Week 8   Projective Geometry
Some Algebraic Methods

© Dana Scott 1993/1997/1998
Notebook  last edited: 14 October 1998

Doing Algebraic Arguments

Generation and bases

Generation by powers

We have used monomials extensively as bases, but the polynomials have many other generating sets — since a vector space of higher dimension always has many choices of bases.  In fact,  powers are adequate.

Theorem. The linear space (ℂ^n)[u, v, w] of homogeneous uvw-polynomials of degree n is generated (as a vector space) by the nth powers of the linear polynomials.

Proof:  We can see from the proof of the Evaluation Theorem that the expression PD[f, g] cannot vanish on all the nth powers of linear uvw-polynomials unless f is indeed identically zero.

Because the two spaces (ℂ^n)[x, y, z]and (ℂ^n)[u, v, w]are dual to each other as vector spaces, this means that the subspace generated by the nth powers of the linear polynomials must span the whole space of nth-degree homogeneous uvw-polynomials; since, otherwise, there would be an f, not zero, where the expression PD[f, g] did vanish on the subspace.  Q.E.D.

As in any vector space, there are many choices of generating sets for the space of nth-degree polynomials, and the proof above does not suggest any particular basis (being a proof by contradiction).  Here follows a neat proof that gives one maximal array of points in what we might call “nth-degree” general position:

Theorem. The powers  (((i u + j v + w))^n),  where the coefficients run 	
	i  =  0, 1,  ...  , n, and  j  =  0, 1,  ...  , i,  
form an independent basis for the space  (ℂ^n)[u, v, w] of homogeneous uvw-polynomials of degree  n.

Proof:  (by Marko Petkovsek). The points in question form a triangular array with the property that for each point  p  in the array,  there exist  n  lines in the plane which together cover all the points of the array except  p.  These lines are obtained by approaching  p  from the outside with lines parallel to the sides of the triangle. We illustrate this for  n = 6.  We need 28 points.

Show[
      Graphics[Join[{Hue[1], PointSize[0.05]}, 
          Point/@Join@@Table[{i, j}, {i, 0, 6}, {j, 0, i}], {
            GrayLevel[0], Point[{3, 1}], Thickness[0.001], 
            Line[{{0, 0}, {6, 0}, {6, 6}, {0, 0}}], Line[{{1, 0}, {6, 5}}], 
            Line[{{4, 0}, {4, 4}}], Line[{{5, 0}, {5, 5}}]}]], 
      AspectRatio -> Automatic];

In order to show that the nth-powers of the points are independent, it is sufficient to find an nth-degree polynomial that vanishes at all points of the array except at one.  (Why?)  Now, to each line there of course corresponds a linear xyz-polynomial which vanishes at points on the line and is non-zero everywhere else.  The product  f  of these linear polynomials corresponding to the  n  lines of a covering which leaves a point  p  exposed,  is a polynomial of degree  n  which vanishes at every point of the array except at p.  Q.E.D.

The argument about independence of n-th powers would apply to any configuration with the property used in the above proof.  Since the dimension of the space of nth degree homogeneous polynomials is known, we can derive an immediate corollary.

Corollary.  A configuration of points in the plane arranged so that removing one point always allows the remainder to be covered with n lines — none passing through the removed point — has cardinality at most (1/2) (n + 1) (n + 2).

Here is another corollary.

Corollary.  An nth-degree polynomial vanishing at all points in such a configuration must be identically zero.  Otherwise said, there is no non-trivial nth-degree curve passing through all of the points.

Comment.  This is still not a very constructive proof, and it would be interesting to have explicit formulae to show how every monomial is a sum of nth powers.  (Petkovesek found one such algorithm.)  One thing that makes this problem somewhat messy is that there is no unique basis of powers and no unique representation.  And what about representing a polynomial as a minimal number of nth powers?

Here is an application to finding a general formula that can be derived using powers.

Corollary.  If n  ≥  m,  and f is a linear xyz-polynomial, and if h is any m-degree uvw-polynomial, then:
		PD[(f^n), h]==(({{n}, {m}})) PD[(f^m), h] (f^(n - m)).

Proof:  Inasmuch as both sides of the equation are linear in h, we only have to prove this for mth powers of of linear polynomials.  (Why?)  Note also that by previous theorems PD[(f^m), h]is the evaluation of h at the coefficients of f.

