Factorizing functions near the zero level set of a function of maximal rank I'm reading Proposition 2.10 in Olver's textbook "Applications of Lie Groups to Differential Equations":

Let $F: M\to \mathbb{R}^l$ be of maximal rank on the subvariety $S = \{x: F(x) = 0\}$. Then a real-valued function $f: M\to \mathbb{R}$ vanishes on $S$ if and only if there exist smooth functions $Q_1(x), \cdots, Q_l(x)$ such that
$$f(x) = Q_1(x)F_1(x)+\cdots+Q_l(x)F_l(x)$$
for all $x\in M$.

The strategy of the proof is to first prove a local version of the result and then bootstrap to a global result over $M$ using a partition of unity. But I don't understand how the ('only if' direction of the) local result follows. Since $F$ has maximal rank along $S$, it follows that $S$ is a regular submanifold. But it's not clear how this gets me any closer to such a factorization via smooth functions. Can anyone clarify this for me?
 A: I couldn't think of a very direct way to prove this. I will sketch one way you can prove it. Let $Y:U\subset M \to \mathbb{R}^n$ and, $X:V \to \mathbb{R}^l$  be coordinate charts given by the rank theorem for $F$. That is
$$X\circ F\circ Y^{-1}(x_1,\ldots,x_n)=(x_1,\ldots,x_l).$$
For simplicity assume that $X$ is the identity. Write $\hat{f}=f\circ Y^{-1}$ and notice that $f=\hat{f}\circ Y$. By Taylor theorem we can write
$$\begin{aligned}
&\hat{f}(\boldsymbol{x})=\sum_{|\alpha| \leq 1} \frac{D^\alpha \hat{f}(\boldsymbol{a})}{\alpha !}(\boldsymbol{x}-\boldsymbol{a})^\alpha+\sum_{|\beta|=2} R_\beta(\boldsymbol{x})(\boldsymbol{x}-\boldsymbol{a})^\beta, \\
&R_\beta(\boldsymbol{x})=\frac{|\beta|}{\beta !} \int_0^1(1-t)^{|\beta|-1} D^\beta \hat{f}(\boldsymbol{a}+t(\boldsymbol{x}-\boldsymbol{a})) d t
\end{aligned},$$
where $\alpha,\beta$ are multi indices (see Wikipedia entire for Taylor's theorem). Notice that because $f$ vanishes on $S$ we have that for $a\in \mathbb{R}^n$ such that $Y(a)\in S$ we get
$$\frac{\partial \hat{f}(a)}{\partial x_i\partial x_j}=\frac{\partial \hat{f}(a)}{\partial x_i}=0,$$
if $i,j >l$. Use this to deduce that all non-zero summands in the formula for $\hat{f}$ have an index $s\le l$ such that $x_s$ divides them. To deduce the claim use this formula to write $f=\hat{f}\circ Y$.
EDIT:
There is one big complication that I missed. We need to show that
$$\int_0^1(1-t)^{|\beta|-1} D^\beta \hat{f}(t(\boldsymbol{x})) d t$$
can also be factored when $x=(x_1\ldots,x_n)$ when some $x_j\ne 0$ for $j\le l$. In that case, we can write $\hat{f}(tx)=x_jt\hat{Q}(tx)$ for $\hat{Q}$ smooth away from $x_j=0$. Now you can differentiate under the integral sign to deduce that
$$\int_0^1 t(1-t)D^\beta \hat{Q}(tx)dt$$
is smooth.
