I'm taking a second course of abstract algebra but I feel like I don't have the enough basis from the first course so it's very difficult the second one (well that's how I feel it).
For example, I have no idea how to solve this problem:
Let the extension $\frac{\mathbb Z_5[x]}{<1+x+x^2>}$: $\mathbb Z_5$ and $ \alpha$=$2+x+<1+x+x^2>$ $\in$$\frac{\mathbb Z_5[x]}{<1+x+x^2>}$.
Find $\alpha^{-1}$.
I've looking on google for this kind of exercises but I only find something like this 'Let $a=1+x^2$ in $\mathbb Z_5[x]$, find the inverse of a'.So clearly the field doesn't has the form of $\frac{\mathbb Z_5[x]}{<1+x+x^2>}$ and a neither respect to $\alpha$.
Can somebody give an idea on how to solve the exercise? or does anyone know about any link where there are solved exercises of this type?
For simplicity, write $f(x)$ to denote the classe of $f(x)$ in $F=\frac{\mathbb{Z}_5[x]}{\langle 1+x+x^2\rangle}$. Thus $1+x+x^2=0$ in $F$. Then $1+x+x^2=0 \Leftrightarrow \\1+x=-x^2 \Leftrightarrow \\2+x=1-x^2=(1+x)(1-x)$. But $1+x=-x^2$, and we have $2+x=1-x^2=-x^2(1-x)=-x^2-x^3 \Leftrightarrow\\ 2=-x(1+x+x^2)=-x\cdot0=0$. In particular, $1=-1$ in $F$. Thus $2+x=x$ in $F$. So to find the inverse of $2+x$ is to find the inverse of $x$ in $F$. Since $1+x+x^2=0$, we have $1=-1=x(1+x)$. That is, the inverse of $2+x=x$ in $F$ is $1+x$.