In the field $\mathbb{Z}_7[x]/\langle x^4+x+1\rangle$, find the inverse of $f(x)=x^3+x+3$. In the field $\mathbb{Z}_7[x]/\langle x^4+x+1\rangle$, find the inverse of $f(x)=x^3+x+3$.
I know how to find the inverses of elements within sets, rings, and fields. I know what to do if the field was just $\mathbb{Z}_7$, but the fact that the field is $\mathbb{Z}_7[x]/\langle x^4+x+1\rangle$ confuses me. I don't know where to start.
 A: The computation here is the same in ${\mathbb Q}$ as in
${\mathbb Z}_7$.
You look for a solution of the form
$$
z=a+bx+cx^2+dx^3 \tag{1}
$$
You then have 
$$
z(x^3+x+1)=dx^6+cx^5+(b+d)x^4+(a+c+3d)x^3+(b+3c)x^2+(a+3b)x+3a=Q(x) \tag{2}
$$
Next, divide the result by $x^4+x+1$ :
$$
Q(x)=(x^4+x+1)(dx^2+cx+b+d)+R(x) \tag{3}
$$ 
where the remainder $R(x)$ equals 
$$
R(x)=(a+c+2d)x^3+(b+2c-d)x^2+(a+2b-c-d)x+(3a-b-d) \tag{4}
$$
Then, solve the system
$$
a+c+2d=b+2c-d=a+2b-c-d=0, \ \ 3a-b-d=1 \tag{5}
$$
This will lead you to the solution
$$
a=\frac{11}{47}, b=\frac{-8}{47}, c=\frac{1}{47}, d=\frac{-6}{47},
z=\frac{11-8x+x^2-6x^3}{47} \tag{6}
$$
A: I would guess that you would first check that $\gcd(x^{4}+x+1,x^{3}+x+3)=1$ in $\mathbb{Z}_{7}[x]$ (else $x^{3}+x+3$ is a zero divisor in the field, thus cannot have an inverse).
Then use the characterization of the gcd as a linear combination to write $$p(x)(x^{4}+x+1)+q(x)(x^{3}+x+3)=1.$$ in $\mathbb{Z}_{7}[x]$. When we project down to the specified field, we kill the $x^{4}+x+1$ term (by the definition of the field), so the desired inverse is the projection of $q(x)$.
