Given the extension $\dfrac{\mathbb Z_7[x]}{\langle 3+x^3\rangle}:\mathbb Z_7$ and $\alpha =x^2+\langle 3+x^3\rangle \in \dfrac{\mathbb Z_7[x]}{\langle 3+x^3\rangle}$, find $\alpha^{-1}$.
I tried to solve it..
Suppose $\alpha^{-1}=a+bx+cx^2$. Then $1=\alpha^{-1}\alpha =(a+bx+cx^2)x^2=4b+4cx+ax^2$ (because $x^3=-3$).
And then, $a=0,4c=0,4b=1.$ But how do I find $b$?
Is my solution correct??
On
The underlying coefficient ring is $\mathbb{Z}_7$, so $4b = 1$ has solution $b = 2$. (More specifically, you are solving $4b \equiv 1 \pmod 7$, which can be done by inspection. Or with the Euclidean algorithm, if you prefer).
It is then easy to check your solution (which is apparently $2x$) by multiplying it by $\alpha$ and verifying that you get $1$.
The solution is correct.
If $g$ is a nonzero element in $\mathbb{Z}_p$, to find its inverse you can use Bézout's Identity. Since $p$ is prime, then $\gcd(g,p) = 1$ so there are integers $x,y$ such that $gx + py = 1$. Note that this equality in $\mathbb{Z}_p$ is equivalent to saying that there is an element $x\in \mathbb{Z}_p$ such that $gx = 1$, i.e. $g$ has an inverse. To compute it you can use the Euclidean Algorithm.