a^{-1}b$ is a unit. But then $a(1\pm a^{-1}b) = a\pm b$ is a unit.
• Nov 20th 2008, 04:51 AM
pkr
So for the 2X2 Matirx would you take B to be the matrix (0 1, 0 0) as that squared = (0 0, 0 0)

This is confusing me slightly, I haven't seen the notation of a "unit" in this context before...What exactly would the conditions be I need to satisfy for A-B to not be a unit?
• Nov 20th 2008, 12:39 PM
ThePerfectHacker
Quote:

Originally Posted by pkr
So for the 2X2 Matirx would you take B to be the matrix (0 1, 0 0) as that squared = (0 0, 0 0)

This is confusing me slightly, I haven't seen the notation of a "unit" in this context before...What exactly would the conditions be I need to satisfy for A-B to not be a unit?

For matrices the identity is the matrix I = (1 0, 0 1).
Therefore a matrix A is a unit if and only if there is B such that AB = BA = I.
• Nov 20th 2008, 12:51 PM
pkr
Many thanks, all sorted now!