Can every irreducible cubic be proved irreducible using the Eisenstein criterion?

Let $k$ be an algebraically closed field, and let $f(X,Y) \in k[X,Y]$. One common way to show that $f$ is irreducible is to regard $f$ an element of $k[X][Y]$ or $k[Y][X]$ and use the Eisenstein criterion. Since $k$ is algebraically closed, the nonzero prime ideals of $k[X]$ are of the form $(X - x)$ for $x \in k$, and so to apply Eisenstein in this way, $f$ would have to look something like

$$X^3 + g_2(Y-a)X^2 + g_1(Y-a)X + h(Y)(Y-a)$$

for $g_1, g_2 \in Tk[T], h \in k[T], h(Y-a) \neq 0$.

Now, $f$ is irreducible if and only if $f(\phi(X,Y))$ is irreducible, where $\phi$ is an affine transformation of the form $X \mapsto cX + dY + e, Y \mapsto fX + gY + h$, and $\textrm{det} \begin{pmatrix} c& d \\ f &g \end{pmatrix} \neq 0$. So it is possible that one may prove $f$ is irreducible by applying a change of variables, and then apply the Eisenstein criterion.

My question is, are there examples of irreducible cubic polynomials for which no change of variables will allow you to conclude irreducibility by the Eisenstein criterion? If such examples exist, I would particularly be interested in an example with $k = \overline{\mathbb{F}_p(t)}$.

Solution 1:

If you're looking at a smooth curve of degree $3$, then no.

This will be an elliptic curve. If $P,Q,R$ are its three points at infinity and $S$ is a point such that $3[S] = [P]+[Q]+[R]$ then the divisor $3[S]-([P]+[Q]+[R])$ is the divisor of the equation of an affine line $ax+by+c = 0$ that goes through a triple point.
Using the coordinates $(x-x_S)$ and $z = ax+by+c$, the equation becomes $(x-x_S)^3 = zg(x-x_S,z)$, and since it's a regular point (I assumed the curve was smooth) you can't have $g(0,0) = 0$ or else both partial derivatives would vanish there.
Then you can apply Eisenstein's criterion there to deduce that the curve is irreducible.

In case $S$ is a point at infinity this may not quite work exactly like this, but since there are $9$ distinct points $S$ and only three points at infinity, you can always find one which is not at infinity.