Versal unfolding

Remarks

Let $\mathcal{E}_n$ be the ring of smooth germs at $0$ on $\R^n$ and $J(f)$ the ideal generated by the partial derivatives of $f$. The versality theorem says that $F$ is versal if and only if $$\mathcal{E}_n = J(f) + \R \cdot 1 + \R \cdot \frac{\partial F}{\partial u_1}(\cdot, 0) + \cdots + \R \cdot \frac{\partial F}{\partial u_r}(\cdot, 0),$$ where $u_1, \ldots, u_r$ are the parameters. Example: for $f = x^3$, $J(f) = (x^2)$ and $\mathcal{E}_1 / J(f)$ has basis the classes of $1$ and $x$; for $F(x, u) = x^3 + u x$ one has $\partial F / \partial u (x, 0) = x$, so $F$ is versal, and universal, since one parameter is needed. Non-example: $G(x, u) = x^3 + u x^2$ is an unfolding of $x^3$, but $\partial G / \partial u (x, 0) = x^2$ lies in $J(f)$, so the class of $x$ is not reached and $G$ is not versal. Indeed, every function $G(\cdot, u)$ has a critical point near $0$, while $x^3 + u x$ has none for $u > 0$. A germ has a versal unfolding with finitely many parameters if and only if $\mathcal{E}_n / J(f)$ is finite-dimensional. If moreover $f$ has a critical point at $0$, a universal unfolding has $\dim_\R \mathcal{E}_n / J(f) - 1$ parameters.