I am trying to understand something which is probably basic for experts so I am sorry if this is not suited for this forum.


Let $\mathcal{O}\_n$ denote the ring of germs at $0 \in \mathbb{R}^n$ of real-analytic functions. It is known that $\mathcal{O}\_n$ is a [Noetherian ring](https://en.wikipedia.org/wiki/Noetherian_ring) (see [1, Ch 3, Thm 3.8]). I understand from reading related bibliography that this should imply (quite easily?) the following result, but I cannot find a reference where it is stated exactly as below.


**Theorem?** Let $f \in \mathcal{O}\_n$ and $\varphi\_1, \dotsc, \varphi\_q \in \mathcal{O}\_n$. Then there exist $g\_1, \dotsc, g\_q \in \mathcal{O}\_n$ such that $f = g\_1 \varphi\_1 + \dotsb + g\_q \varphi\_q$ if and only if there exists a formal solution at the origin (i.e. formal series $\hat{g}\_1, \dotsc, \hat{g}\_q$ in $n$ variables such that $\hat{f} = \hat{g}\_1 \hat{\varphi}\_1 + \dotsb + \hat{g}\_q \hat{\varphi}\_q$ where $\hat{f}$ and $\hat\varphi\_i$ denote the Taylor series of $f$ and $\varphi\_i$ at 0).


I have read statements with $q > 1$ (such as [1, Ch 6, Th 1.1']) where the assumptions must hold in a whole vicinity of the origin (not just at the origin as above) and statements with $q = 1$ (see [2]) where the assumptions hold only at the origin (even in the quasianalytic setting).


My understanding is that the above theorem should follow rather easily from the Noetherianity of the ring of germs at 0, but I somehow failed to prove it by myself. **So, does the above result hold and is it stated somewhere?**


1. Malgrange, *Ideals of differentiable functions*.
2. Jan Nowak, *[On division of quasianalytic function germs](https://doi.org/10.1142/S0129167X13501115)*



Does Noetherianity imply division theorem?
This is a consequence of the noetherianity of $\mathcal{O}\_n$, in some way.


Let me write $A\_n$ for the ring of real formal formal series around the origin, i.e., $A\_n=\mathbb{R}[[t\_1,\dotsc,t\_n]]$. Let $I$ be the ideal of $\mathcal{O}\_n$ generated by $\varphi\_1,\dotsc,\varphi\_q$. Then your question is equivalent to whether
$$ \mathcal{O}\_n\cap IA\_n = I.$$


Let $\mathfrak{m}$ be the maximal ideal of $\mathcal{O}\_n$. Then $A\_n$ is the completion of $\mathcal{O}\_n$ at with respect to the $\mathfrak{m}$-adic topology. The fact that $\mathcal{O}\_n$ is noetherian now means that $A\_n$ is faithfully flat over $\mathcal{O}\_n$ (see [the Stacks project, Lemma 10.97.3](https://stacks.math.columbia.edu/tag/00MC), for instance). This means that the evident map $\mathcal{O}\_n/I\to (\mathcal{O}\_n/I)\otimes\_{\mathcal{O}\_n}A\_n\cong A\_n/IA\_n$ is injective, and it follows that $\mathcal{O}\_n\cap IA\_n = I$.


