Q:

What does it mean to represent elements of an ideal?

Say I have the polynomial $x^9 + 1$
Then: $x^9 + 1 = (x+1)(x^2 + x + 1)(x^6 + x^3 + 1)$
is a complete factorization over $GF(2)$ of  $x^9 + 1$
The dimension of each ideal is: length $n - deg(ideal)$
So for $n=9$, dimension of $(x+1)$ = $9-1=8$
of $(x^2 + x + 1) = 9 - 2 = 7$
of $(x^6 + x^3 + 1) = 9 - 6 = 3$
So let's use $(x^6 + x^3 + 1)$ as the example. The dimension is 3. So there should be $2^3 = 8$ elements. How do I find those elements?

A:

Let’s call $f=x^6+x^3+1$. You want three linearly independent elements of the ideal $(f)$ of the ring $R=\Bbb F_2[x]/(x^9+1)$. Since $(f)$ is just the set of multiples of $f$, you certainly have $1\cdot f$, $xf$, and $x^2f$. Notice that $x^3f=x^9+x^6+x^3=1+x^6+x^3=1\cdot f$, already counted. I’ll leave it to you to show that those three polynomials are $\Bbb F_2$-linearly independent in $R$.

