An approach to polynomials of low degree by way of algebraic geometry

Adam Logan

1. Introduction The purpose of this brief note is to show how some of the basic ideas of algebraic geometry can be used to derive the standard formulas that give the solutions of cubic and quartic polynomials. The interesting history of the discovery of these formulas has been discussed comprehensively in various books and articles and need not be reiterated here; see, for example, [BM, section 15.10]. The results are not new, but we hope that the new viewpoint will illustrate the use of some of the objects and techniques of algebraic geometry.

2. Quadratics As an introduction to our point of view, let us consider a quadratic polynomial ax^{2} + bx + c. Since the factorization of a polynomial is unaffected by scaling, we regard this polynomial as corresponding to the point [a : b : c] of the projective plane ℙ^{2}. Now, we can solve quadratics by completing the square: in other words, we can write

ax^{2} + bx + c = a(x + \frac{b}{2a})^{2} − (\frac{b^{2}}{4a} − c).

This is a difference of squares, and factoring it as such we recover the usual quadratic formula.

Another way to think about this is that we wish to express ax^{2} + bx + c as a linear combination of a square and the constant 1. Geometrically, the constant 1 corresponds to the point [0 : 0 : 1]; a square is a point on the subvariety of ℙ^{2} parametrized by square polynomials, which is the subvariety V (y^{2} − 4xz) (recall that V (f_{1}, …, f_{n}) means the subvariety of ℙ^{k} on which the f_{i} all vanish, or the closed subscheme of ℙ^{k} defined by the ideal (f_{1}, …, f_{n}) of the coordinate ring of ℙ^{k}); and expressing a polynomial ax^{2} + bx + c as a linear combination corresponds to finding the other point of V (y^{2} − 4xz) that meets the line through [a : b : c] and [0 : 0 : 1]. Since V (y^{2} − 4xz) has degree 2, there is one such point other than [0 : 0 : 1], counted with proper multiplicity. This method breaks down when the line through [a : b : c] and [0 : 0 : 1] is the tangent to [0 : 0 : 1], which is the line x = 0. But in this case the polynomial is actually linear.

3. Cubics A direct generalization of this method to cubics is impossible; a general cubic ax^{3} + bx^{2} + cx + d cannot be written in the form (rx + s)^{3} − t^{3}, because there are four parameters in the first expression and only three in the second. However, using the standard identity

x^{3} + y^{3} = (x + y)(x + ωy)(x + ω^{2}y),

where ω is a complex cube root of 1, we can factor any cubic which is the sum of two cubes of linear polynomials. Therefore it is enough to write a cubic in the form a_{1}(x + b_{1})^{3} + a_{2}(x + b_{2})^{3}.

As before, we consider a cubic polynomial to correspond to the point in ℙ^{3} given by its coefficients. We will also need the subvariety C of ℙ^{3} that parametrizes the cubes of linear polynomials. This subvariety is clearly the image of ℙ^{1} under the map [x : y] → [x^{3} : 3x^{2}y : 3xy^{2} : y^{3}]. Proposition 3.1 For a ≠ 0, the tangent line to the point [a^{3} : 3a^{2}b : 3ab^{2} : b^{3}] is the line through this point and [0 : 3a^{2} : 6ab : 3b^{2}]. The tangent line at [1 : 0 : 0 : 0] is the line z = w = 0. In particular, for a ≠ 0 the tangent line at the image of [a : b] parametrizes the polynomials that are multiples of (ax + b)^{2}. Proof. When a ≠ 0 we can safely set a = 1, and then we simply have the parametrization (1 : 3b : 3b^{2} : b^{3}). Differentiating with respect to b, we get the tangent vector (0 : 3 : 6b : 3b^{2}), and homogenizing gives the stated [0 : 3a^{2} : 6ab : 3b^{2}]. The calculation for [1 : 0 : 0 : 0] is similar. The last statement is an immediate consequence.

Theorem 3.2 C is defined by the equations

3xz − y^{2}, 9xw − yz, 3yw − z^{2},

and is of degree 3. Proof. Both parts of this statement can be proved in several different ways. The first part follows from the standard equations for the rational normal curve [x^{3} : x^{2}y : xy^{2} : y^{3}] by a change of coordinates (see, for example, [SD] II.4.5). For the second statement, consider the parametrization. The intersection of the curve with a general plane consists of those points for which 1, 3b, 3b^{2}, b^{3} satisfy a given linear equation; in other words, it is defined by the vanishing of a cubic polynomial, so it has degree 3. More concretely, the intersection with the plane y + z = 0 consists of three points [1 : 0 : 0 : 0], [1 : −3 : 3 : −1], [0 : 0 : 0 : 1], and all of the intersections are transverse.

