Find out if polynomial has inverse in quotient ring Find out if polynomial $x^3-x^2+x-1$ has inverse element in quotient ring ${\displaystyle \mathbb {Z} }_{11}/(x^4+3x^3-3x^2-4x-1)$, if yes find this inverse.
I know that $x^3-x^2+x-1$ has inverse element when $gcd(x^4+3x^3-3x^2-4x-1;x^3-x^2+x-1)$~$1$. I also know that I can use Euclidean algorithm. But I'm not sure about my result, I have got result that the g$cd$~$5$ it means that $x^3-x^2+x-1$ does not have inverse. Am I correct? 
Thanks for any help!      
 A: I did not check the computations, but if you got $5$, then that means that they are relatively prime and therefore that it has an inverse. That's so because asserting that $5$ is a greatest common divisor of two polynomials is the same thing as asserting that $1$ is a greatest common divisor of them. Note that, in $\mathbb{Z}_{11}$, $5\neq0$.
A: The extended Euclidean algorithm over $\mathbb Q$ gives:
$$
232 =  (49 x^3 + 168 x^2 - 75 x - 195)(x^3-x^2+x-1) +(-49 x^2 + 28 x - 37)(x^4+3x^3-3x^2-4x-1)
$$
Mod $11$, this becomes
$$
1 \equiv (5 x^3 + 3 x^2 - 9 x - 8)(x^3-x^2+x-1) \bmod (x^4+3x^3-3x^2-4x-1)
$$
and so the inverse is $5 x^3 + 3 x^2 - 9 x - 8 \equiv 5x^3+3x^2+2x+3$.
A: Since Jose already covered the fact that your approach is valid, I'll present a cool alternative approach. First, construct the companion matrix of $x^4+3x^3-3x^2-4x-1$, call it $A$.
Here $$A = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -3
\end{pmatrix}_.$$
Now consider the matrix, $B = A^3 - A^2 + A - I$. Then $B$ is invertible (recall that the coefficients of the matrix lie in $\mathbb{Z}_{11}$) iff $x^3 - x^2 + x - 1$ is invertible. Even stronger, if you express $B^{-1}$ as a linear combination of $I, A, A^2, A^3$, this tells you the inverse of $(x^3-x^2 + x -1)$!
More explicitly, if $B^{-1} = c_3A^3 + c_2A^2 + c_1A + c_0I$, then $$(x^3-x^2+x-1)^{-1} = c_3x^3+c_2x^2+c_1x+c_0.$$
