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 [Jelonek, 1993, Rodríguez Díaz, 2026, Truong, 2026].