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§4. Multiple or Singular Points 55

In the special case of a nonsingular point w : x : y on C f (k), the polynomial
ϕ(t), given above, will have a simple root for all lines L except the tangent line
(at, bt) where bx − ay is the leading form of f (1, x, y). In other words,

f (1, x, y) = (bx − ay) + f 2 (x, y) +· · · + f d (x, y).

Then in that case the order of the root t = 0of ϕ(t) is ≥ 1. This leads to the following
definition which is used in the study of cubic curves.

(4.7) Definition. A point of inflection (or flex) is a nonsingular point P = w : x : y
on a curve C f such that i(P; L, C f ) ≥ 3 for the tangent line L at P.

In other words, if 1 : 0 : 0 is a flex of the curve C f with a tangent line given by
the equation f 1 (x, y) = bx − ay = 0, then the polynomial f (1, x, y) has the fol-
lowing form:

f (1, x, y) = f 1 (x, y) +· · · + f d (x, y),
where f i (x, y) is homogeneous of degree i with f 1 (x, y) dividing f 2 (x, y).Since
any flex, by projective transformation, can be moved to the origin 1 : 0 : 0, we see
that a nonsingular conic will have no flexes.
For a curve C f of degree d with f (w, x, y) ∈ k[w, x, y] d we consider the con-
dition of f that 1 : r : s is a flex on C f . We expand

f (1, x + r, y + s) = f 0 (r, s) + f 1 (r, s)(x, y) + ··· + f d (r, s)(x, y),

where f i (r, s)(x, y) ∈ k[r, s][x, y] i . Setting x = y = 0, we see that f (1,r, s) =
f 0 (r, s) and hence 1 : r : s ∈ C f if and only if f 0 (r, s) = 0.
Now assume that 1 : r : s ∈ C f .Then1 : r : s is a nonsingular point if and only
if f 1 (r, s)(x, y) = b(r, s)x − a(r, s)y is nonzero in k[r, s][x, y] 1 . A smooth point
1: r : s on C f is a flex if and only if f 1 (r, s)(x, y) divides f 2 (r, s)(x, y) in k[x, y].
In terms of the resultant R(r, s) = R( f 1 (r, s), f 2 (r, s)) ∈ k[r, s] we see that 1 : r : s
is a flex of C f if and only if f (1,r, s) = 0and R(r, s) = 0. This resultant can
2
be calculated as a 3 by 3 determinant for f 2 (r, s)(x, y) = A(r, s)x + B(r, s)xy +
2
C(r, s)y as follows:
2 2
R(r, s) = A(r, s)a(r, s) + B(r, s)a(r, s)b(r, s) + C(r, s)b(r, s) .
Since for degree of f strictly bigger that 2 the curve C f given by f (1, x, y) = 0
and the resultant R(x, y) = 0 have an intersection point 1 : r : s we deduce the
following result.

(4.8) Proposition. Let C f be an algebraic plane curve without singularities of de-
gree d ≥ 3 over an algebraically closed field k. Then C f has at least one flex.