Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mathbb{K}$ be an algebraically closed field, $S=\mathbb{K}[T_0, \dots, T_r]$ and $X=\mathrm{Proj}(S)=\mathbb{P}^r_{\mathbb{K}}$. On page 117, Hartshorne defines the twisting sheaf of Serre to be $\mathcal{O}_X(1)=S(1)^{\sim}$ which is the sheaf associated to $S(1)$ on $X$.

On page 120, he further defines the twisting sheaf for any scheme $Y$. According to this definition, we have $\mathcal{O}_X(1)=g^*(\mathcal{O}(1))$, where $g\colon X=\mathbb{P}^r_{\mathbb{Z}} \times_{\mathrm{Spec}(\mathbb{Z})} \mathbb{P}^r_{\mathbb{K}} \; \rightarrow \;\mathbb{P}^r_{\mathbb{Z}}$ is the natural map.

I'm totally confused by these definitions. Probably this is a stupid question, but do they coincide? And if yes, why?

Thanks a lot in advance!

share|improve this question
    
Dear claudi, $Y$ does not appear in your second definition. –  Bruno Joyal Oct 24 '13 at 19:15
2  
Oh my god, "definitions" in Hartshorne ... –  Martin Brandenburg Oct 24 '13 at 19:47
1  
@ Marie: I know, I'm considering the case $Y=X$. So my question is whether the twisting sheaf $g^*(\mathcal{O}_X(1))$ (second definition) is the same as $S(1)^{\sim}$ (first definition)... –  claudi Oct 24 '13 at 20:01
1  
The twisting sheaf is not defined on any scheme $Y$ (!); it's defined, for any scheme $Y$, on $\mathbb{P}_Y^r = \mathbb{P}_{\mathbb{Z}}^r\times Y$. The question becomes sensitive for $Y = Spec(k)$, because then the $X$es become isomorphic. –  Ben A. Oct 25 '13 at 6:54
    
Thanks Ben A., your answer helped me a lot. –  claudi Oct 27 '13 at 13:43

1 Answer 1

First you need a proposition

Proposition 5.12 (c) Let $S$ be a graded ring and let $X = \text{Proj } S$. Assume that $S$ is generated by $S_1$ as an $S_0$ algebra. Let $T$ another graded ring, generated by $T_1$ as a $T_0$ algebra, let $\phi: S\to T$ a morphism preserving degrees, and let $U\subset Y = \text{Proj } T$ and $f: U\to X$ be the morphism determined by $\phi$. Then $f^*(\mathcal{O}_X(n)\cong \mathcal{O}_Y(n)|_U$

Remember that the induced morphism $f$ is $f(\mathfrak{p}) = \phi^{-1}(\mathfrak{p})$ this is well defined because $\phi$ is a graded morphism. Now let me cite a lemma wich you surely have been using but maybe didn't notice

Lemma. $\text{Proj} \, A[x_0,...,x_n] = \text{Proj} \, \mathbb{Z}[x_0,...,x_n] \times_{\text{Spec }\mathbb{Z}} \text{Spec} \, A$

What you need to show is the following

Claim. If $Z=\text{Spec } A$ then the twisting sheaf $\mathcal{O}(1)$ on $\mathbb{P}^r_Z$ is the same as the $\mathcal{O}(1)$ defined on $\mathbb{P}^r_A$

Proof. We are going to show that satisfies the hypothesis of the previous proposition, so let $S=\mathbb{Z}[x_0,\dots,x_r]$ and let $T=A[y_0,\dots,y_r]$ both with the natural graduation. Consider the canonical morphism $\phi: S\to T$ and let $Y=\text{Proj } T$, in this case use $U=Y$. Now $X= \mathbb{P}_\mathbb{Z}^r$ and by the lemma

$$\mathbb{P}^r_Z = \mathbb{P}^r_\mathbb{Z}\times_{\text{Spec }\mathbb{Z}} \text{Spec } A = \text{Proj} \, \mathbb{Z}[x_0,...,x_r] \times_{\text{Spec }\mathbb{Z}} \text{Spec} \, A = \text{Proj } A[x_0,\dots,x_r] = \text{Proj } T = Y$$

Observe that by definition the induced map $f: Y \to X$ is the same as the natural map $g$ used in your second definition, so by the proposition $\mathcal{O}(1) = g^*(\mathcal{O}_X(1)) = f^*(\mathcal{O}_X(1)) = \mathcal{O}_Y(1)$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.