---
title: Discriminant and behavior at infinity
tags: [nonproperness, discriminant, infinity]
---

# Discriminant and behavior at infinity

The cubic model identifies exactly where inverse branches escape.

## Discriminant hypersurface

For

$$
p(w)=cw^3-2w^2+bw-2a,
$$

the discriminant is $4\Delta(a,b,c)$, where

$$
\boxed{
\Delta=b^2-cb^3-16a+18abc-27a^2c^2.}
$$

Thus $\Delta=0$ is the locus where the binary cubic has a multiple root.

## Explicit escape through every discriminant point

Every point of $\Delta=0$ has a repeated finite root $\rho$ and can be
written

$$
(a,b,c)=\left(\rho^2-c\rho^3, 4\rho-3c\rho^2, c\right).
$$

For $t\ne0$, define

$$
(x,y,z)=
\left(
\frac1t,\,
\rho-t,\,
5t^2-3\rho t-ct^3
\right).
$$

Exact substitution gives

$$
F(x,y,z)=
\left(
\rho^2+\rho t-c\rho^3,
4\rho+2t-3c\rho^2,
c
\right).
$$

As $t\to0$, the source escapes to infinity and the image converges to the
chosen discriminant point.  Hence every point of $V(\Delta)$ is a
nonproperness value.

This family can also be read as a uniform perturbation proof.  If $\rho$ is a
repeated root at $(a_0,b_0,c_0)$, set

$$
(a_\varepsilon,b_\varepsilon,c_\varepsilon)
=(a_0+\varepsilon\rho/2,b_0+\varepsilon,c_0).
$$

Then $\rho$ stays a root and its derivative becomes $\varepsilon$, so its
inverse has $x=2/\varepsilon$ and escapes.

Conversely, let $I=\{Q=0\}$ be the full projective-root incidence inside
$\mathbb A^3\times\mathbb P^1$.  The projection
$I\to\mathbb A^3$ is projective and quasi-finite, hence finite.  Over
$\Delta\ne0$, all roots are simple, so $I=I^{\mathrm{sm}}$ and the
marked-root isomorphism identifies this finite étale map with the restriction
of $F$.
All root-gradient normalizations and the inverse formulas remain bounded in a
neighborhood of a limiting target, so no inverse branch can escape.  Therefore
the nonproperness set is exactly

$$
\boxed{S_F=V(\Delta).}
$$

## The simplest escaping curve

Taking $\rho=0,c=0$, and replacing $t$ by $1/s$, gives

$$
(x,y,z)=(s,-1/s,5/s^2),\qquad F(x,y,z)=(0,2/s,0).
$$

As $|s|\to\infty$, the source escapes and the image tends to the origin.

## Why nonproperness is not an objection

The Keller condition says that $F$ is étale: it has no critical points and is
locally invertible.  It does **not** assume properness.  A proper étale
self-map of complex affine space would be a finite covering and global
topology would force degree one.  The Jacobian conjecture asserted, in effect,
that polynomiality plus the constant-Jacobian condition was already strong
enough to prevent escape at infinity.

This example shows exactly the opposite: the discriminant is where two or three
otherwise finite inverse branches run to infinity.  The escape curve is the
mechanism of the counterexample, not a loophole in it.  This is consonant with
the long-standing focus on nonproperness and ramification at infinity
{cite}`jelonek1993,rodriguez2026,truong2026`.