I am trying to find the inverse of $x-1$ in $F=Z_{5} / (f(x))$
Where $f(x) = x^3 + x + 1$
So performing the EEA
$1 = ((x-1) - (3)(2x-2))$
$1 = ((x-1) - ((x^3 + x + 1) - (x - 1)(x^2 + x + 2)(2x-2))$
$1=((x-1)(1-(x^2+x+2)(2x-2))-(x^3+x+1)(2x-2)$
$1=(x-1)(3x^3+x+2)-(x^3+x+1)(2x-2)$
so the inverse of $(x-1)$ is $(3x^3+x+2)$ but when I multiply the two together I don't get $1$
Could someone help me out? i've performed the EEA multiple times and am still getting the same answer..
WA gives this over $\mathbb Z$, which you can easily check: $$ 3 = (-x^2 - x - 2)(x-1)+(1)(x^3+x+1) $$ In $F$, dividing by $3$ is multiplying by $2$, and so in $F$ we get this: $$ 1 = 2(-x^2 - x - 2)(x-1) $$ Therefore, the inverse of $x-1$ in $F$ is $2(-x^2 - x - 2)=3x^2+3x+1$.