The complement of a polynomial zero set is open and dense

Idea. Openness is continuity. Density means that the zero set $Z=\{p=0\}$ contains no ball, and a polynomial that vanishes on a ball is the zero polynomial.

Open. $p$ is continuous, so $\{p\ne0\}=p^{-1}(\mathbb{R}\setminus\{0\})$ is open.

Dense. A set is dense iff its complement has empty interior. Suppose, for contradiction, that $p=0$ on a ball $B(a,r)$.
- On this ball $p$ is the constant function $0$, so every partial derivative $\partial^\alpha p$ vanishes on $B(a,r)$, in particular at $a$.
- A polynomial equals its Taylor expansion at $a$, which is a *finite* sum $p(x)=\sum_\alpha\frac{\partial^\alpha p(a)}{\alpha!}(x-a)^\alpha$. All coefficients are $0$, so $p\equiv0$, contradicting $p\ne0$.

Hence $Z$ has empty interior, and $\{p\ne0\}=\mathbb{R}^k\setminus Z$ is dense.

Remark. This is the basic tool of the whole section. In $M_n(\mathbb{R})\cong\mathbb{R}^{n^2}$, "polynomial in the entries" covers $\det$, the coefficients of the characteristic polynomial, minors, discriminants and so on. Lesson. To prove a property holds on an open dense set, find a nonzero polynomial whose non-vanishing implies it.