Let $X \hookrightarrow \mathbb{P}^{n}_k$ be a closed subscheme with $k$ an algebraically closed field. Then, the line bundle $\mathcal{O}(1)$ on $\mathbb{P}^n_k$ pulls back to $\mathscr{L} = \mathcal{O}_X(1)$ which is generated by the global sections $x_0, \dots, x_n$. Let $P$ and $Q$ be closed points and suppose $s = \sum a_i x_i$ defines a hyperplane containing $P$ and not $Q$.

Then, $s$ defines a global section of $\mathscr{L}$ and I would like to show that $s \in \mathfrak{m}_P \mathscr{L}_P$ and $s \notin \mathfrak{m}_Q \mathscr{L}_Q$. I do not see how to prove this although it seems intuitive.

For context, this is the situation in Hartshorne's proof of Proposition II.7.3. Please let me know if you need me to clarify any notation.

My Attempt:

I'll let $I$ be the homogenous ideal of $X$. Then $(\mathcal{O}_X)_{P} = \left(k[x_0, \dots, x_n]/I \right)_{(\mathfrak{p})}$ where $\mathfrak{p} = \langle P_ix_j - P_Jx_i\rangle$ is the ideal of the closed point $P = [P_0: P_1: \cdots : P_n]$, since $k$ is algebraically closed. Then, $\mathfrak{m}_P$ is the extension of this ideal in the degree 0 localization $\left(k[x_0, \dots, x_n]/I \right)_{(\mathfrak{p})}$. Then, we know that $\sum a_i P_i =0$. I'm then confused how to write $\sum a_ix_i$ as a product of elements of $\mathfrak{m}_P$ and $\mathscr{L}_P$.

Any help is greatly appreciated.


1 Answer 1


Let $f$ be a linear form with $P,Q\in D(f)$. The sections of $\mathcal{O}_{\Bbb P^n}(1)$ over $D(f)$ are $f\cdot k[\frac{x_0}{f},\cdots,\frac{x_n}{f}]$, so the sections of $\mathscr{L}$ over $X\cap D(f)$ are $f\cdot k[\frac{x_0}{f},\cdots,\frac{x_n}{f}]/I_X$. Next, $s$ vanishes at $P$ iff $s=f\cdot\frac{s}{f}=f\cdot \sum a_i\frac{x_i}{f}$ also vanishes at $P$. Going to stalks, $f\in\mathscr{L}_P$ and $\frac{s}{f}\in\mathfrak{m}_P$, so $s\in \mathfrak{m}_P\mathscr{L}_P$ iff $s$ vanishes at $P$. The claim for $Q$ is proved in exactly the same manner.

  • $\begingroup$ Dear KReiser, thank you very much for the nice answer. I apologize for taking so long to accept the answer and give the bounty; I did not know they were separate selections. Also, I have been moving these past weeks so I have not been as attentive as I would have hoped to be. Either way, I appreciate your insight. $\endgroup$
    – Daniel
    Commented Sep 24, 2021 at 21:50
  • $\begingroup$ Don't worry about it, I've also been a bit busy lately. Glad to help! $\endgroup$
    – KReiser
    Commented Sep 28, 2021 at 18:47

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