Describe the multiplication in the ring $\Bbb Q[x]/(x^2+x+1)$. Describe the multiplication in the ring $\Bbb Q[x]/(x^2+x+1)$. What is the multiplicative inverse of $[x]$?
To figure this out, I found that $x = -\frac {1}{2} + i \sqrt {\frac{3}{4}}$. So there clearly aren't any real solutions to this. But I am trying to figure out what elements of this ring look like in order to find an inverse and to see what multiplication looks like. I am new to field extensions and am a little confused as to how to work with them. Any help is appreciated.
 A: Since you have 
$$
x^2 + x +1 = 0
$$
you see that
$$
1 = - x^2 - x = x (- x -1)
$$
so 
$$
x^{-1} = - x - 1
$$
A: There are a few ways to think about it. 
The first is quite literal. The elements of $\frac{\mathbb{Q}[x]}{(x^2+x+1)}$ are residues of polynomials, $\overline{f(x)}$. In this ring $\overline{f(x)} = \overline{g(x)}$ if and only if $f(x)-g(x)$ is divisible by $x^2+x+1$.
To multiply $\overline{f(x)}$ and $\overline{g(x)}$, multiply them as polynomials in $\mathbb{Q}[x]$, then divide by $x^2+x+1$ and take the remainder.

Another way: $$\frac{\mathbb{Q}[x]}{(x^2+x+1)} \cong \{a+b\alpha |\ a,b \in \mathbb{Q}, \text{ and } \alpha^2 + \alpha + 1 = 0 \}$$
$\alpha$ is some "thing" that satisfies the polynomial $x^2 + x +1$. Then multiplication becomes:
\begin{align*} (a + b\alpha)(c + d\alpha) &= ac + (ad + bc)\alpha + bd\alpha^2\\
&= ac + (ad + bc)\alpha + bd(-\alpha - 1)\\
&= (ac-bd) + (ad + bc - bd)\alpha.
\end{align*}

Per user1952009's suggestion:
How could we express $\alpha^3$ and $\alpha^4$ as something of the form $a+b\alpha$?
\begin{align*}
\alpha^3 &= \alpha \alpha^2\\
&= \alpha(-\alpha-1)\\
&= -\alpha^2 - \alpha\\
&= 1.
\end{align*}
In this cool turn of events, we see that $\alpha^3 = 1$. Another way to see that this would happen is that $x^3-1 = (x-1)(x^2+x+1)$ and since $\alpha$ satisfies $x^2+x+1$, it will then satisfy $x^3-1$.
From this we see that $\alpha^4 = \alpha$ and in fact for $n \in \mathbb{N}$:
$$
\alpha^n = \begin{cases}
1 \ &\text{ if } n \equiv 0 \pmod 3 \\
\alpha \ &\text{ if } n \equiv 1 \pmod 3 \\
-1 - \alpha \ &\text{ if } n \equiv 2 \pmod 3
\end{cases}
$$
A: The ring in question is in fact an algebra over the rationals. As a vector space one can represent the multiplication with $x$ as a matrix with respect to the basis $\{1, x\}$. We have that $x : 1\mapsto x$ and $x: x\mapsto x^2 = -x-1$, so the multiplication by $x$, as a linear transformation can be written as the matrix $$M = \begin{pmatrix} 0 & -1 \\ 1 & -1\end{pmatrix}$$ Note that $M$ satisfies $M^2+M+I = 0$ (with $I$ the identity matrix), and that the minimal polynomial of $M$ is $x^2+x+1$. The ring is now expressed as the $2$ dimensional matrix algebra generated by $M$ with basis $\{I, M\}$. The inverse of $x$ is expressed by $$M^{-1} = \begin{pmatrix} -1 & 1 \\ -1 & 0\end{pmatrix} = (-1).I + (-1).M$$ In this manner the inverse of any element $a.I +b.M$ of the ring can be calculated. Note that the first column gives the coefficients of the element w.r.t. the basis. Note that these statements can be restated for higher degree polynomials.
A: The elements of this ring are of the form $a+bu$ with $u^2=-u-1$ and $a,b\in\mathbb Q$.
Therefore, $(a+bu)\cdot(c+du)=(ac-bd)+(bc+ad-bd)u$.
