Number of elements in the ideal of Ring.

Question

We have $$x^9+1 = (x+1)(x^2+x+1)(x^6+x^3+1)$$ is factorization of irreducible polynomials over $GF(2)$ (Galois field). Then we know that one of its ideal for the ring is $$R = GF(2)[x]/(x^9+1)$$ One of ideals is generated by $$I = (x^2+x+1)(x^6+x^3+1)$$ Then

(i) What is the dimension of this ideal $I$ in $R$?

(ii) Number of elements in this ideal?

(iii) How each element of $I$ looks like?

Any hint or help?

Answer 1

The ring $R$ has the monomial basis $1, x, x^2, \dotsc, x^8$. This is a vector space basis of $R$ as a vector field over $GF(2)$.

You want to find a $GF(2)$ basis for $I$. This can be done by taking your generator $f = (x^2 + x + 1)(x^6 + x^3 + 1)$ and computing the elements $$f, xf, x^2f, x^3f,\dotsc$$ Those clearly generate $I$ over $GF(2)$, so you want to choose a basis from this generating system. To do this, write the elements in the monomial basis $1, x, x^2, \dotsc, x^8$ of $R$, i.e. $x^kf = \sum_i a_{i,k} x^i$ and look at the vectors $$v_k = \left[\begin{matrix}a_{0,k}\\\vdots\\a_{8,k}\end{matrix}\right].$$ For which $l$ does the system $\{v_0,v_1,\dots,v_l\}$ become linearly dependent? Then $\{v_0,\dots,v_{l-1}\}$ is your basis.

This directly gives you the dimension of $I$, so it has $2^l$ elements.

Answer 2

You’re asking for the ideal generated by the product $h(x)=(x^2+x+1)(x^6+x^3+1)$, in the ring $R=\Bbb F_2[x]/(x^9+1)=\Bbb F_2/\bigl((x+1)h(x)\bigr)$. Now by direct inspection, $h(x)=\sum_{j=0}^8x^j$, and you easily check that $xh\equiv h\pmod{(x^9+1)}$. It follows from this that $h^2\equiv h\pmod{(x^9+1)}$, from which you conclude that your ideal is one-dimensional over $\Bbb F_2$: its only elements are $0$ and $h$.