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An algebraic (affine) plane curve of degree ${d}$ over some field ${k}$ is a curve ${\gamma}$ of the form

$\displaystyle \gamma = \{ (x,y) \in k^2: P(x,y) = 0 \}$

where ${P}$ is some non-constant polynomial of degree ${d}$. Examples of low-degree plane curves include

• Degree ${1}$ (linear) curves ${\{ (x,y) \in k^2: ax+by=c\}}$, which are simply the lines;
• Degree ${2}$ (quadric) curves ${\{ (x,y) \in k^2: ax^2+bxy+cy^2+dx+ey+f=0\}}$, which (when ${k={\bf R}}$) include the classical conic sections (i.e. ellipses, hyperbolae, and parabolae), but also include the reducible example of the union of two lines; and
• Degree ${3}$ (cubic) curves ${\{ (x,y) \in k^2: ax^3+bx^2y+cxy^2+dy^3+ex^2+fxy+gy^2+hx+iy+j=0\}}$, which include the elliptic curves ${\{ (x,y) \in k^2: y^2=x^3+ax+b\}}$ (with non-zero discriminant ${\Delta := -16(4a^3+27b^2)}$, so that the curve is smooth) as examples (ignoring some technicalities when ${k}$ has characteristic two or three), but also include the reducible examples of the union of a line and a conic section, or the union of three lines.
• etc.

Algebraic affine plane curves can also be extended to the projective plane ${{\Bbb P} k^2 = \{ [x,y,z]: (x,y,z) \in k^3 \backslash 0 \}}$ by homogenising the polynomial. For instance, the affine quadric curve ${\{ (x,y) \in k^2: ax^2+bxy+cy^2+dx+ey+f=0\}}$ would become ${\{ [x,y,z] \in {\Bbb P} k^2: ax^2+bxy+cy^2+dxz+eyz+fz^2=0\}}$.

One of the fundamental theorems about algebraic plane curves is Bézout’s theorem, which asserts that if a degree ${d}$ curve ${\gamma}$ and a degree ${d'}$ curve ${\gamma'}$ have no common component, then they intersect in at most ${dd'}$ points (and if the underlying field ${k}$ is algebraically closed, one works projectively, and one counts intersections with multiplicity, they intersect in exactly ${dd'}$ points). Thus, for instance, two distinct lines intersect in at most one point; a line and a conic section intersect in at most two points; two distinct conic sections intersect in at most four points; a line and an elliptic curve intersect in at most three points; two distinct elliptic curves intersect in at most nine points; and so forth. Bézout’s theorem is discussed in this previous post.

From linear algebra we also have the fundamental fact that one can build algebraic curves through various specified points. For instance, for any two points ${A_1,A_2}$ one can find a line ${\{ (x,y): ax+by=c\}}$ passing through the points ${A_1,A_2}$, because this imposes two linear constraints on three unknowns ${a,b,c}$ and is thus guaranteed to have at least one solution. Similarly, given any five points ${A_1,\ldots,A_5}$, one can find a quadric curve passing through these five points (though note that if three of these points are collinear, then this curve cannot be a conic thanks to Bézout’s theorem, and is thus necessarily reducible to the union of two lines); given any nine points ${A_1,\ldots,A_9}$, one can find a cubic curve going through these nine points; and so forth. This simple observation is one of the foundational building blocks of the polynomial method in combinatorial incidence geometry, discussed in these blog posts.

In the degree ${1}$ case, it is always true that two distinct points ${A, B}$ determine exactly one line ${\overleftrightarrow{AB}}$. In higher degree, the situation is a bit more complicated. For instance, five collinear points determine more than one quadric curve, as one can simply take the union of the line containing those five points, together with an arbitrary additional line. Similarly, eight points on a conic section plus one additional point determine more than one cubic curve, as one can take that conic section plus an arbitrary line going through the additional point. However, if one places some “general position” hypotheses on these points, then one can recover uniqueness. For instance, given five points, no three of which are collinear, there can be at most one quadric curve that passes through these points (because these five points cannot lie on the union of two lines, and by Bézout’s theorem they cannot simultaneously lie on two distinct conic sections).

