My question is: Let $f(x) = x^2 + 1 \in \mathbb{Z}_3[x]$. I have to determine the splitting field of this polynomial. OK, so is intuitive to think that $i,-i$ are roots. OK, but does it make any sense? Can I assure that $i,-i$ are indeed roots? I know that the splitting field is $Z_3(\alpha)$ with $\alpha^2 = -1$, can I assure that $\alpha = i?$. On my opinion I think that this isn't a absurd. Note that $x^2 + 1 = 0 \in Z_3[x]$ is the same that $x^2 + 1 \equiv 0 \pmod 3$. It has solution on $Z[i]$, the Gauss integers, so makes sense to say that $\alpha = i$ in same way. So, is this true?
Thanks a lot.
Since $x^2+1$ is irreducible quadratic in $\mathbb Z_3[x]$, we get its splitting field by adjoining just one root, yielding $\mathbb Z_3[x]/\langle x^2+1\rangle$. But, remembering that $\mathbb Z_3$ is itself a quotient, this is the same as $$ (\mathbb Z/\langle 3\rangle)[x]/\langle x^2+1\rangle \cong \mathbb Z[x]/\langle 3\rangle/\langle x^2+1\rangle \cong \mathbb Z[x]/\langle 3, x^2+1\rangle \cong \mathbb Z[x]/\langle x^2+1\rangle/\langle 3\rangle = \mathbb Z[i]/\langle 3\rangle $$
So the splitting field is a quotient of the Gaussian integers, and the root of $x^2+1$ is exactly the residue class of $(\pm)i$, as you have intuited.