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.  
 A: 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.
A: You have already found the inverse, it is: $X^2 + I$.
You know that:
$(X^3 + X + 1)(X + 1) + (X^2 + X + 1)X^2 = [X^4 + X^3 + X^2 + (X+X) + 1] + [X^4 + X^3 + X^2]$
Now in $F_2[X]$, for any element, we have $f(x) + f(x) = (1 + 1)(f(x)) = 0(f(x)) = 0$.
So in the product above, we continue:
$[X^4 + X^3 + X^2 + (X+X) + 1] + [X^4 + X^3 + X^2] = [X^4 + X^3 + X^2 + 1] + [X^4 + X^3 + X^2]$
$= (X^4 + X^4) + (X^3 + X^3) + (X^2 + X^2) + 1 = 0 + 0 + 0 + 1 = 1$.
This shows your gcd calculation is correct.
Now take your calculation mod $I$:
$[(X^3 + X + 1) + I][(X + 1) + I] + [((X^2 + X + 1) + I][X^2 + I] = 1 + I$
$[0 + I][(X + 1) + I] + [((X^2 + X + 1) + I][X^2 + I] = 1 + I$
($(X^3 + X + 1) + I = 0 + I = I$, because $X^3 + X + 1$ is in $I$).
$[0 + I] + [((X^2 + X + 1) + I][X^2 + I] = 1 + I$
(because $0 + I = I$ is an ideal, and "absorbs" anything we multiply it by)
$[((X^2 + X + 1) + I][X^2 + I] = 1 + I$
(since $0 + I = I$ is the additive identity of the quotient ring, and we can omit it in any sum).
This last equation shows explicitly that $(X^2 + X + 1) + I$ is indeed a unit.
