A line meets a parabola in two points. Two circles in two. A cubic and a line in three. Multiply the degrees of the curves and you get the count.

Three failures of the naive count

Take a line of degree 1 and a parabola of degree 2. Substitute $y = mx + c$ into $y = x²$ and you get a quadratic in x — two roots, two intersections. $1 · 2 = 2$. Clean. Now break it three ways: (a) two parallel lines $y = x$ and $y = x + 1$ — substitute and you get $0 = 1$, a contradiction; zero intersections, not the predicted $1 · 1 = 1$. (b) The line $y = 0$ is tangent to $y = x²$ at the origin — they meet at one point, not two. (c) Two disjoint unit circles $x² + y² = 1$ and $(x − 3)² + y² = 1$ share zero points, not four.

The product rule is right often enough to look like a theorem, and wrong often enough to look like a guess. Either it’s a guess, or the plane is the wrong stage. We take the second option.

Fix #1 — the projective plane

A line is a direction plus an offset. Parallel lines share a direction. Suppose we add a new point to the plane for each direction — call it the point at infinity for that direction. Then any two parallel lines, sharing a direction, share that one extra point. Two non-parallel lines still meet at their usual one place. So every two distinct lines now meet at exactly one point. Add up all the points at infinity (one per direction) and you’ve added one extra line — the line at infinity. The plane plus that line is the projective plane $ℝℙ²$.

The clean way to write a projective point is $[X : Y : Z]$ — three numbers, not all zero, up to overall scale (so $[2 : 2 : 0], [1 : 1 : 0], [5 : 5 : 0]$ are the same point). Affine (ordinary, finite) points are $[x : y : 1]$. Points at infinity are $[X : Y : 0]$: the direction $(X, Y)$. The two parallel lines from §1 both pass through $[1 : 1 : 0]$ — the “direction (1, 1)” point. Failure (a) is fixed.

Fix #2 — multiplicity

A tangent line meets a parabola at one geometric point but a double algebraic point. Substitute $y = 0$ into $y = x²$: the equation becomes $x² = 0$, with $x = 0$ as a repeated root. Polynomials remember what pictures forget. Define the intersection multiplicity of two curves at a point as the order to which the eliminated polynomial vanishes there. A transverse crossing is multiplicity 1; a simple tangency is 2; a triple tangency 3; and so on. Bezout’s count is multiplicity-counted, never raw.

In the widget above, switch to the tangent preset and turn on show multiplicity: the tangent point earns its $×2$ label, and the visible 3 jumps to 4. Failure (b) fixed.

Fix #3 — complex coordinates

The disjoint circles $x² + y² = 1$ and $(x − 3)² + y² = 1$ seem to meet nowhere. Subtract them: $−6x + 9 = 0$, so $x = 3/2$. At that x the first circle gives $y² = 1 − 9/4 = −5/4$ — no real $y$, but two perfectly good complex ones: $y = ± i \sqrt{5} / 2$. The two circles really do meet at two points $(3/2, ± i \sqrt{5} / 2)$ — they just live in $ℂ²$, not $ℝ²$.

In the widget, the disjoint preset shows zero real intersections — turn on show complex, and four open dots appear in the side panel: the four Bezout-promised intersections, projected onto the complex x-plane. Failure (c) fixed.

# The disjoint-circles example, hand-eliminated.
# C1: x² + y² − 1     = 0
# C2: (x − 3)² + y² − 1 = 0
# Subtract: −6x + 9 = 0  →  x = 3/2.
# Plug back into C1: y² = 1 − 9/4 = −5/4  →  y = ± i √5 / 2.
import cmath
x = 3 / 2
ys = (cmath.sqrt(1 - x**2), -cmath.sqrt(1 - x**2))
[(x, y) for y in ys]
# → [(1.5, 1.118j), (1.5, -1.118j)]   (two complex intersections)

The statement

Three fixes in: projective plane , complex coordinates, multiplicity . The product rule is now a theorem.

Why it works — the resultant

Eliminate one variable. Two polynomials $f(x, y), g(x, y)$, viewed as polynomials in $y$ with coefficients depending on $x$, share a root iff their resultant in $y$ vanishes. If $f$ has $y$-degree $d$ and $g$ has $y$-degree $e$, the resultant is a polynomial in $x$ of degree $d · e$ — and its $d · e$ complex roots (with multiplicity) are the $x$-coordinates of the $d · e$ intersections. Two conics: a quartic in $x$. Four roots. Four intersections. The widget above is exactly this quartic, solved.

# Sylvester resultant of two conics in y → quartic in x.
import numpy as np

def resultant_y(c1, c2):
    """Each conic c = (a, b, c, d, e, k) for a x² + b xy + c y² + d x + e y + k.
       In y: A_i y² + B_i(x) y + C_i(x), with A_i = c_i, B_i = b_i x + e_i,
       C_i = a_i x² + d_i x + k_i.  Resultant in y of two quadratics is
       (A1 C2 − A2 C1)² − (A1 B2 − A2 B1)(B1 C2 − B2 C1)."""
    a1, b1, cc1, d1, e1, k1 = c1
    a2, b2, cc2, d2, e2, k2 = c2
    P = np.polynomial.Polynomial
    A1, A2 = cc1, cc2
    B1, B2 = P([e1, b1]), P([e2, b2])
    C1, C2 = P([k1, d1, a1]), P([k2, d2, a2])
    return (A1*C2 - A2*C1)**2 - (A1*B2 - A2*B1)*(B1*C2 - B2*C1)

# General preset: x²/4 + y² − 1 = 0   and   x² + y²/4 − 1 = 0
c1 = (1/4, 0, 1,    0, 0, -1)
c2 = (1,   0, 1/4,  0, 0, -1)
sorted(np.round(resultant_y(c1, c2).roots(), 4))
# → [(-0.8944+0j), (-0.8944+0j), (0.8944+0j), (0.8944+0j)]
#   (each root with multiplicity 2; the two y-values per x come back from c1)

Where this shows up — same count, two pillars

Bezout begins as a counting theorem for intersections, then becomes the bookkeeping that makes elliptic-curve addition reliable. The product $d \cdot e$ is the same number on both sides; only what gets counted differs.

graphics : how many points two visible conics actually share — once
         multiplicity, projective points, and complex roots are counted.
finance  : the third intersection of an elliptic curve with a chord — the
         hidden third point that defines elliptic-curve addition.

Curve intersections — two conics in the plane meet in exactly four points, once you accept the three fixes (projective plane, multiplicity, complex roots). The widget runs the count: tangent contact contributes 2 to the multiplicity, parallel asymptotes contribute points-at-infinity, and disjoint conics meet at four complex points. Bezout decides what “two conics meet” means.

Bitcoin signature — an elliptic curve has degree 3; a chord has degree 1; Bezout says they meet in $3 \cdot 1 = 3$ points. Two of those points are the chord’s endpoints; the third is the sum under elliptic-curve addition. The signature scheme is built on this third-point construction iterated millions of times — the count has to be exact, every time.

Same theorem, different ledger: graphics asks “where do these two curves meet”; finance asks “what is the third point on this chord”. Bezout guarantees the third point exists and is unique, and that is why ECDSA is well-defined.