The basis of $\mathbb{Z}_{2}/\left \langle x^{2}+x+1 \right \rangle$ is $\left \{ [1], [x] \right \}$ 
Show that $A=\left \{ [1], [x] \right \}$  is a basis of $\mathbb{Z}_{2}/\left \langle x^{2}+x+1 \right \rangle$ over $\mathbb{Z}_{2}$

The first thing I did was see that the elements of the ring were $[0], [1] ,[x]$ and $[x+1]$. Then to show that it's a basis, it must be proved that the set A span the ring and that the elements of A are linearly independent.
To see that it span the ring, I must see that it generates the elements $[0]$ and $[x + 1]$, so $[x + 1] = 1\cdot [x] + 1 \cdot [1]$, but $[0] = 0 + p(x)$ , ($p (x) \in \mathbb{Z}_{2}[x]$)
I don't know if I can say that $[0]=0\cdot [x]+p(x)\cdot [1]$, because $[1] = (1 + q (x))$ and thus $[0] = p (x) + p (x) q (x)$ which becomes a polynomial again
I appreciate any comments
 A: Something a bit more general is true. Take $k$ a field (such as $\mathbb{Z}_2$) and $$f = a_nX^n + a_{n-1}X^{n-1} \ldots + a_1 X + a_0 \in k[X]$$ with $a_n \neq 0$. Then $B = \{[1], [X], \ldots, [X]^{n-1}\}$ is a $k$-basis for $k[X]/(f)$ as a $k$-vector space.
To see this: pick $[g] \in K[X]/(f)$. Since $k$ is a field, we know that $k[X]$ has a division algorithm. Namely, there exist unique $q,r \in k[X]$ with $r = 0$ or $\deg r < n$ such that $g = qf + r$; and so $[g] = [qf]+[r] = [r]$. But since either $r = 0$ or $r$ has degree less than $n$, we have
$$
r = b_0 + b_1 X + \ldots + b_{n-1}X^{n-1}
$$
with $b_1, \ldots, b_{n-1} \in k$ (some or even all of these may be zero). Consequently,
$$
[g] = [r] = b_0[1] + b_1[X] + \cdots +b_{n-1}[X^{n-1}]
$$
and thus $B$ is a set of generators. Finally, to see that $B$ is linearly independent, suppose that
$$
0 = b_0[1] + b_1[X] + \cdots +b_{n-1}[X^{n-1}] = [b_0 + b_1 X + \ldots + b_{n-1}X^{n-1}]
$$
for some $b_i \in k$. This means that there exists $g$ such that
$$
b_0 + b_1 X + \ldots + b_{n-1}X^{n-1} = gf.
$$
But that's impossible unless $b_0 = \cdots = b_{n-1} = 0$ and $g = 0$, as otherwise the right hand side has degree $\deg gf \geq \deg f = n$.
