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I have result: measure of the set of critical values of $f$ is zero (by Sard's theorem), where $f: \mathbb{R^n} \rightarrow \mathbb{R}$ are polynomial functions. How do you show that the set of critical values of $f$ is finite?

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@copper.hat It isn't obvious from what you wrote. How do reduce the problem to finding the zeros of some single variable polynomial? The critical points which are the zeros of the differential can certainly be infinite, but it is the critical values that are finite. –  Matt Dec 16 '12 at 21:34
    
@Matt: You are correct, I didn't see the $n$ on the $\mathbb{R}^n$. It is not as immediate as I thought. –  copper.hat Dec 16 '12 at 22:44
    
Thanks @copper.hat, but I cannot solve this problems. :( –  user52523 Dec 18 '12 at 16:05
    
Sorry, I was too quick off the mark. I have spent some time looking at this, but have made no headway at all. Do you have any other hints? –  copper.hat Dec 18 '12 at 16:15
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All I meant by this was assume for example $f(x,y)=Ax^2+By^2+Cxy+Dx+Ey+F$ has two distinct critical values: $f(x_0, y_0)=G_0$ and $f(x_1, y_1)=G_1$. You get a 6x6 linear system to solve for the coefficients of $f$ using that $Df(x_0, y_0)=0$ and $Df(x_1, y_1)=0$. If you suppose there are too many critical values the system is overdetermined and possibly inconsistent which will show no choice of coefficients could produce a degree $d$ polynomial with a huge number of critical values. This would be really tough to get in full generality though. –  Matt Dec 21 '12 at 15:03

1 Answer 1

$$V=crit(f)_\Bbb C=(\cap_{i=1}^n\{x\in \Bbb C^n|\frac {df}{dx_i}(x)=0\})$$

is an algebraic variety. Hence it is a finite union of irreducible algebraic varieties . If a variety is irreducible, it is easy to see it is connected in the Zariski topology. So in particular, $V$ has a finite number of connected components in the Zariski topology. An algebraic variety (in an algebraically closed field) is connected in the Zariski topology iff it is connected in the analytic topology (see here). Since $Df=0$ on $V$ and $f$ is smooth, we have $f$ is constant on each connected (in the analytic sense) component of $V$. Therefore $f$ only takes finitely many values on $V$. But $crit(f)_\Bbb R\subset crit(f)_\Bbb C$, so $f$ only takes finitely many values when restricted to its real critical set. So $f(crit(f)_\Bbb R)$ is finite as desired.

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