Now in case h == (g^m), where g is linear, then by a previous theorem we have:
		PD[(f^n), (g^m)] == 
        ((n! PD[f, g]^m (f^(n - m)))/((n - m)!))				
But when m == n, we have:
		PD[(f^m), (g^m)] == m! PD[f, g]^m				
We thus derive:
		PD[(f^n), (g^m)] == 
        ((n! PD[(f^m), (g^m)] (f^(n - m)))/((n - m)! 
            m!))				
This establishes what we wanted for mth powers.  Q.E.D.

Invariance of polydifferentiation

We can consider our study as involving the vector space (ℂ^n)[x, y, z] of n-degree homogeneous xyz-polynomials on the one hand (with complex coefficients in ℂ), the vector space  (ℂ^m)[u, v, w] of m-degree homogeneous uvw-polynomials on the other hand.  These vector spaces are closed under addition and scalar multiplication.  The operation of multiplication can be considered as an operation which adds degrees:
	· : (ℂ^n)[x, y, z]× 
            (ℂ^m)[x, y, z]⟶ 
              (ℂ^(n + m))[x, y, z], and
	· : (ℂ^n)[u, v, w]× 
            (ℂ^m)[u, v, w]⟶ 
              (ℂ^(n + m))[u, v, w]

Warning.  Do not forget that to make sense of these remarks the exceptional 0 polynomial has to be considered here as having every degree.

Polydifferentiation, however, subtracts degrees.  In case m ≤ n, we have:
	PD:(ℂ^n)[x, y, z]× (ℂ^m)[u, v, w]⟶ (ℂ^(n-m))[x, y, z].
But, in case m ≥ n, we have:
	PD:(ℂ^n)[x, y, z]× (ℂ^m)[u, v, w]⟶ (ℂ^(m - n))[u, v, w].
And then, if  m == n, we have:
	PD : (ℂ^n)[x, y, z]× 
            (ℂ^n)[u, v, w]⟶ 
              ℂ.
And this last case is how we consider having dual vector spaces.

When M is a non-singular matrix, then the corresponding linear transformations map polynomials in this way:
	TP[M] : 
          (ℂ^n)[u, v, w]⟶ 
            (ℂ^n)[u, v, w], and
	TL[M] : 
          (ℂ^n)[x, y, z]⟶ 
            (ℂ^n)[x, y, z].

?TP

Warning.  In each case above we have a non-singular linear transformation of the indicated space, but keep in mind that not all linear transformations are of the form TP[M] or TL[M] when we consider degrees n > 1.  (Why?)

Comment.  The question of whether two higher-degree polynomials are projectively equivalent by a TP[M] or TL[M] is an interesting and sometimes difficult question.

However, the projective transformations as represented by matrices do preserve polydifferentiation.  We can say just as projective transformations are algebra automorphisms, they are also differentiation automorphisms.  Specifically, we have:

Theorem.  (i)  If M is a non-singular matrix, m ≤ n, and f is an n-degree xyz-polynomial, and g is an m-degree uvw-polynomial, then we have:
		TL[M][PD[f, g]] == PD[TL[M][f], TP[M][g]].
(ii)  In case m ≥ n, there is a dual result with TP[M] on the left.  
(iii)  In case m == n, then there is no transformation on the left, as scalars are always left fixed.

Proof:  As we already know, the result is true for linear (= first-degree) polynomials by the definition of the transformations based on the fact that the two linear spaces of linear polynomials are dual to each other under PD as dot product.

Now, projective transformations are defined on higher-degree polynomials by substitution, and they are therefore automorphisms of the ring of polynomials.  Because, the desired formula in the theorem is bilinear in f and g, we need only prove it for powers of linear polynomials.  In this case it follows from the linear result by invoking a formula proved before:
	TL[M][PD[(l^n), (p^m)]]
		== ((n!)/((n - m)!)) PD[l, p]^m (((TL[M][l]))^(n - m)) 
		== ((n!)/((n - m)!)) PD[TL[M][l], TP[M][p]]^m TL[M][l]^(n - m)
		== PD[TL[M][l]^n, TP[M][p]^m]

Q.E.D.

Comment.  A consequence of this important result is the observation that any (homogeneous) relationships we define by combining algebra with polydifferentiation must have geometric significance, because the relationship must be preserved by all projective transformations.  This is a very considerable generalization of the principle that projective transformations preserve the incidence of points and lines.

