Suppose the inverse of $ 2x-1 $ is $ 2x-1 $, then
$ (2x-1)(2x-1)=1 $
$ 4x^2+2x+2x+1=1 $
$ 4x^2+4x+1=1 $, but in $ Z_4[x] $, 4=0. so,
$ 0x^2+0x+1=1 $
$ 1=1 $
Therefore, $ 2x-1 $ has an inverse in $ Z_4[x] $ and specifically, that inverse is $ 2x-1 $
Did you mean to put 2x+1? -Sarah
