Binary cubics and the complete fiber description

The map has an explicit three-valued algebraic inverse. This chapter makes the projective-root interpretation precise.

Throughout this chapter the base field is \(\mathbb C\) (the same arguments work over any algebraically closed field of characteristic zero).

Finite roots

For a target \((a,b,c)\), dehomogenize its binary cubic at \(V=1\):

\[ p_{a,b,c}(w)=cw^3-2w^2+bw-2a. \]

Every preimage with \(x\ne0\) has

\[ w=y+\frac1x,\qquad p_{a,b,c}(w)=0,\qquad p'_{a,b,c}(w)=\frac2x. \]

Conversely, let \(w\) be any simple root and put

\[ h=\frac{p'_{a,b,c}(w)}2. \]

Then \(h\ne0\), and the unique corresponding preimage is

\[ \boxed{ x=\frac1h,\qquad y=w-h,\qquad z=5h^2-3wh-ch^3.} \]

Substitution into the rational identities of the previous chapter returns \((A,B,C)=(a,b,c)\).

The root at infinity

The projective point \([U:V]=[1:0]\) is a root precisely when \(c=0\). It is always simple because \(Q_V(1,0)=-2\). Its corresponding source point is

\[ \boxed{(x,y,z)=(0,b,a-4b^2).} \]

Indeed, \(F(0,y,z)=(z+4y^2,y,0)\).

Why every simple projective root gives one point

At any simple root \([u_0:v_0]\), Euler’s identity for the homogeneous cubic implies that its gradient is a nonzero scalar multiple of \((v_0,-u_0)\). There is a unique rescaling \((U,V)=s(u_0,v_0)\) for which

\[ Q_U(U,V)=2V,\qquad Q_V(U,V)=-2U. \]

When \(V\ne0\), the preceding affine formula reconstructs \((x,y,z)\). When \(V=0\), the explicit infinity formula does. This is inverse to the marked-root construction from the source.

More precisely, let

\[ I^{\mathrm{sm}}= \{(a,b,c,[U:V]):Q_{a,b,c}(U,V)=0, [U:V]\text{ is simple}\}. \]

Then

\[ \mathbb A^3\longrightarrow I^{\mathrm{sm}},\qquad (x,y,z)\longmapsto(F(x,y,z),[1+xy:x]) \]

is an isomorphism, not merely a bijection on complex points. The finite-root formula gives its regular inverse on \(V\ne0\). On \(U\ne0\), put

\[ s=V/U,\qquad k=1-bs+3as^2. \]

Simplicity is exactly \(k\ne0\), and a regular inverse on this chart is

\[ \boxed{ x=\frac{s}{k},\qquad y=b-3as,\qquad z=ak^3-y^2k(k+3).} \]

At \(s=0\) this reduces to the root-at-infinity formula.

Fiber cardinalities and image

A nonzero binary cubic over \(\mathbb C\) has one of three root patterns:

• three simple projective roots;

• one double root and one simple root;

• one triple root.

Only simple roots produce source points. Therefore the corresponding fiber cardinalities are exactly \(3\), \(1\), and \(0\).

The triple-root cubics are parametrized by

\[ \boxed{ (a,b,c)=\left(\frac{\rho^2}{3}, 2\rho, \frac{2}{3\rho}\right), \qquad \rho\ne0.} \]

Consequently the set-theoretic image on complex points is

\[ F(\mathbb A^3(\mathbb C)) =\mathbb A^3\setminus \left\{\left(\frac{\rho^2}{3},2\rho,\frac{2}{3\rho}\right): \rho\in\mathbb C^*\right\}. \]

Because \(F\) is dominant, its scheme-theoretic image—and the Zariski closure of its point-set image—is still all of \(\mathbb A^3\). The omitted curve is also the reduced singular locus of the discriminant hypersurface.

For example \((1/3,2,2/3)\) is omitted because

\[ p(w)=\frac23(w-1)^3. \]

Independently, a Gröbner basis gives

\[ (F_1-1/3,F_2-2,F_3-2/3)=(1). \]

Generic degree and monodromy

The generic cubic has three distinct roots, so the map has geometric and function-field degree \(3\):

\[ [\mathbb C(x,y,z):\mathbb C(F_1,F_2,F_3)]=3. \]

There is also a direct function-field proof. Over \(k=\mathbb C(b,c)\), solve the root equation for the remaining coefficient:

\[ a=h(w)=\frac{cw^3-2w^2+bw}{2}. \]

The rational map \(h:\mathbb P^1\to\mathbb P^1\) has degree three, so

\[ [k(w):k(h(w))]=3. \]

The inverse formulas show that \(w\) generates the full source function field over \(\mathbb C(a,b,c)\); hence its cubic is the minimal polynomial and the map has geometric degree three.

The discriminant factor \(\Delta\) is not a square. Viewed as a quadratic in \(a\), its own discriminant is

\[ -4(3bc-4)^3, \]

which is nonsquare in \(\mathbb C(b,c)\). Thus \(\Delta\) is irreducible and has odd valuation along its divisor. The degree-three extension is generically nonnormal (and has trivial automorphism group), while its degree-six normal closure—and the geometric monodromy—has group \(S_3\).

“Generically three-to-one” is the correct phrase. The exact fiber sizes vary on the discriminant and omitted loci.