Now, fix a point P = [a : b : c : d] of ℙ^{3} not lying on C. We need to find a secant line of C through P; in other words, a line through P that meets C in two points. To do this, we project C to ℙ^{2} away from P. A line through P meeting C in two points will map to a singular point of the image. Such a singular point must exist, because the image of C in ℙ^{2} is an irreducible curve of degree 3 (because C is) and geometric genus 0 (because it is the image of a curve with geometric genus 0). But curves of degree 3 have arithmetic genus 1, so there must be one singular point contributing 1 to the arithmetic genus: in other words, either a node or a cusp. A node would be the image of a secant line; a cusp, that of a tangent.

For a general point of ℙ^{3}, the projection is given by the equations bx − ay, cx − az, dx − aw. (Of course these equations are not usable for a = 0; it would be simple matter to write equations for this case, but simpler still to point out that if a = 0 we have a quadratic, not a cubic.)

By a computer calculation, the image of the projection is the subscheme of ℙ^{2} defined by

\begin{matrix}(ad^{2} − c^{3}/27)r^{3} + (−acd + bc^{2}/9)r^{2}s + (−2abd + 2ac^{2}/3)r^{2}t + (abd − b^{2}c/9)rs^{2} & + \\ (3a^{2}d − abc/3)rst + (−3a^{2}c + ab^{2})rt^{2} + (−a^{2}d + b^{3}/27)s^{3} + (a^{2}c − ab^{2}/3)s^{2}t, \end{matrix}

where r, s, t are the coordinates on ℙ^{2}. This curve must be singular, and indeed it has a singularity at

[9a^{2}c^{2} − 6ab^{2}c + b^{4} : 27a^{2}cd − 9ab^{2}d − 3abc^{2} + b^{3}c : 27a^{2}d^{2} − 9abcd + ac^{3} + b^{3}d] = [f : g : h].

This indicates that the line L through P and [0 : f : g : h] meets the curve C twice. (We need not worry about tangents, for it was shown above that if [a : b : c : d] lies on a tangent to C then the corresponding polynomial has a square factor.)

To complete the solution of the cubic, it suffices to indicate how to find the points of intersection of L with C and to express [a : b : c : d] as a linear combination of them. A Groebner basis for the ideal defining L ∩ C is given by

(b^{2}d^{2} − 2bc^{2}d/3 + c^{4}/9)x + (−3a^{2}d^{2} + abcd − ac^{3}/9 − b^{3}d/9)z + (3a^{2}cd − ab^{2}d − abc^{2}/3 + b^{3}c/9)w,
(bd − c^{2}/3)y + (−3ad + bc/3)z + (3ac − b^{2})w,
(bd − c^{2}/3)z^{2} + (−9ad + bc)zw + (9ac − 3b^{2})w^{2}.

The first two of these equations are linear, so the rational points are defined over the field obtained by adjoining the square root of the discriminant of the last polynomial (which is −3 times the usual discriminant of the original cubic at^{3} + bt^{2} + ct + d).

Dividing through by w^{2} and solving the quadratic, we find that

z/w = \frac{9ad − bc ± \sqrt{81a^{2}d^{2} − 54abcd + 12ac^{3} + 12b^{3}d − 3b^{2}c^{2}}}{2(bd − c^{2}/3)} .

Let these values be ρ_{1}, ρ_{2}. The point on the rational normal cubic with z/w = ρ_{1} corresponds to the polynomial (tρ_{1}/3 + 1)^{3}, so we have

at^{3} + bt^{2} + ct + d = k_{1}(tρ_{1}/3 + 1)^{3} + k_{2}(tρ_{2}/3 + 1)^{3}.