For cubic curves, the situation is more complicated still. Consider for instance two distinct cubic curves ${\gamma_0 = \{ P_0(x,y)=0\}}$ and ${\gamma_\infty = \{P_\infty(x,y)=0\}}$ that intersect in precisely nine points ${A_1,\ldots,A_9}$ (note from Bézout’s theorem that this is an entirely typical situation). Then there is in fact an entire one-parameter family of cubic curves that pass through these points, namely the curves ${\gamma_t = \{ P_0(x,y) + t P_\infty(x,y) = 0\}}$ for any ${t \in k \cup \{\infty\}}$ (with the convention that the constraint ${P_0+tP_\infty=0}$ is interpreted as ${P_\infty=0}$ when ${t=\infty}$).

In fact, these are the only cubics that pass through these nine points, or even through eight of the nine points. More precisely, we have the following useful fact, known as the Cayley-Bacharach theorem:

Proposition 1 (Cayley-Bacharach theorem) Let ${\gamma_0 = \{ P_0(x,y)=0\}}$ and ${\gamma_\infty = \{P_\infty(x,y)=0\}}$ be two cubic curves that intersect (over some algebraically closed field ${k}$) in precisely nine distinct points ${A_1,\ldots,A_9 \in k^2}$. Let ${P}$ be a cubic polynomial that vanishes on eight of these points (say ${A_1,\ldots,A_8}$). Then ${P}$ is a linear combination of ${P_0,P_\infty}$, and in particular vanishes on the ninth point ${A_9}$.

Proof: (This proof is based off of a text of Husemöller.) We assume for contradiction that there is a cubic polynomial ${P}$ that vanishes on ${A_1,\ldots,A_8}$, but is not a linear combination of ${P_0}$ and ${P_\infty}$.

We first make some observations on the points ${A_1,\ldots,A_9}$. No four of these points can be collinear, because then by Bézout’s theorem, ${P_0}$ and ${P_\infty}$ would both have to vanish on this line, contradicting the fact that ${\gamma_0, \gamma_\infty}$ meet in at most nine points. For similar reasons, no seven of these points can lie on a quadric curve.

One consequence of this is that any five of the ${A_1,\ldots,A_9}$ determine a unique quadric curve ${\sigma}$. The existence of the curve follows from linear algebra as discussed previously. If five of the points lie on two different quadric curves ${\sigma,\sigma'}$, then by Bezout’s theorem, they must share a common line; but this line can contain at most three of the five points, and the other two points determine uniquely the other line that is the component of both ${\sigma}$ and ${\sigma'}$, and the claim follows.

Now suppose that three of the first eight points, say ${A_1,A_2,A_3}$, are collinear, lying on a line ${\ell}$. The remaining five points ${A_4,\ldots,A_8}$ do not lie on ${\ell}$, and determine a unique quadric curve ${\sigma}$ by the previous discussion. Let ${B}$ be another point on ${\ell}$, and let ${C}$ be a point that does not lie on either ${\ell}$ or ${\sigma}$. By linear algebra, one can find a non-trivial linear combination ${Q = aP + bP_0 + cP_\infty}$ of ${P,P_0,P_\infty}$ that vanishes at both ${B}$ and ${C}$. Then ${Q}$ is a cubic polynomial that vanishes on the four collinear points ${A_1,A_2,A_3,B}$ and thus vanishes on ${\ell}$, thus the cubic curve defined by ${Q}$ consists of ${\ell}$ and a quadric curve. This curve passes through ${A_4,\ldots,A_8}$ and thus equals ${\sigma}$. But then ${C}$ does not lie on either ${\ell}$ or ${\sigma}$ despite being a vanishing point of ${Q}$, a contradiction. Thus, no three points from ${A_1,\ldots,A_8}$ are collinear.