Polynomial independence

Independence notions for points

Definition.  A sequence of k points is said to be n-independent if, and only if, the space of n-degree curves passing through them has projective dimension (1/2) n (n + 3) - k.

The above is the geometric definition of independence, but we can translate the condition into an algebraic test at once:

Proposition.  Points (Subscript[p, 1]), ... , (Subscript[p, k]) are n-independent if, and only if, the nth powers  
(Subscript[p, 1]^n), ... , (Subscript[p, k]^n)  are linearly independent as uvw-polynomials.

Here are some results about indepencence.

Theorem. If k ≤ n+1, then any sequence of k distinct points is n-independent.

Proof:  We argue by induction on n.  The basis case of n == 1 is obvious.

Now assume the result for n-1 and argue for n.  Suppose the nth powers   (Subscript[p, 1]^n), ... , (Subscript[p, k]^n) are linearly dependent.  This means that we have a relationship:
 		  (Subscript[a, 1]) (Subscript[p, 1]^n)  +   ...  + (Subscript[(a ), k]) (Subscript[p, k]^n)  ==  0
with not all coefficients equal to 0.  Indeed, at least two of the coefficients must be non-zero.  Let m be a line passing through (Subscript[p, k]) and not passing through any of the other points.  By differentiation of the above equation by the linear form m, we cancel the last term and find:
        (Subscript[a, 1]) n PD[m,  (Subscript[p, 1])] (Subscript[p, 1]^(n - 1))  +  ... + (Subscript[(a ), k - 1]) n PD[m,  (Subscript[p, k - 1])] 
            (Subscript[p, k - 1]^(n - 1))  ==  0,
Since at least one of the coefficients in this equation is non zero, this means that the shorter sequence is not (n-1)-independent, which is a contradiction.  Q.E.D.

Theorem. If k  ≤  2 n + 1, then any sequence of k distinct points, no three of which are collinear, is n-independent.

Proof: We argue again by induction on n.  The case of n == 1 is obvious.

Now assume the result for n-1 and argue for n.  Suppose the nth powers   (Subscript[p, 1]^n), ... , (Subscript[p, k]^n) are linearly dependent.  This means that we have a relationship:
 		  (Subscript[a, 1]) (Subscript[p, 1]^n)  +   ...  + (Subscript[(a ), k]) (Subscript[p, k]^n)  ==  0,
with not all coefficients equal to 0.  Indeed, by the previous theorem, at least (n+2) of the coefficients must be non-zero.  Since we are considering a case where n ≥ 1, this means that at least three coefficents are non zero.  If we let m be the line passing through (Subscript[p, k]) and (Subscript[p, k - 1]), then—by assumption—it does not pass through any of the other points.  By differentiation of the equation by m, we cancel the last two terms and find:

(Subscript[a, 1]) n PD[m,  (Subscript[p, 1])] (Subscript[p, 1]^(n - 1))  +  ... + (Subscript[(a ), k - 2]) n PD[m,  (Subscript[p, k - 2])] 
            (Subscript[p, k - 2]^(n - 1))  ==  0,

This means that the shorter sequence is not (n-1)-independent and is of length at most 2 n - 1, which is a contradiction.  Q.E.D.

Theorem. If  k ≤ ((5 n)/2) + 1 , then any sequence of k distinct points, no three of which are collinear and no six are conconic, is n-independent.

Proof:  We argue again by induction on n.  The cases n == 1 and n == 2 are obvious.  Now assume the result for n-2 and argue for n.  Suppose the nth powers   (Subscript[p, 1]^n), ... , (Subscript[p, k]^n) are linearly dependent.  This means that we have a relationship:
 		  (Subscript[a, 1]) (Subscript[p, 1]^n)  +   ...  + (Subscript[(a ), k]) (Subscript[p, k]^n)  ==  0,
with not all coefficients equal to 0.  Indeed, by the previous theorem, at least (2n+2) of the coefficients must be non-zero.  Thus, k has to be larger than 8.

Take f to be any non-zero conic passing through the last five points of the sequence,  (Subscript[p, k - 4]), ... , (Subscript[p, k]) .  By dimensions of the dual space of the space of quadratic uvw-polynomials, such an f must exist.  By assumption, it does not pass through any of the other points.  By differentiation of the equation by f, we find (after dropping a common numerical factor):

(Subscript[a, 1]) PD[f,  ((Subscript[p, 1])^2)] (Subscript[p, 1]^(n - 2))  +  ...  + (Subscript[(a ), k - 5]) PD[f,  ((Subscript[p, k - 5])^2)] 
            (Subscript[p, k - 5]^(n - 2))  ==  0,

This means that the shorter sequence of length at most ((5 n)/2) + 1 - 5 == ((5 (n - 2))/2) + 1 is not (n-2)-independent, which is a contradiction.  Q.E.D.

