Quotient Ring of $\mathbb{F}_{2}[x]$ I had the following part of a problem that I am stuck on.

Let $x^{2}+x+1$ be an element of the polynomial ring $E=\mathbb{F}_{2}[x]$ and use the bar notation $\overline{E}$ to denote the passage to the quotient ring $^{\mathbb{F}_{2}[x]}/_{\left\langle x^{2}+x+1\right\rangle}$. Prove that $\overline{E}$ has four elements: $\overline{0}, \overline{1}, \overline{x}$, and $\overline{x+1}$.

My Attempt: If $\mathbb{F}_{2}$ is the field of two elements, then $\mathbb{F}_{2}=\{0,1\}$. Now, the ideal generated by the given polynomial is $$\left\langle x^{2}+x+1\right\rangle=\left\lbrace f(x)(x^{2}+x+1)\:\big|\:f(x)\in\mathbb{F}_{2}[x]\right\rbrace,$$ so a typical element of the quotient ring is something of the form $\overline{a}x+\overline{b}+\left\langle x^{2}+x+1\right\rangle$, where $a,b\in\mathbb{F}_{2}$. Then we have $$\frac{\mathbb{F}_{2}[x]}{\left\langle x^{2}+x+1\right\rangle}=\left\lbrace \overline{1}x+\overline{1}+\left\langle p(x)\right\rangle,\overline{0}x+\overline{0}+\left\langle p(x)\right\rangle,\overline{0}x+\overline{1}+\left\langle p(x)\right\rangle,\overline{1}x+\overline{0}+\left\langle p(x)\right\rangle\right\rbrace,$$ where $p(x)=x^{2}+x+1$. But then this set should just simplify to $$\left\lbrace\overline{x+1},\overline{0},\overline{1},\overline{x}\right\rbrace$$ since we are modding out by $\left\langle x^{2}+x+1\right\rangle$.
Does my proof look correct? I think everything ties together nicely, but I just want to make sure that my reasoning is sound. Thanks in advance for any comments and help!
 A: Sounds okay. If you want to furthermore justify this, observe that $\mathbb{F}_{2}\left[x\right]$ are polynomials over a field. Thus this is an Euclidean domain, and you can divide with residue each element $f\left(x\right)\in\mathbb{F}_{2}\left[x\right]$ by $x^2+x+1$
$$f\left(x\right)=\left(x^2+x+1\right)q\left(x\right)+r\left(x\right)$$
Here $\deg{r}<2$, so $r\left(x\right)=ax+b$ for some $a,b\in\mathbb{F}_{2}$. From here it is easy to see that
$$f\left(x\right)\equiv ax+b$$
in the quotient ring.
A: You have the right idea. The crux of the issue is the statement

a typical element ... is ... of the form $\overline{a}x+\overline{b}+\left\langle x^{2}+x+1\right\rangle$

This is a true statement. And for someone who well understands the subject material, it doesn't deserve any further comment.
However, I can't tell if you assert this because you actually have good reason to know it to be true, or if you are making an unjustified leap of logic.
In fact, that you go into so much detail for the rest of the argument makes me suspicious you don't realize everything important about the problem is encoded in the assertion above. I expect that either you have made an unjustified leap, or you find the exercise so obvious that you were struggling to find something worth elaborating upon.

There is a minor technical error. If you're realizing the quotient ring by cosets, then the typical element is of the form
$ax+b+\left\langle x^{2}+x+1\right\rangle$, and this element is equal to $\overline{a} \overline{x} + \overline{b}$. The expression $\overline{a}x+\overline{b}+\left\langle x^{2}+x+1\right\rangle$ doesn't realy make sense because $\overline{a}$ is not strictly a member of $\mathbb{F}_2[x]$.
