I want to know how to go about describing a ring after adjoining an element that satisfies a certain relation.
As an example, I'm considering the ring obtained from Z3 by adjoining an element a satisfying $a^2 + a + 1=0$
So we consider $Z3[x]/x^2 + x + 1$
The way I'm thinking about it, is that we want to find all the divisors of $x^2 + x + 1$ first. We know that the polynomial has linear factors since it's reducible. Namely $(x-1)$. We also know all elements in this ring will have deg. 0 or 1. I think we should have elements $0, 1, 2$. But how can I get the rest of the elements in the ring?
Also any general tips on how to approach this type of problem with arbitrary rings/polynomials?
By the division algorithm, any polynomial $f(x) \in (\mathbb{Z}/3\mathbb{Z})[x]$ can be written as $$ f(x) = q(x) (x^2 + x + 1) + r(x) \equiv r(x) \pmod{x^2+x+1} $$ for some $q(x),r(x) \in (\mathbb{Z}/3\mathbb{Z})[x]$, where either $\deg(r(x)) < 2$ or $r(x) = 0$. This shows, as you stated, that every element of $(\mathbb{Z}/3\mathbb{Z})[x]/(x^2+x+1)$ has a representative of degree $0$ or $1$. Such a representative is of the form $a + bx$ with $a,b \in \mathbb{Z}/3\mathbb{Z}$. Does this help you see what the other elements of $(\mathbb{Z}/3\mathbb{Z})[x]/(x^2+x+1)$ look like? How many are there?
On the other hand, the factorization of $x^2 + x + 1$ over $\mathbb{Z}/3\mathbb{Z}$ gives you a different sort of information. We know $(x-1)(x^2+x+1) = x^3 - 1 = (x-1)^3$, so $x^2 + x + 1 = (x-1)^2$. Sending $x$ to $y+1$, we have $$ \frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x-1)^2} \cong \frac{(\mathbb{Z}/3\mathbb{Z})[y]}{(y^2)} \, . $$ This shows us that the quotient ring is isomorphic to the ring of dual numbers. In particular, it has nontrivial nilpotent elements: $\overline{y}^2 = \overline{y^2} = 0$, but $\overline{y} \neq 0$.