# Set of critical points of polynomial: why finite

Let $P$ be a $C^\infty$ real constant coefficients polynomial defined on $\mathbb{R}^n$ and let $Z(P)$ be its set of critical values, that is $Z(P)=P(\{\xi\in\mathbb{R}^n: \nabla P (\xi)=0\})$. I read on my book that this is obviously finite. I get that is of Lebesgue measure zero by simply applying Morse-Sard's theorem but I am missing the last piece: how come it's exactly finite?

• When you say a polynomial defined on $\mathbb{R}^n$, does it mean a multivariate polynomial with $n$ variables? – ziyuang Jan 8 '14 at 11:33
• It's false for constant polynomials... Also, for $P(x,y)=x^2$, the zero set of the gradient is the whole line $x=0$. – Hans Lundmark Jan 8 '14 at 11:48
• Exactly, $P:\mathbb{R}^n\to \mathbb{R}$, polynomial with constant coefficients. – blackdragone Jan 8 '14 at 11:49
• Hans, maybe you didn't notice that $Z(P)$ is equal to $P(...)$. So if $P(x_1,...,x_n)=C$ (constant), then you would get $Z(P)=\{C\}$ or in the other case you mention, you get $Z(P)=\{0\}$. – blackdragone Jan 8 '14 at 11:51
• Let $V = \{ \xi : \nabla P(\xi) = 0\}$. $P$ is constant on each connected component of $V$. $V$ has only finitely many components, if memory doesn't deceive me. – Daniel Fischer Jan 8 '14 at 11:54