I want to find a fifth root of unity modulo $81$ using a suggested method from the book (I can't come up with any other good method anyway). It is given that $x^4+x+2 \in \mathbb{F}_3[x]$ is irreducible, and that given $\alpha$ such that $\alpha^4 + \alpha +2=0$
$$ \mathbb{F}_3(\alpha) = \mathbb{F}_3[x] / (x^4+x+2) \cong \mathbb{F}_{81} $$
So first thing is, above is given, but I'm not entirely sure how $\mathbb{F}_3[x] / (x^4+x+2) \cong \mathbb{F}_{81}$ holds. Also, above notation seems to hint that (and I think it should be) $\mathbb{F}_3[x] / (x^4+x+2)$ is a field extension of $\mathbb{F}_3$. But saying that it is isomorphic to $\mathbb{F}_{81}$ is clearly wrong since latter is not a field? (unless I'm making a stupid mistake again...)
Anyway, I'm suggested to consider above, and find a fift primitive root of $\mathbb{F}_{81}$ in its unit subgroup (group of unit elements) in terms of $\alpha$. So obviously the problem wants me to look at $\mathbb{F}_3[x] / (x^4+x+2)$ instead of $\mathbb{F}_{81}$.
First thing I notice is that if I can find non-identity element $a$ in $\mathbb{F}_{81}$ such that $a^5=1$, then I'm basically done since $5$ is a prime. That, translating this into the quotient ring, is equivalent to finding polynomials $g,q$ such that
$$ (g(x))^5 = q(x) f(x) + 1 $$
where $f(x)=x^4+x+2$.
But I didn't have clear idea from this point, and just wasted some time on trying some particularly nice form of $g$ like $(x+1)$, but didnt have much success. I'm not even sure if I'm on the right track.
Any helps appreciated!
You're confusing $\mathbf F_{81}$ (the field with 81 elements) with $\mathbf Z/81\mathbf Z$. The former is the splitting field of any irreducible (over $\mathbf F_3$) polynomial of degree 4.
Remember that for any prime $p$ and any $n\ge 1$, there is only one field with $p^n$, up to isomorphism. In particular, $\mathbf F_{81}$ is the splitting field of the 5th cyclotomic polynomial, $x^4+x^3+x^2+x+1$. This proves the existence of primitive 5th root of unity in $\mathbf F_{81}$.
Now, taking into account we're in characteristic $3$ and the relation $\alpha^4=1-\alpha $, we can compute the order of $\alpha$ in the multiplicative group $\mathbf F_{81}^\times$. By Lagrange's theorem, it must be a divisot of $80$.
Indeed, one checks its order is $80$, so that the order of $a^k$ is $\,\dfrac{80}{80\wedge k}$. If we want $\alpha^k$ to be of order$5$, $k$ must be chosen so that $\,80\wedge k=16$. The simplest is $k=16$, so that $$\alpha^{16}=-1+\alpha -\alpha^3$$ is a (primitive) 5th root of unity in $\mathbf F_{81}$.