In a similar vein, suppose next that six of the first eight points, say ${A_1,\ldots,A_6}$, lie on a quadric curve ${\sigma}$; as no three points are collinear, this quadric curve cannot be the union of two lines, and is thus a conic section. The remaining two points ${A_7, A_8}$ determine a unique line ${\ell = \overleftrightarrow{A_7A_8}}$. Let ${B}$ be another point on ${\sigma}$, and let ${C}$ be another point that does not lie on either ${\ell}$ and ${\sigma}$. As before, we can find a non-trivial cubic ${Q = aP + bP_0+cP_\infty}$ that vanishes at both ${B, C}$. As ${Q}$ vanishes at seven points of a conic section ${\sigma}$, it must vanish on all of ${\sigma}$, and so the cubic curve defined by ${Q}$ is the union of ${\sigma}$ and a line that passes through ${A_7}$ and ${A_8}$, which must necessarily be ${\ell}$. But then this curve does not pass through ${C}$, a contradiction. Thus no six points in ${A_1,\ldots,A_8}$ lie on a quadric curve.

Finally, let ${\ell}$ be the line through the two points ${A_1,A_2}$, and ${\sigma}$ the quadric curve through the five points ${A_3,\ldots,A_7}$; as before, ${\sigma}$ must be a conic section, and by the preceding paragraphs we see that ${A_8}$ does not lie on either ${\sigma}$ or ${\ell}$. We pick two more points ${B, C}$ lying on ${\ell}$ but not on ${\sigma}$. As before, we can find a non-trivial cubic ${Q = aP + bP_0+cP_\infty}$ that vanishes on ${B, C}$; it vanishes on four points on ${\ell}$ and thus ${Q}$ defines a cubic curve that consists of ${\ell}$ and a quadric curve. The quadric curve passes through ${A_3,\ldots,A_7}$ and is thus ${\sigma}$; but then the curve does not pass through ${A_8}$, a contradiction. This contradiction finishes the proof of the proposition. $\Box$

I recently learned of this proposition and its role in unifying many incidence geometry facts concerning lines, quadric curves, and cubic curves. For instance, we can recover the proof of the classical theorem of Pappus:

Theorem 2 (Pappus’ theorem) Let ${\ell, \ell'}$ be two distinct lines, let ${A_1,A_2,A_3}$ be distinct points on ${\ell}$ that do not lie on ${\ell'}$, and let ${B_1,B_2,B_3}$ be distinct points on ${\ell'}$ that do not lie on ${\ell}$. Suppose that for ${ij=12,23,31}$, the lines ${\overleftrightarrow{A_i B_j}}$ and ${\overleftrightarrow{A_j B_i}}$ meet at a point ${C_{ij}}$. Then the points ${C_{12}, C_{23}, C_{31}}$ are collinear.

Proof: We may assume that ${C_{12}, C_{23}}$ are distinct, since the claim is trivial otherwise.

Let ${\gamma_0}$ be the union of the three lines ${\overleftrightarrow{A_1 B_2}}$, ${\overleftrightarrow{A_2 B_3}}$, and ${\overleftrightarrow{A_3 B_1}}$ (the purple lines in the first figure), let ${\gamma_1}$ be the union of the three lines ${\overleftrightarrow{A_2 B_1}}$, ${\overleftrightarrow{A_3 B_2}}$, and ${\overleftrightarrow{A_1 B_3}}$ (the dark blue lines), and let ${\gamma}$ be the union of the three lines ${\ell}$, ${\ell'}$, and ${\overleftrightarrow{C_{12} C_{23}}}$ (the other three lines). By construction, ${\gamma_0}$ and ${\gamma_1}$ are cubic curves with no common component that meet at the nine points ${A_1,A_2,A_3,B_1,B_2,B_3,C_{12},C_{23},C_{31}}$. Also, ${\gamma}$ is a cubic curve that passes through the first eight of these points, and thus also passes through the ninth point ${C_{31}}$, by the Cayley-Bacharach theorem. The claim follows (note that ${C_{31}}$ cannot lie on ${\ell}$ or ${\ell'}$). $\Box$

