# Commutative Ring

• Nov 19th 2008, 11:50 AM
pkr
Commutative Ring
• Nov 19th 2008, 05:41 PM
ThePerfectHacker
We will first prove that if $\displaystyle b^2 = 0$ then $\displaystyle 1+b$ is a unit. This is simple, $\displaystyle (1+b)(1-b) = 1 - b^2 = 1$. Thus, $\displaystyle 1+b,1-b$ are inverses of eachother. Say that $\displaystyle a$ is a unit. Since $\displaystyle b^2 = 0$ it means $\displaystyle a^{-2}b^2 = 0$ and so $\displaystyle (a^{-1}b)^2 = 0$ since the ring is commutative. Using the above result it means $\displaystyle 1 \pm a^{-1}b$ is a unit. But then $\displaystyle a(1\pm a^{-1}b) = a\pm b$ is a unit.
• Nov 20th 2008, 03:51 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, 11:39 AM
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, 11:51 AM
pkr
Many thanks, all sorted now!