Re: Curve implicitization from parametrization

john_ramsden@xxxxxxxxxxxxxx writes:

>Tony wrote:
>>
>> How does one prove that parametrized curve of degree n
>> (in 2D or 3D) can be implicitized into polynomial of
>> the same degree?
>>
>> For example, that from:
>>
>> x=At^3 + Bt^2 + Ct + D
>> y=Et^3 + Ft^2 + Gt + H
>>
>> follows F(x,y)=0 where F(x,y) is cubic.
>>
>> And how does one go about finding F(x,y) in such
>> form starting from the parametric form?
>
>Find F as an "eliminant" (as I think it's called) by
>constructing a set of linear equations in powers of t:
[details omitted]

This is of course correct "in 2D". The situation is much
more vexed in "3D"; the nature of the vexation varies a bit
depending on what field K you're working over. I'll assume
that K is either R, the real numbers, or C, the complex numbers.
Let G = {(x,y,z) : x = p(t), y = q(t), z = r(t)}, where p, q,
and r are rational functions (or polynomials) with coefficients
in K. Given a polynomial F in three variables, let V_F =
{(x,y,z) : F(x,y,z) = 0}.

In case K = C, G can never have the form V_F (as John certainly
knows; it's hard for me to judge the likelihood that Tony knows,
so I'll write as though he doesn't, intending no offence in case
he does). Here, G really does have "complex dimension" 1 (and
this can be detected locally at every point of G), V_F really
does have "complex dimension" 2 (also detectable locally at
every point of G), and something of complex dimension 1 cannot
be equal to something of complex dimension 2.

In case K - R, G *does* have the form V_F, but only by a trick,
and *never* for F of the low degree Tony wants. For instance,
if (p(t),q(t),r(t)) = (t,0,0), where the maximum degree of the
component functions is 1, then we could take F(x,y,z) = x^2+y^2+z^2
which is of degree 2 > 1, and obviously we can't take F of degree
1 because then V_F would be a plane.

In general, for K either C or R, there exist polynomials F_1,..,
F_n, where I *believe* that it has been proved that n = 3 is
sufficient, such that *as a set* G is the intersection of
V_{F_1},..., V_{F_n}; in the jargon of algebraic geometry,
this is the assertion that G is a "set-theoretic complete
intersection". It follows that for K = R, the *set* G is
V_F, where F is the sum of the squares of the F_j. However,
it is not always the case that G is such an intersection *as
an algebraic variety* (this would fail if, for instance, n
were 2 but there existed points of G at which the intersection
of V_{F_1} and V_{F_2} was locally not transversal): in fact,
the "twisted cubic" G, parametrized by (t,t^2,t^3), is not
such an "ideal-theoretic complete intersection" although it
is a set-theoretic complete intersection. I believe that
it is still an open question which "monomial space curves"
(generalizations of the twisted cubic, parametrized by
monomials of degrees k, m, n with gcd(k,m,n) = 1) are
ideal-theoretic complete intersections; it may even be
unknown which are set-theoretic complete intersections
(and surely the classification for arbitrary rational
space curves G is incomplete and may remain so forever).

Lee Rudolph
.

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