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Let $Q_1, Q_2\subset \mathbb{P}^4$ be two degree two smooth hypersurfaces, and $S:=Q_1\cap Q_2$ be their complete intersection. If $S$ is smooth, then it is known that $S$ is a del Pezzo surface of degree 4. Now let $c_1, c_2\subset C$ be two conics.

I'm wondering that is there any criterion for $\operatorname{Hom}(I_{c_1/S}, I_{c_2/S})\neq 0$? Here $I_{c_i/S}$ is the ideal sheaf of $c_i$ in $S$.

When $S$ is smooth, then using some theory of curves on del Pezzo surface, we can prove that $\mathcal{O}_S(-c_1)\cong \mathcal{O}_S(-c_2)$. But when $S$ is singular, is there any similar result? For example, if there's a quadric $Q_3$ in the pencil spaned by $Q_1$ and $Q_2$, such that $\langle c_1\rangle\cup \langle c_2\rangle\cup S\subset Q_3$, where $\langle c_i\rangle$ is the projective plane spaned by $c_i$, then can we conclude that $\operatorname{Hom}(I_{c_1/S}, I_{c_2/S})\neq 0$?

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  • $\begingroup$ When $S$ is smooth it is not true in general that $\mathcal{O}_S(-c_1) \cong \mathcal{O}_S(-c_2)$. In fact, there are 10 linear systems of conics on $S$, and if you take conics from different linear systems, their ideal sheaves are not isomorphic. $\endgroup$
    – Sasha
    Nov 3, 2021 at 9:34
  • $\begingroup$ Oh thanks! Maybe I need to double-check my computations. Is it true under the assumption that $<c_1>$ and $<c_2>$ are both in $Q_3\in <Q_1, Q_2>$? $\endgroup$
    – Kim
    Nov 3, 2021 at 9:58
  • $\begingroup$ Still not true in general --- in this case $Q_3$ must be singular, and if it is a cone over a smooth quadric surface $\bar{Q}_3$, planes on $Q_3$ correspond to lines on $\bar{Q}_3$, and if two lines belong to different rulings, conics are not linearly equivalent. $\endgroup$
    – Sasha
    Nov 3, 2021 at 10:07
  • $\begingroup$ Oh I see, you mean if $<c_1>$ and $<c_2>$ in $Q_3$ that correspond to two non-linearly equivalent lines on $\overline{Q_3}$, then $<c_1>$ and $<c_2>$ are also not linearly equivalent on $Q_3$, and thus $c_1$ and $c_2$ are not linearly equivalent on $S$? $\endgroup$
    – Kim
    Nov 3, 2021 at 10:16

1 Answer 1

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If $f \colon I_{c_1} \to I_{c_2}$ is a non-trivial morphism then it is an isomorphism. Indeed, $f$ is injective because both $I_{c_1}$ and $I_{c_2}$ are torsion free, and since the Hilbert polynomials of $I_{c_1}$ and $I_{c_2}$ are equal, the Hilbert polynomial of the cokernel is zero, hence the cokernel vanishes.

Thus, $\mathrm{Hom}(I_{c_1},I_{c_2}) \ne 0$ if and only if $c_1$ and $c_2$ are linearly equivalent Weil divisors.

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