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Let $\mathcal{L}$ be an invertible sheaf on a projective $A$-scheme $X$.

Then we can always find two very ample invertible sheaves such that $$\mathcal{L}= \mathcal{M} \otimes \mathcal{N}^*$$

(here * stands for the dual).

By a projective $A$-scheme I mean a scheme X that is isomorphic to a closed subscheme of $\mathbb{P}^n_A$, for some $n.$

I have tried using the line bundle associated with the closed embedding to the projective space, which would be very ample, but I don't know how to do anything meaningful with it.

Any help, please?

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    $\begingroup$ Just trying to rewrite @orangeskid idea. $\mathcal{P}^n$ is very ample for all n. Also for large enough n $\mathcal{P}^n \otimes \mathcal{L}$ is generated by global sections and now using Exercise 7.2(d) chapter 2 in hartshorne, we get that $\mathcal{P}^{n+1} \otimes \mathcal{L}$ is very ample. And now $\mathcal{L}$ is the difference of two very ample line bundles. $\endgroup$ – random123 Oct 17 '17 at 13:03
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I have a feeling for a proof: consider the line bundle associated to the projective embedding $\mathcal{O}(1)$. Then for every line bundle $\mathcal{L}$, the bundle $\mathcal{O}(n) \otimes \mathcal{L}$ will be very ample if $n$ is large enough. So now you have two very ample bundles $\mathcal{O}(n)$ and $\mathcal{O}(n) \otimes \mathcal{L}$, and you take the quotient of them to get $\mathcal{L}$.

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