On page 150 of Algebraic Geometry by Hartshorne, line 4 of paragraph 2, it is said that $\mathcal{O}(1)$ is generated by the global sections $x_1, \ldots, x_n$. How to show that $\mathcal{O}(1)$ is generated by the global sections $x_1, \ldots, x_n$?

By definition, $\mathcal{O}(1) = \widetilde{S(1)}$, $S=k[x_0, \ldots, x_n]$. Let $U_i=\{(x_0, \ldots, x_n): x_i \neq 0\}$ be the open covering of $\mathbb{P}^{n}_k$. Then $\mathcal{O}(1)(U_i)$ is the set of functions $s: U_i \to \sqcup_{p \in U_i} S(1)_{p}$ such that $s(p) \in S(1)_p$ and $s$ satisfies some properties. I think that $S(1)_n = S_{n+1}$. Therefore $S(1)_0 = S_1 = \sum_{i=0}^{n} k x_i$. But how to show that $\mathcal{O}(1)$ is generated by the global sections $x_1, \ldots, x_n$? Thank you very much.


To say that $\mathcal{O}(1)$ is generated by the global sections $x_0,\ldots,x_n$ means we have a surjective morphism of $\mathcal{O}_X$-modules

$$\mathcal{O}_X^{\oplus(n+1)} \to \mathcal{O}(1) \to 0$$ where the global section $e_i$ of $\mathcal{O}_X(X)^{\oplus( n+1)}$ maps to $x_i\in \mathcal{O}(1)(X)$. Thus it is enough to compute global sections of $\mathcal{O}(1)$. Do you know how to do this?

The answer is: $$\mathcal{O}(1)(X) = \{\text{linear polynomials in $x_0,\ldots,x_n$}\}.$$

How to compute this

Use the standard affine open cover of $\Bbb{P}^n_k$ by $U_0,\ldots,U_n$. Recall that $\mathcal{O}(1)(U_i) = k[x_0,\ldots,x_n]_{(x_i)}$ where $(x_i)$ means the degree zero component of the localization at $\{1,x_i,x_i^2,\ldots\}$.

Now say you had a global section $s$. What does $s_i:= s|_{U_i}$ look like? Say it is $f_i(x_0,\ldots,x_n)/x_i^{k_i}$ where $f_i$ is homogeneous of degree $k_i$. Since the polynomial ring over a field is a UFD we may assume the denominator and numerator for every $i$ have no common factor. Now what does it mean to now say $$s_i|_{U_i \cap U_j} = s_j|_{U_i \cap U_j}?$$

Can you deduce your original $f_i$'s had to be linear? If you're confused then just do the case of $X =\Bbb{P}^1$; I have always found that very illuminating.

  • $\begingroup$ thank you very much. How to show that $f_i$'s are linear? I think that in line 7 you mean $$ \mathcal{O}(1) (\mathbb{P}^n)= \{\text{linear polynomials in }x_0, \ldots, x_n\}. $$ $\endgroup$ – LJR Dec 25 '13 at 16:40
  • $\begingroup$ @LJR Try doing it yourself first (if you're stuck do $\Bbb{P}^1$) and if all else doesn't work ping me and i'll edit my answer. $\endgroup$ – user38268 Dec 25 '13 at 23:34
  • $\begingroup$ Let $s \in \mathcal{O}(1)(\mathbb{P}^1)$. Then $s|_{U_0}=g/x_0^d$, $s|_{U_1}=g'/x_1^{\ell}$ for some $g, g'$, $\deg g = d+1, \deg g' = \ell+1$. Let $(1: x_1)\in U_0 \cap U_1$. Suppose that $g(x_0, x_1) = \sum_{i+j=d+1} k_{ij} x_0^i x_1^j$ and $g'(x_0, x_1) = \sum_{i+j=\ell+1}k'_{ij}x_0^i x_1^j$. Then $s(1, x_1) = g(1, x_1)=\sum_{i+j=d+1}k_{ij}x_1^j$ and $s(1, x_1)=g'(1,x_1)/x_1^\ell=\sum_{i+j=d+1}k'_{ij} x_1^j/x_1^\ell=\sum_{i+j=\ell+1}k'_{ij}x_1^{j-\ell}$. Therefore the degree of $x_1$ in $s(1, x_1)$ is less or equal to $1$ and greater or equal to $0$. $\endgroup$ – LJR Jan 5 '14 at 10:25
  • $\begingroup$ Similarly, the degree of $x_0$ in $s(x_0,1)$ is less or equal to 1 and greater or equal to $0$. Since $s(x_0, x_1)$ is of degree 1, $s(x_0, x_1)$ is a linear combination of $x_0, x_1$. $\endgroup$ – LJR Jan 5 '14 at 10:27
  • $\begingroup$ $s(1, x_1), s(x_0, 1)$ are polynomials of degree less or equal to 1. Therefore $s(x_0, x_1)$ is a polynomial. Since $s \in \mathcal{O}(1)(\mathbb{P}^1)$, $s$ is of degree 1. $\endgroup$ – LJR Jan 5 '14 at 14:58

Just in case there is someone who wants to think about it in another way. (This is how I thought about it.)

The situation is the following. Let $A$ be a ring, $\mathbb{P}^n_A= \text{Proj}(S)$ with $S=A[x_0,\dots,x_n]=\bigoplus_{d\geq 0}S_d$ (standard grading) be the $n$-th projective space over $A$ and let us for short denote by $\mathscr{O}=\mathscr{O}_{\mathbb{P}^n_A}$ the structure sheaf of $\mathbb{P}^n_A$.

What is meant by "$\mathscr{O}(1)$ is generated by the global sections $\{x_i\}_i$" is that for every point $P \in \mathbb{P}^n_A$, the stalk $\mathscr{O}(1)_P$ is generated over $\mathscr{O}_P$ by the germs $\{{x_i}_P \}_i$ of the global sections at $P$. But here one has to be carefully (at least at the first attempt): the $x_i$ are not "really" global sections of the twist $\mathscr{O}(1)$. At first we have to clarify how they give rise to global sections of $\mathscr{O}(1)$.

By definition, \begin{align*} \mathscr{O}(1)=S(1)^{\sim}=(\bigoplus_{d\geq 0}S_{d+1})^{\sim}=(S_+)^{\sim}. \end{align*} Then $x_i$ gives a map \begin{align*} X_i \colon \mathbb{P}^n_A &\longrightarrow \coprod_{P \in \mathbb{P}^n_A} (S_+)_{(P)}, \\ P & \longmapsto \big(\frac{x_i}{1},P\big), \end{align*} (here note that $\frac{x_i}{1}$ is a degree zero element in the localization $(S_+)_P$, since $x_i$ is a degree zero element in the grading of $S_+$) which is a global section of $\mathscr{O}(1)$.

Now using the natural isomorphisms \begin{align*} (\mathscr{O}(1))_P \cong (S_+)_{(P)} \;\;\; \text{and} \;\;\; \mathscr{O}_P \cong S_{(P)}, \end{align*} we see that it is enough to show that the elements $\frac{x_i}{1} \in (S_+)_{(P)}$ (corresponding to the germs of the $X_i$ at $P$) generate $(S_+)_{(P)}$ over $S_{(P)}$.

But $\{ x_i \}_i$ generates $S_+$ over $S$, hence we get that $\{ \frac{x_i}{1} \}_i$ generates $(S_+)_{(P)}$ over $S_{(P)}$ as desired.


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