Malgrange preparation theorem

Remarks

This is the smooth analogue of the Weierstrass preparation theorem for holomorphic functions. The hypothesis says that $t \mapsto F(t, 0)$ vanishes to order exactly $k$ at $0$. It is necessary: since $q(0, 0) \ne 0$ and $a_i(0) = 0$, the function $F(t, 0)$ vanishes at $t = 0$ to the same order as $t^k$. In the smooth case the proof is much harder than in the holomorphic case, because there is no Cauchy integral formula. Unlike the holomorphic case, the factorisation need not be unique. A closely related form is the division theorem: every smooth $G$ near $(0, 0)$ can be written as $G = Q F + \sum_{i < k} r_i(x) t^i$ with smooth $Q$ and $r_i$. Mather gave a form for modules of germs. Let $\varphi \colon (\R^n, 0) \to (\R^p, 0)$ be a smooth map germ, and let $A$ be a finitely generated module over the ring $\mathcal{E}_n$ of smooth germs on $\R^n$. Let $\mathfrak{m}_p$ be the maximal ideal of the ring $\mathcal{E}_p$ of smooth germs on $\R^p$. Then $A$ is finitely generated over $\mathcal{E}_p$ through $\varphi$ if and only if $A / \varphi^*(\mathfrak{m}_p) A$ is a finite-dimensional real vector space. Example: for $F(t, x) = t^2 - x$ the theorem holds with $q = 1$, $a_1 = 0$ and $a_0(x) = -x$.

Going deeper

For a student

The preparation theorem is the main algebraic tool of the local theory of smooth maps. In Mather's module form it turns infinitesimal conditions into finite-dimensional linear algebra. It is used to prove that infinitesimal stability implies stability, and results on finite determinacy and versal unfoldings.