Re: Uniqueness of implicit functions

From: Dave Rusin (rusin_at_math.niu.edu)
Date: 12/14/04 

Date: 14 Dec 2004 05:08:08 GMT

In article <sutx3xu8to3k@legacy>,
Mikhail V. Sokolov <mmm_spam@rambler.ru> wrote:
>Consider two implicit functions:
>f(x,y)=0,
>g(x,y)=0.
>What can be said about relation between them, if it is known that
>f(x,y)=0 iff g(x,y)=0.

Maybe I don't understand what an "implicit function" is. If you
mean, "what can be said about two functions f,g : R^2 --> R knowing
that they vanish at the same points?" the answer might be "Not much";
I would say, for example, that f(x,y) = (sin(x) - cos(y) + xy)^2 + 1
and g(x,y) = ( exp(xy) + x - y )^2 + 1 are very different, but
they do vanish at the same points... .

But that's assuming you meant x and y to be real. If you allow them to be
complex the situation is different. IN particular, when f and g are both
_polynomials_ and they vanish at the same points (x,y) in C^2, then by
Hilbert's Nullstellensatz, the ideals (f) and (g) have the same radical
in the ring C[x,y], so that f and g have the same irreducible divisors
(possibly to different powers, e.g. f = x^2y and g = x y^3.)

>For example, if they are differentiable, then
>f1(x,y)*g2(x,y) = f2(x,y)*g1(x,y), where f1,f2,g1,g2 are derivatives
>of corresponding function by the first (second) variable.

Fails in my example, above. More generally, if the factorizations are
f = product( (p_i)^(n_i) ) and g = product( (p_i)^(m_i) ) , then
f_1 / f = sum( (n_i) (p_i)_1 / (p_i) ) and similarly for f_2 / f,
so that f_1 / f_2 =
   sum( (n_i) (p_i)_1 / (p_i) )/ sum( (n_i) (p_i)_2 / (p_i) )
Of course g_1 / g_2 is almost identical; one need only change the n_i
to m_i. So we don't have f_1 / f_2 = g_1 / g_2 unless f and g
are powers of the same irreducible.

I am told that the ring of analytic functions behaves in a very similar way;
certainly the previous example suggests ways that your conjecture could fail.

dave

PS -- I tried to reach you much earlier but you did not provide a way
to determine a valid email address.