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Let $R$ be a, say, noetherian ring and $M$ an $R$-module. The Wikipedia article on 'locally free sheaf' tells me that the following two statements are equivalent:

1. The module $M$ is locally free (Edit: this means there is an open cover $\{U_i\}$ of $Spec R$ such that every $\tilde{M}_{|U_i}$ is free as an ${\mathcal{O}_{Spec R}}{|U_i}$-module.)
2. $M_p$ is a free $R_p$-module for every prime ideal $p$ of $R$.

I see that these two things are equivalent if $M$ is finitely generated but I cannot see this in general, even if $R$ is noetherian. Am I missing something or is there a mistake on Wikipedia?

If the latter case is true, has anybody an example of an $R$-module $M$ over a noetherian $R$ such that $M_p=(R_p)^{n_p}$ for every prime ideal $p$ of $R$ and such that $M$ is not locally free?

Let $R$ be a, say, noetherian ring and $M$ an $R$-module. The Wikipedia article on 'locally free sheaf' tells me that the following two statements are equivalent:

1. The module $M$ is locally free (Edit: this means there is an open cover $\{U_i\}$ of $Spec R$ such that every $\tilde{M}_{|U_i}$ is free as an ${\mathcal{O}_{Spec R}}{|U_i}$-module.)
2. $M_p$ is a free $R_p$-module for every prime ideal $p$ of $R$.

I see that these two things are equivalent if $M$ is finitely generated but I cannot see this in general, even if $R$ is noetherian. Am I missing something or is there a mistake on Wikipedia?

If the latter case is true, has anybody an example of a (non-finitely generated) $R$-module $M$ over a noetherian $R$ such that $M_p=(R_p)^{n_p}$ for every prime ideal $p$ of $R$ and such that $M$ is not locally free?

Let $R$ be a, say, noetherian ring and $M$ an $R$-module. The Wikipedia article on 'locally free sheaf' tells me that the following two statements are equivalent:

1. The module $M$ is locally free, i.e. there is an open cover $\{U_i\}$ of $Spec R$ such that every $\tilde{M}_{|U_i}$ is free as an ${\mathcal{O}_{Spec R}}{|U_i}$-module.
2. $M_p$ is a free $R_p$-module for every prime ideal $p$ of $R$.

I see that these two things are equivalent if $M$ is finitely generated but I cannot see this in general, even if $R$ is noetherian. Am I missing something or is there a mistake on Wikipedia?

If the latter case is true, has anybody an example of an $R$-module $M$ over a noetherian $R$ such that $M_p=(R_p)^{n_p}$ for every prime ideal $p$ of $R$ and such that $M$ is not locally free?

Let $R$ be a, say, noetherian ring and $M$ an $R$-module. The Wikipedia article on 'locally free sheaf' tells me that the following two statements are equivalent:

1. The module $M$ is locally free (Edit: this means there is an open cover $\{U_i\}$ of $Spec R$ such that every $\tilde{M}_{|U_i}$ is free as an ${\mathcal{O}_{Spec R}}{|U_i}$-module.)
2. $M_p$ is a free $R_p$-module for every prime ideal $p$ of $R$.

I see that these two things are equivalent if $M$ is finitely generated but I cannot see this in general, even if $R$ is noetherian. Am I missing something or is there a mistake on Wikipedia?

If the latter case is true, has anybody an example of an $R$-module $M$ over a noetherian $R$ such that $M_p=(R_p)^{n_p}$ for every prime ideal $p$ of $R$ and such that $M$ is not locally free?

Let $R$ be a, say, noetherian ring and $M$ an $R$-module. The Wikipedia article on 'locally free sheaf' tells me that the following two statements are equivalent:

1. The module $M$ is locally free.
2. $M_p$ is a free $R_p$-module for every prime ideal $p$ of $R$.

I see that these two things are equivalent if $M$ is finitely generated but I cannot see this in general, even if $R$ is noetherian. Am I missing something or is there a mistake on Wikipedia?

If the latter case is true, has anybody an example of an $R$-module $M$ over a noetherian $R$ such that $M_p=(R_p)^{n_p}$ for every prime ideal $p$ of $R$ and such that $M$ is not locally free?

Let $R$ be a, say, noetherian ring and $M$ an $R$-module. The Wikipedia article on 'locally free sheaf' tells me that the following two statements are equivalent:

1. The module $M$ is locally free, i.e. there is an open cover $\{U_i\}$ of $Spec R$ such that every $\tilde{M}_{|U_i}$ is free as an ${\mathcal{O}_{Spec R}}{|U_i}$-module.
2. $M_p$ is a free $R_p$-module for every prime ideal $p$ of $R$.

I see that these two things are equivalent if $M$ is finitely generated but I cannot see this in general, even if $R$ is noetherian. Am I missing something or is there a mistake on Wikipedia?

If the latter case is true, has anybody an example of an $R$-module $M$ over a noetherian $R$ such that $M_p=(R_p)^{n_p}$ for every prime ideal $p$ of $R$ and such that $M$ is not locally free?