# Freeness of stalk Implies locally free

Let $A$ be a Noetherian ring, and $M$ a finitely generated $A$ module. Suppose that $\mathfrak { p } \in M$ such that $M_{\mathfrak{p}}$ is free. Show that there is a $f \in A \setminus \mathfrak{p}$ such that $M_{f}$ is free over $A_{f}$.

P.S. Some related questions are 1) Flatness and Local Freeness 2) Locally free sheaves, though, both of these don't answer the specific question that I have above, in that I am just looking around one prime (so my module is not projective etc). I have seen this in Vakil but I can't find it at the moment. I will post my proof of the fact above below but I would like to see what are some other ways to do it.

Suppose that $M$ is generated by $b_{1}, \ldots, b_{k}$ over $A$, and $M_{\mathfrak{p}}$ has a basis given by $\beta_{1}, \ldots, \beta _{ n} \in M_{ \mathfrak{p}}$. Let $\beta_{i} = m_{i} / s_{i}$ for $m_{i} \in M$, $s_{i} \in A \setminus \mathfrak{p}$. There exist $a_{ij} \in A$, $t_{ij} \in A \setminus \mathfrak{p}$ for $1 \leq i \leq n$, $1 \leq j \leq k$ such that $$b_{j} = \sum_{ i = 1 } ^ { n } \frac{a_{ij} } { t_{ij} } \frac { m_{i} } { s_{i } }$$ because $\beta_{i}$ form a basis. Let $g = \prod _{ i, j } t_{ij} \cdot \prod _ { i } s_{i}$ Consider the sequence $$0 \to I \to A _{g} ^ { n } \xrightarrow { \varphi } M_{g} \to 0$$ where the map $A_{g}^{n} \to M_{g}$ sends the the $e_{i}$ to $m_{i}$. Localizing this sequence at $\mathfrak { p }$ kills $I$, and since $I$ is finitely generated, one can localize at an element $h$ such that $I_{h} = 0$. The element $f = gh$ then works.
• You should perhaps be careful because the natural map $\alpha:M\to M_\mathfrak{p}$ is not necessarily injective. Of course $\ker \alpha_\mathfrak{p}=0$, so there must be an element $a\in A\setminus \mathfrak{p}$ such that $\ker \alpha_a=0$, so replacing $M$ by $M_a$ reduces to this case. – Meow Dec 24 '18 at 16:10