Question.  Are these the best results?

Curves through given points

Here is an important immediate corollary of the proofs of independence.

Theorem.  Given five points, no three collinear, there is exactly one conic passing through them.

Question. Given such five points, how do you find the conic passing through them (as a polynomial)?

We will answer this question later when we study conics more.  Our independence results do not have as sharp a consequences for cubics as we derived for conics, however.

Theorem. Any sequence of 8 distinct points—no three of which are collinear and no six conconic—determines a unique pencil of cubics.

As we shall see, there are configurations of 9 points—no three of which are collinear and no six conconic—which are not 3-independent.  The reason for the trouble here is that two distinct cubics generally intersect in 9 points.

Problem. How should these theorems be generalized to higher degrees?

Defining Contact, Multiplicity, and Tangency

Degree of contact and multiplicity

Degree of contact

So far we have been using very algebraic definitions, and we have to check in many examples that they agree with geometric intuition.

Suppose, for example, that f is a homogeneous xyz-polynomial of degree n representing a curve and p is a linear form representing a point.  Suppose further that the point lies on the curve:  
		PD[f, (p^n)] == 0.
Consider a line through the point p.  If q is a point on the line distinct from q, then the points “around” p can be represented as linear combinations p + t q.   The question is to determine what degree of contact  the line has with the curve (aside from cutting the curve at p).   Here is the algebraic definition we are going to use.

Definition.  Let the point p lie on a curve f == 0, and let the point q be another point on a line l.   The line l has degree of contact m at  p  with f if, and only if t == 0 is an m-fold root of the equation:
		PD[f,(((p+t q))^n)]==0.

Clearly t == 0 is a root of this equation, because p is on f.  If the line “cleanly” cuts the curve, then 0 is but a simple root.  If the line has closer contact with the curve, then t == 0 is a multiple  root.  The exact multiplicity of this root in this equation is taken as the definition of degree of contact.  We have to investigate, however, how well this definition works out.  The first result shows the answer only depends on the line and not on the choice of the extra point on the line.

Theorem.  The definition of degree of contact does not depend on the choice of q as the other point on l.

Proof:  We illustrate the argument with curves of degree 6 by doing some symbolic computation.  First we declare f to be such a curve and p and q to represent points.

\(ClearAll[f, p, q, Subscript]; 
f /: DegP[f] = 6; 
\(p /: DegP[p] = 1; 
    \)\)q /: DegP[q] = 1;

Next we look at PD[f,(((p+t q))^6)], which we can do by applying the well known Binomial Theorem:

\(g\_1 = Expand[\((p + t\ q)\)\^6]
\)(Collect[#1, t]&)[PD[f, (Subscript[g, 1])]]

For t == 0 to be a triple root of the equation PD[f, (((p + t q))^6)] == 0, we see that the first three terms must vanish.  Let us declare that this is so:

f /: PD[f, (p^6)] = 0; f /: PD[f, (p^5) q] = 0; f /: PD[f, (p^4) (q^2)] = 0;

Carrying out the computation of  PD[f, (((p + t q))^6)] again, we see they are gone, allowing us to factor the result:

(Collect[#1, t]&)[PD[f, (Subscript[g, 1])]] // Factor

Now, we replace q by another point on l – different from p, namely a p + q – and do the computation again:

\(g\_2 = Expand[\((p + t\ \((a\ p + q)\))\)\^6]; 

    \((Collect[#1, t]&)\)[PD[f, g\_2]]
\)% // Factor

We find that (t^3) has the same coefficient as before.  So if t == 0 is exactly a triple root of the first equation based on q, then it is also exactly a triple root of the second equation based on a p  +  q.  Q.E.D.

Here is an example with a 7th-degree polynomial.

