An irreducible and not separable polynomial

Question

How can I show that $Y^3 - X$ is irreducible and not separable polynomial in $({\mathbb{Z}}_3(X))[Y]$? I am defining ${\mathbb{Z}}_3$ as $\mathbb{Z} / (3)$. It looks a strange field for me and my attemps do not reach any result. If $F$ is a field, I define $F(X)$ as $$ F(X) = \left\{\frac{p(X)}{q(X)} : p(X) , q(X) \in F[X] \quad \mbox{ and } \quad q(X) \not \equiv 0\right\}\mbox{,} $$ being $F[X]$ the set of polynomials with coefficients in $F$. I have seen the next argument to show the not separability: if $\alpha$ is a root of $Y^3 - X$, then $$ Y^3 - X = Y^3 - {\alpha}^3 = {(Y - \alpha)}^3\mbox{.} $$ The first step is clear but I don't understand the second step. What are we applying here?

Answer 1

Irreducible: cubic is irreducible iff it has no roots. Use the fact that $\Bbb Z_3[X]$ is a graded ring to show that there is no roots: If $(f/g)^3 - X = 0$ where $f, g \in \Bbb Z_3[X]$, then $f^3 = Xg^3$. Now take the valuation (degree) of both sides.

Inseparable: $D(Y^3-X) = 3Y^2 = 0$

---

$Y^3 - X = (Y-\alpha)^3$ so you have multiple roots. That's like the definition of inseparability.