Double roots of polynomials

Question

If $f\in R[x]$ is a real univariate polynomial of degree $2d$ then $f$ can have at most $d$ double roots. What about generalizations to multivariate polynomials? Suppose that for $f\in R[x_1,\ldots,x_n]$ of degree $d$ the set $S:=\{\xi\in R^n : f(\xi)=0 \frac{\partial f}{\partial x_i}(\xi)=0, i\in\{1,\ldots,n\}\}$ is finite. What is a bound on $|S|$?

Answer 1

This is not a complete answer but a bit too long for a comment and only a sketch. In dimension $n \geq 2$ and for a degree $d \geq 2$ there is a polynomial with exactly ${d + n - 2 \choose n}$ such points. This gives a lower bound for $|S|$. Here is how such a polynomial can be constructed. Let $\ell_1, \ldots, \ell_{d+n-2}$ be affine functions on $\mathbb{R}^n$ (linear plus constant term) in general position and such that $\frac{\partial \ell_k}{\partial x_1} \neq 0$ for all $k$. In general position means that any subset of $n$ functions together with the constant function $1$ is linearly independent. Now let $P$ be the polynomial $$P=\frac{\partial^{n-2}}{\partial x_1^{n-2}} \prod_{k=1}^{d+n-2} \ell_k$$ then $\deg(P) = d$ and $S$ is the set of intersections of any $n$ hyperplanes $\ell_k = 0$. So $|S| = {d+n-2 \choose n}$.

Note that for $n=2$ this is just the product of lines and $S$ is where any two lines intersect. For $n>2$ the differential operator makes sure that all partial derivatives vanish only if enough hyperplanes meet in a point. An example in $\mathbb{R}^3$ would be $$\ell_1=x_1, \ell_2=x_1+x_2, \ell_3 = x_1+x_3$$ and $$P=\frac{\partial}{\partial x_1}x_1 (x_1+x_2)(x_1+x_3) = 3 x_1^2+x_2 x_3+2 x_1x_2+2 x_1x_3$$ has a double point only for $x_1=x_2=x_3=0$.