Show that the polynomial $f = x^2 + 1$ is irreducible in $\mathbb{Z}_3[x]$. How many elements does $\mathbb{Z}_3[x]/(f)$ have? Show that the polynomial $f = x^2 + 1$ is irreducible in $\mathbb{Z}_3[x]$. How many elements does $\mathbb{Z}_3[x]/(f)$ have? Write out all of the elements of this field, and find the inverse of each nonzero element.
I am trying to show that the polynomial $f = x^2 + 1$ is irreducible in $\mathbb{Z}_3[x]$. To do so, I am using the following theorem: (Let $F$ be a field, $f \in F[x]$ of degree 2 or 3) If $f$ has no roots, then $f$ is irreducible. So, since $f(0), f(1)$ and $f(2)$ are all not $\equiv$ 0(mod 3), $f$ has no roots and thus $f$ is irreducible. Now how would I calculate how many elements $\mathbb{Z}_3[x]/(f)$ has? Would I need to find all of the polynomials in $\mathbb{Z}_3[x]$ that don't have roots? I would appreciate any feedback!

The elements of $\mathbb Z_3[x]/(f)$ are $a+bx$ with $a,b\in\mathbb Z_3$.
There are $3$ possibilities for $a$ and $3$ for $b$, so $9$ elements altogether.
Note that in the quotient ring $x^2\equiv-1$.  
The multiplicative inverse of $x$ is $2x$.  The inverse of $1+x$ is $2+x$.
I will leave the rest as an exercise for the reader.
