# Is twisting operation an intrinsic property of projective schemes?

Let $X=Proj(T),\quad Y=Proj(S),\quad T,S$ are graded rings. Suppose $(X,\mathcal{O}_X)\cong (Y,\mathcal{O}_Y)$. Is it always ture $\mathcal{O}_{X}(n)\cong \mathcal{O}_{Y}(n)$?

This problem comes out when I try to justify one step in a proof of Hartshorne:

Let $X$ be a projective scheme over a noetherian ring $A$("projective" here means there is a close immersion $i:X\to \mathbb{P}_{A}^{r}$), and let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_X$-module, then $i_{\star}(\mathcal{F}(n))=(i_{\star}\mathcal{F})(n)$.

I try to consider the simplest case where $\mathcal{F}=\mathcal{O}_X$. Although I know there exist a homogenous ideal $I\subset A[x_1 \dots x_r]$, s.t. $X\cong Proj(A[x_1 \dots x_r]/I)$, I know nothing about $\mathcal{O}_{X}(n)$. For one thing, $I$ may not be unique, for another $X$ can isomorphism to a complete different $Proj(S)$. The question thus can also be phrased as: Does it make sense to say twisting operation without knowing the graded ring of a projective scheme.

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I think the answer is no even when you restrict the question to projective schemes. See Ch. I Ex. 3.9. – Zhen Lin Jul 24 '11 at 8:50
Thank you for pointing out that exercise. It is an example of$Proj S\cong Proj T$ with S not isomorphism to T, but do they have different twisting sheaves? So far, I cannot check this. – Li Zhan Jul 25 '11 at 9:05

The Serre twist is more of a property of "embedding in projective space" than an intrinsic one. If $X$ is a scheme, say of finite type over a noetherian ring $A$ (one can loosen these hypotheses), one would say that a line bundle $\mathcal{L}$ on $X$ is very ample if there is an immersion of $X$ into some projective space over $A$ (Hartshorne assumes some $\mathbb{P}^n_A$; EGA assumes more general projectivizations of vector bundles) such that $\mathcal{L}$ is the pull-back of $\mathcal{O}(1)$. So your question equates to asking whether a very ample line bundle on a projective scheme is uniquely determined. The answer is almost always no: if $\mathcal{L}$ is very ample, so is any positive tensor power $\mathcal{L}^{\otimes d}$.
This corresponds to the fact that $\mathrm{Proj} A$ is isomorphic to $\mathrm{Proj} A^{(d)}$ (elements whose homogeneous components have degree divisible by $d$), but that this isomorphism (the $d$-uple embedding) pulls $\mathcal{O}(1)$ on $\mathrm{Proj} A^{(d)}$ back to $\mathcal{O}(d)$ on $\mathrm{Proj} A$.
@Li Zhan: An ample line bundle on a projective scheme is a way of "presenting" the scheme as $\mathrm{Proj} A$ for a graded ring $A$. Different ample line bundles will give different "presentations." In particular, this "presentation" determines $\mathcal{O}(1)$ (as just the line bundle itself). – Akhil Mathew Jul 28 '11 at 2:20