I have a question about the proof of what is sometimes called the generalized Malgrange preparation theorem. This proof is in both Brocker and Lander's "Differentiable Germs and Catastrophes" and Guillemin and Golubitsky's "Stable Mappings and Their Singularities". It is on page 107 in Guillemin and Golubitsky, and page 57-58 in Brocker and Lander. The result boils down to the following:
Let $C^\infty_0(\mathbb R^n)$ denote germs of smooth functions at $0$ in $\mathbb R^n$. Let $f:\mathbb R^{n+1} \to \mathbb R^n$ be the projection $f(t,x) = x$. We obtain a homomorphism $f^* : C^\infty_0(\mathbb R^n) \to C^\infty_0 (\mathbb R^{n+1})$, $f^* g(t,x) = g(f(t,x)) = g(x)$. Let $A$ be a finitely generated $C^\infty_0(\mathbb R^{n+1})$ module. Then $A$ is finitely generated as a $C^\infty_0(\mathbb R^n)$ module (with multiplication via $f^*$) if and only if the real vector space $A / (f^*m(n) A)$ is finite dimensional. Here $m(n) = \{f \in C^\infty_0(\mathbb R^n) : f(0) = 0\}$, i.e. the unique maximal ideal of $C^\infty_0(\mathbb R^n)$.
Some straightforward manipulation allows you to conclude the following: let $a_1,...,a_k$ generate $A$ as a $C^\infty_0(\mathbb R^{n+1})$ module and span $A/(f^* m(n) A)$ as a $\mathbb R$ vector space. Then any $a \in A$ can be written $$a = \sum_j c_j a_j + \sum_j z_j a_j,$$ where $c_j \in \mathbb R$ and $z_j \in f^*m(n) C^\infty_0(\mathbb R^{n+1})$. Then, in both Brocker and Lander and Guillemin and Golubitsky, the following is claimed: we can write $$ta_i = \sum_j (c_{ij} + z_{ij}) a_j,$$ where $c_{ij} \in \mathbb R$ and $z_{ij} \in f^* m(n) C^\infty_0(\mathbb R^{n+1})$.
My issue is this: it seems to me that $c_{ij}$ depends on $t$. How do we conclude that $c_{ij}$ can be chosen independently of $t$?
$ta_i$ is an element of $A$ so you can also write
$ta_i=\sum_jc_{ij}+\sum_jz_{ij}a_j$ where $c_{ij}$ is a constant and $z_{ij}\in f^*m(n)C^{\infty}_0(R^{n+1})$.