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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.

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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.

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    $\begingroup$ 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. $\endgroup$ – Meow Dec 24 '18 at 16:10

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