A homogeneous polynomial \(f\) of degree \(d\) in three variables \(x, y, z\) has \(\binom{d+2}{2}\) coefficients. Let these coefficients represent a point in \(\PP^N\), where \[ N = \binom{d+2}{2} - 1 = \frac{1}{2} d(d+3). \]
Show that this gives a correspondence between points of \(\PP^N\) and algebraic sets in \(\PP^2\) that can be defined by an equation of degree \(d\). The correspondence is one-to-one except in some cases where \(f\) has a multiple factor.
Show that under this correspondence the irreducible nonsingular curves of degree \(d\) correspond one-to-one to the points of a nonempty Zariski-open subset of \(\PP^N\).
Hint: Use elimination theory (5.7A) applied to the homogeneous polynomials \(\frac{\partial f}{\partial x_0}, \ldots, \frac{\partial f}{\partial x_n}\), and use (Ex. 5.5, 5.8, 5.9).
Solution
Let \[ V_d=k[x,y,z]_d, \] so \(\dim_kV_d=\binom{d+2}{2}=N+1\) and the parameter space of nonzero degree-\(d\) forms up to scalar is \(\PP(V_d)\cong\PP^N\).
1. A point \([f]\in\PP(V_d)\) determines the algebraic set \(Z(f)\subseteq\PP^2\), and every algebraic set cut out by one degree-\(d\) equation occurs in this way.
Replacing \(f\) by a nonzero scalar multiple does not change its zero set, so the assignment is well-defined on projective coefficient space. Conversely, by definition, an algebraic set in \(\PP^2\) which can be defined by one homogeneous equation of degree \(d\) is \(Z(f)\) for some nonzero \(f\in V_d\), hence comes from the point \([f]\).
2. If \(f\) is square-free, then \([f]\) is the unique point of \(\PP(V_d)\) defining \(Z(f)\); nonuniqueness can occur only when repeated irreducible factors are present.
Factor \[ f=c\prod_{i=1}^r p_i^{a_i}, \] where the \(p_i\) are pairwise nonassociate irreducible homogeneous polynomials. Over the algebraically closed field \(k\), the projective Nullstellensatz says that the reduced homogeneous ideal of \(Z(f)\) has the same irreducible hypersurface factors \(p_i\). Thus if another degree-\(d\) form \(g\) satisfies \(Z(g)=Z(f)\), then \[ g=c'\prod_{i=1}^r p_i^{b_i} \] with all \(b_i\ge1\).
If \(f\) is square-free, then every \(a_i=1\) and \[ d=\sum_i\deg p_i. \] Since \(g\) also has degree \(d\) and every \(b_i\ge1\), \[ d=\sum_i b_i\deg p_i\ge\sum_i\deg p_i=d, \] so equality forces every \(b_i=1\). Thus \(g\) is a scalar multiple of \(f\) and \([g]=[f]\), even if no square-freeness hypothesis was imposed on \(g\). Hence every failure of injectivity requires repeated factors. Such failures do occur: the distinct cubic forms \(x^2y\) and \(xy^2\) define the same union of the two coordinate lines. This proves part (a).
3. The set of coefficient points whose degree-\(d\) form has a singular projective zero is Zariski closed in \(\PP^N\).
Write the universal degree-\(d\) form as \[ F_A(x,y,z)=\sum_{i+j+k=d}A_{ijk}x^iy^jz^k, \] where the \(A_{ijk}\) are homogeneous coordinates on the coefficient space. A point \(P=[x:y:z]\in\PP^2\) is singular on \(Z(F_A)\) exactly when \[ F_A(P)=\frac{\partial F_A}{\partial x}(P) =\frac{\partial F_A}{\partial y}(P) =\frac{\partial F_A}{\partial z}(P)=0. \] Including \(F_A\) itself is necessary in characteristics dividing \(d\), where Euler’s relation does not recover \(F_A(P)=0\) from the partial derivatives alone.
Apply the elimination theorem to these four homogeneous polynomials in \(x,y,z\) with indeterminate coefficient coordinates \(A_{ijk}\). It produces homogeneous polynomial conditions in the \(A_{ijk}\) whose simultaneous vanishing is equivalent to the existence of a common nonzero solution \((x,y,z)\). Therefore the singular forms constitute a projective algebraic subset \[ \Delta_d\subseteq\PP^N. \] Its complement \(U_d=\PP^N\setminus\Delta_d\) is Zariski open.
4. The open set \(U_d\) is nonempty for every \(d>0\), and every point of \(U_d\) defines an irreducible nonsingular plane curve of degree \(d\).
The card a nonsingular plane curve of each degree in each characteristic gives an explicit degree-\(d\) homogeneous form whose three partial derivatives and defining equation have no common projective zero. Its coefficient point therefore belongs to \(U_d\), so \(U_d\) is nonempty.
If \([f]\in U_d\), then \(Z(f)\) is nonsingular by construction. It is also irreducible. Indeed, if \(f=gh\) with \(g,h\) nonconstant homogeneous forms, then the two positive-degree plane curves \(Z(g)\) and \(Z(h)\) meet in \(\PP^2\). At any point of intersection, \[ \partial(gh)=g\,\partial h+h\,\partial g=0 \] for every first partial derivative, and \(gh=0\) as well, producing a singular point of \(Z(f)\). This contradicts \([f]\in U_d\). Thus \(f\) is irreducible and defines an irreducible nonsingular curve.
5. Irreducible nonsingular degree-\(d\) curves correspond one-to-one with the points of \(U_d\).
Step 4 shows that every point of \(U_d\) gives such a curve. Conversely, the defining form of an irreducible nonsingular degree-\(d\) plane curve has no singular projective zero, so its coefficient point lies in \(U_d\).
If two points of \(U_d\) define the same curve, their defining forms are irreducible and hence square-free. Step 2 then shows that the two forms differ by a scalar, so the two points of \(\PP^N\) are equal. This gives the required bijection and proves part (b).
6. Q.E.D.
Steps <1>1–<1>2 prove the coefficient-space correspondence and its possible nonuniqueness, while steps <1>3–<1>5 identify the irreducible nonsingular curves with the nonempty Zariski-open set \(U_d\).