4
$\begingroup$

I am a little bit confused with the idea of an invertible ideal sheaf. I cannot convince myself that there are invertible ideal sheaves on a scheme $X$ non isomorphic to the structure sheaf of $X$. I would like to know where my reasoning fails.

Given an invertible ideal sheaf $\mathcal{I} \subset \mathcal{O}_X$ it comes equipped with the inclusion morphism into the structure sheaf. If the localization $\mathcal{I}_x \cong \mathcal{O}_{X,x}$ how is possible that the inclusion morphism does not induce an isomorphism between stalks?

$\endgroup$
4
  • 1
    $\begingroup$ The point is that an isomorphism $I_x \cong O_{X,x}$ is abstract, it is not induced by the embedding $I \to O_X$. In fact, if $I$ is an invertible ideal, the morphism $I_x \to O_{X,x}$ induced by the embedding is zero for some points $x \in X$. $\endgroup$
    – Sasha
    Dec 19, 2016 at 10:41
  • $\begingroup$ Hi @Sasha, I knew the reason must be that but I couldn't find myself a nice example. Could you add an example as an answer so I can accept it? Thanks for your time :) $\endgroup$
    – Abellan
    Dec 19, 2016 at 10:44
  • $\begingroup$ One moment in the affine case if $I \subset A$ is an invertible ideal if we consider some maximal ideal $I \subset \mathfrak{m}$ then no element of $I_{\mathfrak{m}}$ will be mapped to $1 \in A_{\mathfrak{m}}$ right? $\endgroup$
    – Abellan
    Dec 19, 2016 at 11:06
  • 1
    $\begingroup$ The simplest example is $I = O(-1)$ on $P^1$, with the embedding $O(-1) \to O$ corresponding to a point $x_0 \in P^1$. In homogeneous coordinates $(u:v)$ on $P^1$ if $x_0 = (u_0:v_0)$ then this map is given by $v_0 u - u_0 v \in \Gamma(P^1,O(1)) = Hom(O(-1),O)$. Then the map $I_x \to O_{X,x}$ is an isomorphism for $x \ne x_0$ and is zero for $x = x_0$. $\endgroup$
    – Sasha
    Dec 19, 2016 at 11:16

1 Answer 1

0
$\begingroup$

Consider another example : Let $X=Spec A$ where $A=k[x,y]/y^2=x^3-x$ and $k$ some algebraically closed field. $X$ is a nonsingular elliptic curve, hence $A$ is a Dedekind domain and every prime ideal is maximal. We want to show that every maximal ideal is invertible i.e. $A_p \cong m_p$. Let $m$ be the maximal ideal and $p$ any prime ideal. If $m \not\subset p$ then $(A-p) \cap m \neq \emptyset$ so $A_p \cong m_p$. In case $m\subset p$ we have $m =p$ (since every prime ideal is maximal) and $A_m$ is a DVR hence $mA$ is a principal ideal i.e invertible.

Hence we have the isomorphism of stalks, but the ideal is not isomorphic to the ring.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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