Equating coefficients of 1 and t gives k_{1} + k_{2} = d, k_{1}ρ_{1} + k_{2}ρ_{2} = c, so k_{1} = (c − dρ_{2})/(ρ_{1} − ρ_{2}) and k_{2} = (dρ_{1} − c)/(ρ_{1} − ρ_{2}). We have now expressed the general cubic as a sum of two cubes, as desired. Let κ = ^{3} \sqrt{k_{1}/k_{2}}. Then

\begin{aligned}at^{3} + bt^{2} + ct + d & = k_{2}((κtρ_{1}/3 + κ)^{3} + (tρ_{2}/3 + 1)^{3}) \\ & = k_{2}(κtρ_{1}/3 + κ + tρ_{2}/3 + 1)(κtρ_{1}/3 + κ + ω(tρ_{2}/3 + 1))(κtρ_{1}/3 + κ + ω^{2}(tρ_{2}/3 + 1)), \end{aligned}

from which we observe that the roots are

\frac{−3(κ + 1)}{κρ_{1} + ρ_{2}}, \frac{−3(κ + ω)}{κρ_{1} + ωρ_{2}}, \frac{−3(κ + ω^{2})}{κρ_{1} + ω^{2}ρ_{2}}.

Clearly these do not depend on the choice of κ from the three cube roots of k_{1}/k_{2}. Nor does interchanging ρ_{1} and ρ_{2} affect this list: that would switch k_{1} and k_{2} and thus change κ to 1/κ. Performing these two operations together does not affect these quantities. Also, it is easily verified that they are indeed the roots of the original equation.

4. Quartics Now let us apply some further techniques to study the general quartic equation. The original solution, due to Ferrari, relies on writing the quartic as a difference of two squares, one of a quadratic and one of a linear polynomial. In other words, we wish to find a line through a general point that meets both the 2-dimensional variety parametrizing squares of quadratics and the 1-dimensional variety parametrizing squares of linear polynomials.

Let us count the dimensions. The family of lines in ℙ^{4} has dimension 6; to meet a variety of codimension d is d − 1 conditions. Since we need our line to meet varieties of codimension 4, 3, 2, we are imposing six many conditions, so we expect that in general finitely many lines satisfy them.

Let us determine the expected number. If the varieties of codimension 2 and 3 were of degree 1, then we would be counting lines through a point, meeting a given line and a given plane. The lines through a point meeting a given line sweep out a plane, which meets a given plane in one point, so we would expect one. Now, in fact the variety of codimension 2 is of degree 4, and that of codimension 3 is of degree 2, so we expect eight. However, there is an excess intersection because the varieties do not intersect properly: every square of a linear polynomial is also the square of a quadratic polynomial, and so, putting b_{0} = 0, b_{1} = a_{0}, b_{2} = a_{1}, we find that the squares of these polynomials are already linearly dependent and that there is always a 1-dimensional intersection, which we must account for.

To deal with this problem, we use the formula of example 9.1.7 of [F]:

(X_{1} · . . . · X_{r})^{Z} = ∑_{i=1}^{r}(K_{X_{i}} · Z)_{X_{i}} − (r − 1)(K_{Y} · Z)_{Y} − K_{Z},

where the X_{i} are subvarieties of Y whose intersection has expected dimension of 0 but which actually meet in the curve Z and where (X_{1} · . . . · X_{r})^{Z} is the contribution to the intersection number from the intersection along Z.

This requires restating the problem so that the solutions that we are interested in are not contained in the excess intersection component. Thus, we need two projective spaces of degree 4, one to contain the squares of linear polynomials and the other to contain the squares of quadratic polynomials. Let V_{1} be the subvariety of ℙ^{4} whose points correspond to squares of linear polynomials and V_{2} the subvariety of squares of quadratic polynomials. Following the notation of [F], let Y = ℙ^{4}×ℙ^{4}, and let X_{1} = V_{1}×ℙ^{4} and X_{2} = ℙ^{4}×V_{2}. In addition, fixing a point P of ℙ^{4} corresponding to the quartic we are interested in, let X_{3} be the subvariety of Y of pairs of points (p_{1}, p_{2}) such that P, p_{1}, p_{2} are collinear. We assume that P does not lie in the plane of V_{1}, for if so then P corresponds to a quadratic polynomial.