The same argument gives the closely related theorem of Pascal:

Theorem 3 (Pascal’s theorem) Let ${A_1,A_2,A_3,B_1,B_2,B_3}$ be distinct points on a conic section ${\sigma}$. Suppose that for ${ij=12,23,31}$, the lines ${\overleftrightarrow{A_i B_j}}$ and ${\overleftrightarrow{A_j B_i}}$ meet at a point ${C_{ij}}$. Then the points ${C_{12}, C_{23}, C_{31}}$ are collinear.

Proof: Repeat the proof of Pappus’ theorem, with ${\sigma}$ taking the place of ${\ell \cup \ell'}$. (Note that as any line meets ${\sigma}$ in at most two points, the ${C_{ij}}$ cannot lie on ${\sigma}$.) $\Box$

One can view Pappus’s theorem as the degenerate case of Pascal’s theorem, when the conic section degenerates to the union of two lines.

Finally, Proposition 1 gives the associativity of the elliptic curve group law:

Theorem 4 (Associativity of the elliptic curve law) Let ${\gamma := \{ (x,y) \in k^2: y^2 = x^3+ax+b \} \cup \{O\}}$ be a (projective) elliptic curve, where ${O := [0,1,0]}$ is the point at infinity on the ${y}$-axis, and the discriminant ${\Delta := -16(4a^3+27b^2)}$ is non-zero. Define an addition law ${+}$ on ${\gamma}$ by defining ${A+B}$ to equal ${-C}$, where ${C}$ is the unique point on ${\gamma}$ collinear with ${A}$ and ${B}$ (if ${A,B}$ are disjoint) or tangent to ${A}$ (if ${A=B}$), and ${-C}$ is the reflection of ${C}$ through the ${x}$-axis (thus ${C, -C, O}$ are collinear), with the convention ${-O=O}$. Then ${+}$ gives ${\gamma}$ the structure of an abelian group with identity ${O}$ and inverse ${-}$.

Proof: It is clear that ${O}$ is the identity for ${+}$, ${-}$ is an inverse, and ${+}$ is abelian. The only non-trivial assertion is associativity: ${(A+B)+C = A+(B+C)}$. By a perturbation (or Zariski closure) argument, we may assume that we are in the generic case when ${O,A,B,C,A+B,B+C,-(A+B), -(B+C)}$ are all distinct from each other and from ${-((A+B)+C), -(A+(B+C))}$. (Here we are implicitly using the smoothness of the elliptic curve, which is guaranteed by the hypothesis that the discriminant is non-zero.)

Let ${\gamma'}$ be the union of the three lines ${\overleftrightarrow{AB}}$, ${\overleftrightarrow{C(A+B)}}$, and ${\overleftarrow{O(B+C)}}$ (the purple lines), and let ${\gamma''}$ be the union of the three lines ${\overleftrightarrow{O(A+B)}}$, ${\overleftrightarrow{BC}}$, and ${\overleftrightarrow{A(B+C)}}$ (the green lines). Observe that ${\gamma'}$ and ${\gamma}$ are cubic curves with no common component that meet at the nine distinct points ${O, A, B, C, A+B, -(A+B), B+C, -(B+C), -((A+B)+C)}$. The cubic curve ${\gamma''}$ goes through the first eight of these points, and thus (by Proposition 1) also goes through the ninth point ${-((A+B)+C)}$. This implies that the line through ${A}$ and ${B+C}$ meets ${\gamma}$ in both ${-(A+(B+C))}$ and ${-((A+B)+C)}$, and so these two points must be equal, and so ${(A+B)+C=A+(B+C)}$ as required. $\Box$

One can view Pappus’s theorem and Pascal’s theorem as a degeneration of the associativity of the elliptic curve law, when the elliptic curve degenerates to three lines (in the case of Pappus) or the union of one line and one conic section (in the case of Pascal’s theorem).