Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

We know by the standard Implicit Function Theorem that

If $f:\mathbb R^4\rightarrow\mathbb > R^2$ is a polynomial (or in fact any continuously differentiable function), then there is a point $a\in\mathbb > R^2$ such that $f^{-1}(a)$ is at least two-dimensional.

Now imagine that instead of $\mathbb R^2$ we have an algebraic surface $S\subset\mathbb R^3$, i.e. the zero set of a trivariate polynomial. It's reasonable that the statement still holds. A general statement along these lines would be something like

If $A,B$ are two algebraic sets such that $\dim(A)>\dim(B)$, and if $f:A\rightarrow B$ is a polynomial mapping, then there is a point $b\in > B$ such that $f^{-1}(b)$ is of dimension at least $\dim(A)-\dim(B)$.

Is it a well-known theorem? Any reference for it?

share|improve this question
    
You hypotheses are too weak, just take $f$ to be a constant map, or $(x_1,x_2,x_3,x_4)\mapsto (x_1,0)$. –  Fernando Muro Oct 10 '11 at 13:54
    
thx. of course i wanted to say 'at least 2-dimensional'. corrected now. –  filipm Oct 10 '11 at 13:56
    
Also, you probably meant $\dim(A)-\dim(B)\dots$ –  Thierry Zell Oct 10 '11 at 14:42
    
@Thierry: yes, i corrected it. –  filipm Oct 10 '11 at 15:13

2 Answers 2

up vote 3 down vote accepted

This is a well-known theorem. I imagine it has been routinely used for many years, so tracking down a historical reference would be hard. The following argument will work for any $f$ which is definable over the real field, i.e. whose graph is a semi-algebraic set.

The graph $\Gamma \subset A\times B$ of your mapping has dimension $\dim(A)$, since the restriction to $\Gamma$ of the projection $\pi_A: A\times B \to A$ is 1-to-1. Now do a cylindrical algebraic decomposition of $A\times B$ adapted to $\Gamma$ and the projection $\pi_B$. Then, any cell in $f(B)$ is the image of a cell of $\Gamma$. In particular, this is true for all the $\dim(A)$-dimensional cells that appear in $\Gamma$; if $C$ is such a cell, any point $b$ in the image $\pi_B(C)$ verifies $\dim(f^{-1}(b)) \geq \dim(A)-\dim(\pi_B(C)) \geq \dim(A)-\dim(B)$.

So not only can you find such a $b$, but there should be quite a few of them.

share|improve this answer

I think Qing Liu's "Algebraic geometry and algebraic curves" (especially 4.3) answers your question (and gives appropriate conditions).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.