Let $f(x) =x^3+x+1 \in \mathbb{Q} [x] $. If ideal $I=(f(x)) $, find the inverse of $x^2+x+1$ in quotient $\mathbb{Q} [x] /I$. Let $f(x) =x^3+x+1 \in \mathbb{Q} [x] $. If ideal $I=(f(x)) $, find the inverse of $x^2+x+1$ in quotient $\mathbb{Q} [x] /I$.
I am having trouble with this. What exactly is the quotient $\mathbb{Q} [x] /I$ here? I guess it's polynomials with degree 2 or less which in multiplication satisfy the relation $ x^3+x+1=0$. How do I find the inverse? I guess I have to find a polynomial of degree 2 or less so that when multiplied by $x^2+x+1$ I get something divisible by $ x^3+x+1$. 

A polynomial in the quotient ring can be represented by $ax^2+bx+c.$
We want $(ax^2+bx+c)(x^2+x+1)\equiv 1 \pmod {x^3+x+1}$.
Note that $\color{blue}{x^3\equiv-x-1}$ and $\color{green}{x^4\equiv -x^2-x}$.
Thus $(ax^2+bx+c)(x^2+x+1)\equiv \color{green}{ax^4}+\color{blue}{(a+b)x^3}+(a+b+c)x^2+(b+c)x+c$
$\equiv \color{green}{-ax^2-ax}\color{blue}{-(a+b)x-(a+b)}+(a+b+c)x^2+(b+c)x+c$
$\equiv (b+c)x^2+(c-2a)x+(c-a-b)\equiv1,$
so we want $b+c=0, c-2a=0, $ and $c-a-b=1$.
Can you solve this system of equations for $a, b, $ and $c$?
Hint: $c=-b=2a$.
