I got (1 - 2x) as the multiplicitave inverse, since that multiplied by (2x + 1) is 1. ... which i guess is the same as the original function. --Jcromer 20:50, 1 April 2009 (UTC)
The unity of $ \scriptstyle Z_4[x] $ is 1, so we're looking for some polynomial $ \scriptstyle g(x) $ such that $ \scriptstyle(2x+1)\cdot g(x)\ =\ 1 $. So, just divide 1 by $ \scriptstyle(2x+1) $ modulo 4:
$ \scriptstyle1+2x $
$ \scriptstyle1+2x)\overline{1+0x+0x^2} $
$ \scriptstyle-\underline{1+2x} $ $ \scriptstyle2x+0x^2 $ $ \scriptstyle-\underline{2x+0x^2} $ $ \scriptstyle0 $
From this it's clear that $ \scriptstyle1\ =\ (2x+1)\cdot(2x+1) $, so the multiplicative inverse of $ \scriptstyle(2x+1) $ is itself.
- --Nick Rupley 01:44, 2 April 2009 (UTC)
