http://i453.photobucket.com/albums/q...188/math10.jpg

Help would be appreciated.

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- Nov 19th 2008, 11:50 AMpkrCommutative Ring
http://i453.photobucket.com/albums/q...188/math10.jpg

Help would be appreciated. - Nov 19th 2008, 05:41 PMThePerfectHacker
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 AMpkr
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 AMThePerfectHacker
- Nov 20th 2008, 11:51 AMpkr
Many thanks, all sorted now!