Showing that a polynomial is a unit in a quotientring

Question

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.

Answer 1

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.

Answer 2

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.