Basis of subspace that consists of polynomials that are divisible by $x^2+x+\overline{1}$

Question

We consider $\mathbb{Z}_2=\{\overline{0}, \overline{1}\}$ and $\mathbb{Z}_2[x]$.

To show that there is no $n\in \mathbb{Z}_2$ with $n^2=n+\overline{1}$, do we do the following?

For $n\equiv 0$ we have $n+1\equiv 1\not\equiv n^2$

For $n\equiv 1$ we have $n+1\equiv 0\not\equiv n^2$

That means that there is no $n\in \mathbb{Z}_2$ such that $n^2=n+\overline{1}$, right?

This also means that $x^2+x+1\equiv x^2-x-1$ has no root in $\mathbb{Z}_2$, right?

We consider the subspace $U\subseteq \mathbb{Z}_2[x]$ that consists of polynomials that are divisible by $x^2+x+\overline{1}$. How can we find a basis of $U$ ? Could you give me a hint?

We have that $U=\{p(x)\cdot (x^2+x+\overline{1})\}$, or not? But how can we find a basis with this information?

Answer 1

Yes, the space for which you want to find a basis is$$\{p(x)(x^2+x+1)\mid p(x)\in\mathbb{Z}_2[x]\}.$$So, a basis will be$$\{x^2+x+1,x(x^2+x+1),x^2(x^2+x+1),x^3(x^2+x+1),\dots\}$$

Answer 2

The polynomial $h(x)=x^2+x+1$ has no root in $\mathbb{Z}_2$ because $h(0)=1$ and $h(1)=1$. (Use overlines if you so prefer.)

The $\mathbb{Z}_2$-linear map $$ \varphi\colon\mathbb{Z}_2[x]\to U,\qquad \varphi(f)=fh $$ is bijective. So it transforms a basis in a basis. A basis of the domain is $\{1,x,x^2,\dotsc\}$.