Factorization of polynomials over $\mathbb{Z}_3$ I have been given these two polynomials
$$f(t)=t^3+2t+1 \text{ & }g(t)=t^3+t^2-t+2$$
the problem says, decide if both factorization fields are isomorphic. For the second polynomial I got that
$$g(t)=(t-1)(t-2)^2$$
and the first one is irreducible since it has degree 3 and no roots in $\mathbb{Z}_3$, so I guess that's enough to say that both fields are not isomorphic since the second one is $\mathbb{Z}_3$ and the first one isn't. Although I would like to explicitely compute $f$ factorization field just to see how it goes but I don't know where to start.
 A: The splitting field of $f$ is $\mathbb{Z}_3[\alpha]$, where $\alpha^3+2\alpha+1=0$, because $$t^3+2t+1=(t- \alpha) (t-\alpha+1) (t-\alpha-1)$$
Indeed, long division gives
$$
t^3+2t+1=(t- \alpha) (t^2+\alpha t+\alpha^2-1)
$$
and $\alpha^2-1=(\alpha+1)(\alpha-1)$ suggests that the roots of $t^2+\alpha t+\alpha^2-1$ are $\alpha+1$ and $\alpha-1$, which is true because $(\alpha+1)+(\alpha-1)=2\alpha=-\alpha$.
A: let $\alpha$ is a root of $f$ in $\mathbb{E}\,$(An extended field of $\mathbb{Z}_3$). we have 
$$f(\alpha)^3=(\alpha^3+2\alpha+1)^3=\alpha^9+2\alpha^3+1=f(\alpha^3)=0$$
$$f(\alpha)^9=(\alpha^3+2\alpha+1)^9=\alpha^{18}+2\alpha^9+1=f(\alpha^9)=0$$
therefore $\alpha \,\, , \alpha^3$ and $\alpha^9$ are roots of $f$ in $\mathbb{E}\,$
we have
$$\mathbb{E}=\mathbb{Z}_3(\alpha)=\{a+x_1\alpha+x_2\alpha^2|a\,,\,x_i\in \mathbb{Z}_3\}$$
A: Not sure whether you were asking about this but here comes anyway. As explained by others the splitting field of $f$ has 27 elements. It is $\Bbb{Z}_3[\alpha]$, where $\alpha$ is a zero of $f(x)$, in other words it satisfies the equation $\alpha^3-\alpha+1=0$. Alternatively you can think that I just denote the coset 
$x+\langle f(x)\rangle\in\Bbb{Z}_3[x]/\langle f(x)\rangle$ by $\alpha$. To keep the notation simple! Note that because $0=3$ in this field we also have $-1=2$ and therefore $2x=-x$ in the polynomial ring.
To show that $K=\Bbb{Z}_3[\alpha]$ is the splitting field of $f(x)$ you need either a bit of luck or to know about the Galois theory of the extension $K/\Bbb{Z}_3$. The latter implies that the other zeros of $f(x)$ are 
$$
\alpha^3=\alpha^3-0=\alpha^3-f(\alpha)=\alpha-1
$$
and
$$
\alpha^9=(\alpha-1)^3=\alpha^3-1^3=(\alpha-1)-1=\alpha+1.
$$
If you are knew to this it is a good exercise to check by hand that $\alpha\pm1$ are also zeros of $f(x)$.