\(ClearAll[f, p, q, Subscript]; 

    f = y\^2\ z\^5 - x\^7 - x\^6\ z - x\^4\ z\^3; 
\(p = w; \)\)q = u;

Expand[PD[f, (((p + t q))^7)]]

The degree of contact with the x-axis (= PE[p,q]) is 4, and here is a way of arranging the test for this outcome:

(PD[f, #1]&)/@{(p^7), (p^6) q, (p^5) (q^2), (p^4) (q^3), (p^3) (q^4)}

And here we change the choice of q:

q = (p/5) + q; 
    (PD[f, #1]&)/@{(p^7), (p^6) q, (p^5) (q^2), (p^4) (q^3), (p^3) (q^4)}

And we see that the test gives the same answer for a degree of contact.  In general, we can say:

Corollary.  Let the point p lie on a curve f == 0, and let the point q be any other point on a line l.   The line l has degree of contact m at  p  with f if, and only if, 
		PD[f,(p^(n - i)) (q^i)]==0 whenever 0 ≤ i < m, and
		PD[f, (p^(n - m)) (q^m)] ≠ 0.

Multiplicity

Two different lines through a point on a curve can have different degrees of contact with the curve.  It can even happen that all lines through the point have higher degree of contact than 1.  However, the degree of contact cannot be higher than the degree of the curve itself.  (Why?)  We can make this observation into a definition.

Definition. A point on a curve is a m-tuple point if, and only if, every line through the point has degree of contact at least m, and one line has degree of contact exactly m.

Here is a test for determining the multiplicity of a point on a curve.

Theorem. If f is a curve of degree n and 0 < m ≤ n, then a point p is an m-tuple point of f  if, and only if, 
		PD[f, (p^(n - m + 1))] == 0,   but   PD[f, (p^(n - m))] ≠ 0.

Proof:  We will work this out for the general 6th-degree curve f.  Suppose p is a triple point on this curve.  This means that every line through p has degree of contact 3.  Consider a line passing through a an arbitrary point q.   As in the previous theorem, we may conclude that, for t == 0  to be a triple root of the equation 
		PD[f,(((p+t q))^6)]==0,
it is necessary and sufficient that 
		PD[f, (p^6)] == PD[f, (p^5) q] == PD[f, (p^4) (q^2)] == 0.
Now we already know that the first quantity vanishes because p is on the curve.  So there is nothing new here.  But to have PD[f, (p^5) q] vanish for all points  q means that PD[PD[f, (p^5)], q] vanishes for all q.  By the dual pairing property, this means that the linear xyz-polynomial PD[f, (p^5)] must vanish itself.

Next, we see that the quadratic polynomial PD[f, (p^4)] has to vanish on all (q^2).  By the Generation Theorem, this means that it must also be the zero polynomial.

Finally, since t == 0 is not a quadruple root for some choice of q, it follows that the cubic polynomial PD[f, (p^3)] cannot be identically zero.  Q.E.D.

Let us consider an example used previously.

ClearAll[f, p]; f = (y^2) (z^5) - (x^7) - (x^6) z - (x^4) (z^3); p = w; (PD[f, #1]&)/@{(p^7), (p^6), (p^5), (p^4), (p^3)}

We have just shown that p is a double point on this curve.  Now we modify the curve a little.

ClearAll[f, p]; p = w; \(f = y\^3\ z\^4 - x\^7 - x\^6\ z; 

    \((PD[f, #1]&)\)/@{p\^7, p\^6, p\^5, p\^4, p\^3}\)

For the new curve, p is a triple point.  And we try another variation:

ClearAll[f, p]; p = w; \(f = y\^3\ z\^4 - x\^7 + x\ y\^2\ z\^4; 

    \((PD[f, #1]&)\)/@{p\^7, p\^6, p\^5, p\^4, p\^3}\)

Again a triple point.  We can graph the real points on the last curve by setting z == 1 and then by using ImplicitPlot:

ImplicitPlot[(y^3) + x (y^2) == (x^7), {x, -2, 2}, 
      Axes -> False];

This shows why the point p is triple very clearly.  We can also check the degree of contact of the x-axis and the y-axis with the curve.

\(q = u + w; r = v + w; 

    \(PD[f, #]&\)/@{p\^7, \(p\^6\) q, \(p\^5\) q\^2, \(p\^4\) q\^3, 
        \(p\^3\) q\^4, p\^2\ q\^5, p\ q\^6, q\^7}
\)PD[f, #]&/@{(p^7), (p^6) r, (p^5) (r^2), (p^4) (r^3), 
        (p^3) (r^4), (p^2) (r^5), p (r^6), (r^7)}

So!  The x-axis has a very high degree of contact witht he curve, but the y-axis has only degree of contact equal to 3.

