Showing that a polynomial is a unit in a quotientring In this exercise I have a polynomial ring over a finite field $F_2$:$(F_2[X],+,*)$ There is then given an ideal : I $=<X^3+X+1>$.
I am then trying to show that $X^2+X+1 + I$ is a unit in $F_2[X]/I$ by using the extended Euclidean algorithm on $X^3+X+1$ and $X^2+X+1$.
I tried using this algorithm and then got that  $(X^3+X+1)*r(X)+(X^2+X+1)*s(X)=gcd(X^3+X+1,X^2+X+1) => (X^3+X+1)*(X+1)+(X^2+X+1)*X^2=1$
I can't really figure out if this is right, and if it is what I am suppose to do from here to show that the given expression is an unit.  

What you have is correct: Since $+1 = -1$, you can write it as
$$
(x^3 + x + 1)(x+1) - (x^2 + x + 1)x^2 = x^4 + x^3 + x^2 + 2x + 1 - x^4 - x^3 - x^2 = 1
$$
since $2x=0$
When you apply the quotient map $\pi : F_2[x] \to F_2[x]/(x^3+x+1)$ to this expression, you get
$$
\pi(x^2+x+1)\pi(x^2) = \pi(1)
$$
Hence, $\pi(x^2+x+1)$ is invertible in the quotient.
