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Let $ X $ be a Noetherian integral regular scheme and $ \mathcal{F} $ be a torsion-free coherent sheaf. (One definition of torsion-free is that the natural map $ \mathcal{F} \rightarrow \mathcal{F} \otimes_{\mathcal{O}_X} \mathcal{K}_X $ is injective, where $ \mathcal{K}_X $ is the constant sheaf of the fraction field of $ X $. In general, care has to be taken when the scheme is not integral.)

Question 1: Is it true that $ \mathcal{F} $ always embeds into a locally free sheaf?

Some things I can do: If $ X $ is a smooth projective variety say over the complex numbers, then this is true. Indeed, being torsion-free implies that $ \mathcal{F} \rightarrow \mathcal{F}^{\vee \vee} $ is an injection. Now by Serre's theorem, a high enough twist of $ \mathcal{F}^{\vee} $ is globally generated, that is, there is a surjection $ \mathcal{O}_X^p \rightarrow \mathcal{F}^{\vee} (n) $ for $ n $ a large number. Tensoring by $ \mathcal{O}_X(-n) $ and dualizing gives an injection $ \mathcal{F}^{\vee \vee} \rightarrow \mathcal{O}_X(n)^p $ and now just compose the two injections to finish the proof.

This proof can be adapted if we replace projective by quasi-projective.

Question 2: Is it true that $ \mathcal{F} $ always embeds into a locally free sheaf of rank equal to the rank of $ \mathcal{F} $?

This is easily true if $ X $ is affine and in that case, we even have $ \mathcal{F} \rightarrow \mathcal{O}_X^{rk (\mathcal{F})} $.

But I'm not sure of a complete answer and I would like one in the greatest generality, if possible some answers for the case of non-integral schemes too.

This was posted on MSE without an answer.

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    $\begingroup$ The closed complement of every open affine (with its reduced scheme structure) is a Cartier divisor in your regular Noetherian scheme. So there are "enough locally free sheaves." $\endgroup$ Commented May 29, 2022 at 10:53

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