Points of tangency and inflection

Tangency

The example of the 7th-degree curve we have just seen, suggests this definition:

Definition. If a line at a m-tuple point of a curve has degree of contact greater than m, then it is said to be tangent  at that point.

Example 1.  Here is an illustration of a tangent to a cubic:

ClearAll[f, p]; f = (x^3) - (y^2) z; p = u + v + w; (PD[f, #1]&)/@{(p^3), (p^2), p, 1}

First, we see that the point p is on the curve, and it is a simple point.  Here is another point q, and the parameterization of the points on the line they determine.

q = -v + 2 w; Collect[PD[f, (((p + t q))^3)], {t}]

The degree of contact of the line with the curve is 2.  So the line joining p and q is a tangent.  Having two points on it, we can calculate it this way:

PE[p,q]

We see that this form for the tangent represents the same line as PD[f, (p^2)].   Here is a picture of the real part of the cubic and its tangent.

ImplicitPlot[{(y^2) == (x^3), 3 x == 2 y + 1}, {x, -2, 2}];

Question.  Is there a theorem here about finding tangents?

Example 2.  A second illustration concerns a double point of a cubic:

ClearAll[f, p, q]; f = (x^3) + (y^2) z + x y z; PD[f, (w^2)]

The point w  is a double point.  We can draw a picture.

ImplicitPlot[(x^3) + (y^2) + x y == 0, {x, -1, 1}];

Now, look at the first derivative:

Factor[PD[f, w]]

Now the two lines only intersect the cubic at the unique point w.  This means they have degree of contact greater than 2.  Thus both lines are tangent to the curve.  Here is a calculation that also shows multiple roots.

Solve[{f == 0, y == 0, z == 1}, {x, y, z}]Solve[{f == 0, y == 0, z == 0}, {x, y, z}]Solve[{f == 0, x + y == 0, z == 1}, {x, y, z}]Solve[{f == 0, x + y == 0, z == 0}, {x, y, z}]

And here is a picture of the situation of the two tangent lines.

ImplicitPlot[{(x^3) + (y^2) + x y == 0, y == 0, x + y == 0}, {x, 
        -1, 1}, Axes -> False, 
      PlotStyle -> {Red, Green, Blue}];

Question.  Is there a theorem here about finding tangents?

Inflection points

Definition.  A simple point (1-tuple point)  on a curve is an inflection point if the tangent to the curve at the point has degree of contact 3 or more.

Theorem.  A point p is an inflection point on a curve f of degree n if, and only if, we have these three conditions satisfied: 
	(i)  PD[f, (p^n)] == 0, 
	(ii)  PD[f, (p^(n - 1))] ≠ 0, 
	(iii)  PD[f, (p^(n - 2))]factors into linear factors.

Proof:  Let  p  be an inflection point of  f, and q a point on the tangent of f at p.  Since degree of contact of the tangent at p is at least 3, we have PD[f, (p^(n - 2)) (q^2)] == 0.  But this means that q lies on the conic PD[f, (p^(n - 2))].  As q was any point on the tangent, this conic contains the entire tangent, and so the conic factors.

Conversely, assume that PD[f, (p^(n - 2))] factors.   As PD[f, (p^(n - 1))] ≠ 0 by assumption, p is not a double point of this conic.  If the factors were equal,  all points would be double; so there must be two distinct factors.

Let q be the intersection of the corresponding lines. It is a double point of the conic, so PD[f, (p^(n - 2)) q] == 0. Differentiating this equation with p and q, respectively, we get 
		PD[f, (p^(n - 1)) q] == PD[f, (p^(n - 2)) (q^2)] == 0.  
By assumption,  PD[f, (p^n)] == 0 as well, and so the line PE[p,q] has degree of contact at least 3.  Q.E.D.

Here is an example about cubics (since a conic cannot have inflection points).  The cubic (x^3)- y (z^2) has only one inflection point.  The point is  w.

(PD[(x^3) - y (z^2), #]&)/@{(w^3), (w^2), w}

The line represented by y is the tangent.  And we see the degree of contact of the line with the curve is 3.

(PD[(x^3) - y (z^2), #]&)/@{(w^3), (w^2) u, w (u^2), (u^3)}

The (real part of the) graph is well known:

ImplicitPlot[(x^3) == y, {x, -1, 1}];

But note that it is not especially clear from the picture that the x-axis actually has higher degree contact with the curve than an ordinary tangent at some other point.  You have to do your algebra to make sure!