Well, first we note (using that $x^2 + 1$ is degree 2) since it has no roots (which we can check by plugging in $-1, 0$, and $1$) it is irreducible, so since $\mathbb{Z}_3[x]$ is a PID we have $\mathbb{Z}_3[x]/(p)$ is a field.
Now we see that, since we are "moding out" by a two degree polynomial (ideal) we have that elements of $\mathbb{Z}_3[x]/(p)$ can only be $0$ and $1$ degree polynomials; i.e. elements of the form $a + bx$; $a,b \in \mathbb{Z}_3$.
2
Bumbble Comm
On
Elements of the field are the following cosets $0,1,2,x,2x,x+1,x+2,2x+1,2x+2.$
Well, first we note (using that $x^2 + 1$ is degree 2) since it has no roots (which we can check by plugging in $-1, 0$, and $1$) it is irreducible, so since $\mathbb{Z}_3[x]$ is a PID we have $\mathbb{Z}_3[x]/(p)$ is a field.
Now we see that, since we are "moding out" by a two degree polynomial (ideal) we have that elements of $\mathbb{Z}_3[x]/(p)$ can only be $0$ and $1$ degree polynomials; i.e. elements of the form $a + bx$; $a,b \in \mathbb{Z}_3$.