Nondegenerate zero of a vector field

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$ (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) for the canonical smooth tangent-bundle structure.

Let $M$ be a smooth $n$-manifold, $X$ a smooth vector field on $M$ and $p\in M$ a zero of $X$ (A smooth vector field is a smooth section of the tangent bundle). View $X$ as a smooth section $X:M\to TM$ (Smoothness of a section is equivalent to smooth local components). The zero section $0_M$ is a smooth embedding (The zero section is a smooth embedding), so its differential $d(0_M{)}_p:T_{p}M\to T_{(p,0)}TM$ is injective (The differential of a smooth map), and the vertical quotient at $p$ is $N_p:=T_{(p,0)}TM/d(0_M{)}_p(T_{p}M)\mathrm{.}$ Since $X$ is a section, both $d{X}_p$ and $d(0_M{)}_p$ are right inverses of the projection $d{\pi }_{(p,0)}$; hence the class of $d{X}_p$ in $\mathrm{Hom}(T_{p}M,N_p)$ is the vertical derivative $D{X}_p\in \mathrm{End}(T_{p}M),$ read through the canonical identification $N_p\cong T_{p}M$: explicitly, the difference $d{X}_p-d(0_M{)}_p$ takes values in the vertical space $\ker d{\pi }_{(p,0)}$, and $D{X}_p$ is that difference followed by the canonical isomorphism from the vertical space $V_{(p,0)}=\ker d{\pi }_{(p,0)}$ to $T_{p}M$. In an induced tangent-bundle chart $\tilde{\phi }$ over a chart $\phi$ with $\phi (p)=0$ the section reads $u\mapsto (u,X_{\phi }(u))$ and $D{X}_p$ corresponds to the ordinary derivative $D{X}_{\phi }(0)$ of the chart representative $X_{\phi }$ (The induced tangent bundle chart, Local and global frames of a vector bundle); a change of chart conjugates $D{X}_p$ by an isomorphism, so invertibility and the determinant $\det D{X}_p$ are independent of the chart (The tangent bundle as a disjoint union).

The zero $p$ is nondegenerate when $D{X}_p$ is invertible. A nondegenerate zero is isolated: in a chart, $X_{\phi }(0)=0$ and $D{X}_{\phi }(0)$ is invertible, so $X_{\phi }$ is a diffeomorphism near $0$ by the inverse function theorem (The smooth inverse function theorem on manifolds) and its only zero near $0$ is $0$ itself. In particular a nondegenerate zero is an isolated zero and $X$ has no other zero in some neighbourhood of $p$. The vertical derivative itself requires no further choice once the smooth tangent-bundle structure is supplied.