Unfortunately X_{3} is singular at the point (P, P), so the formula cannot be used directly. In addition, it is not as easy as one might hope to determine the canonical divisor of a desingularization of X_{3}. Luckily there is a way around this. Proposition 4.1 X_{1} ∩ X_{3} is isomorphic to the blowup of the cone over V_{1} at its vertex; in particular it is nonsingular. Further, V_{2} is isomorphic to ℙ^{2}, so X_{2} is also nonsingular. Proof. For the first statement, choose a ℙ^{3} containing V_{1} and coordinates x_{0}, …, x_{3} on it. By definition, the cone over V_{1} is the subvariety of ℙ^{4} defined by the equations of V_{1}, viewed as equations in ℙ^{4}. Blowing this up at the vertex (0, 0, 0, 0, 1), we obtain the subvariety of ℙ^{4} × ℙ^{3} in which the first component is either (0, 0, 0, 0, 1) or a point whose first five coordinates are the homogeneous coordinates of a point on V_{1}, while the second component is any point of V_{1} in the first case or the corresponding point in the second case. If we change coordinates to send V_{1} to where it had been and (0, 0, 0, 0, 1) to P, the claim becomes obvious.

For V_{2}, there is an obvious map from ℙ^{2} to V_{2} given by squaring: more precisely, the map takes (x_{0} : x_{1} : x_{2}) to (x_{0}^{2} : 2x_{0}x_{1} : 2x_{0}x_{2} + x_{1}^{2} : 2x_{1}x_{2} : x_{2}^{2}). This can be viewed as the 2-uple embedding of ℙ^{2} in ℙ^{5} followed by projection from [0 : 0 : 1 : −2 : 0 : 0], which is not on the secant variety, and hence is an isomorphism. Alternatively, it is not difficult to write explicit inverses for the map on coordinate patches that cover all of V_{2}.

It follows that X_{1} ∩ X_{3} is isomorphic to the Hirzebruch surface F_{2} (see, for example, [Har, example V.2.11.4]). The exceptional section S of self-intersection −2 is the curve {(p_{1}, P) : p_{1} ∈ V_{1}}. On the other hand, a fibre F is given by {(p_{0}, q)} for fixed p_{0}. The canonical divisor meets S with multiplicity 0 and F with multiplicity −2, by the adjunction formula, so it is −2S − 4F. Note that S is a curve of degree (2, 0) while F has degree (0, 1). Proposition 4.2 The excess intersection component E = {(x, x) : x ∈ V_{1}} contributes 5 to the intersection calculation. Proof. From example 9.1.7 of [F], the number is equal to the degree of K_{X_{1}∩X_{3}} · E + K_{X_{2}} · E − K_{Y} · E − K_{E}. To evaluate the first term, we compute S · E and F · E. Clearly S · E is empty, because P ∉ V_{1}, while F · E = 1, the intersection being the point (p_{0}, p_{0}). Combining this with the formula for K_{X_{1}∩X_{3}} above, we find that K_{X_{1}∩X_{3}} · E has degree −4. Since V_{2} is isomorphic to ℙ^{2}, its canonical divisor is −3 times the class of the image of a line. But V_{1} as a subvariety of V_{2} is also the image of a line, so the intersection of V_{1} with K_{V_{2}} is −3. The canonical class of ℙ^{4} × V_{2} being (K_{ℙ^{4}} × V_{2}) + (ℙ^{4} × K_{V_{2}}), its intersection with E is −10 − 3 = −13, since the canonical class of ℙ^{4} is O(−5). That means that K_{Y} = K_{ℙ^{4}} × ℙ^{4} + ℙ^{4} × K_{ℙ^{4}} has intersection −5 · 2 − 5 · 2 = −20 with E. Finally, E is a rational curve, so K_{E} has degree −2. Thus the total is −4 − 13 − (−20) − (−2) = 5.

Therefore there are 8 − 5 = 3 nondegenerate points of intersection. Hence we recover Ferrari’s result that solving a quartic equation can be reduced to solving a cubic.

5. References

[BM] Boyer, C., A history of mathematics (revised by U. Merzbach). New York, Toronto: Wiley, 1989.

[F] Fulton, W., Intersection theory (second edition). Berlin, Heidelberg: Springer-Verlag, 1998.

[Har] Hartshorne, R., Algebraic geometry. GTM 52. New York: Springer-Verlag, 1977.

[SD] V. V. Shokurov and V. I. Danilov, Algebraic Geometry I. Encyclopaedia of Mathematical Sciences 23. Translated by D. Coray and V. N. Shokurov Berlin, Heidelberg: Springer-Verlag, 1994.