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


Yea, he or she did mean that. Look at the line:

$ 4x^2+2x+2x+1=1 $ from that you can see he or she multiplied $ (2x+1)(2x+1)=1 $

or it would look like $ 4x^2-2x-2x+1=1 $


This helped a lot! Thanks...Neely


I believe that (-2x+1) is the inverse, if we first assume (2x-1) is the inverse, we'll get $ (2x-1)(2x+1) = 4x^2 - 1 $

and since 4 = 0, we'll get the above equation to equal -1 mod 4. So, -(2x-1) = (-2x+1) is the inverse.

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