# Invertible ideal sheaf.

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?

• 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 $x \in X$. Dec 19, 2016 at 10:41
• 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 :) Dec 19, 2016 at 10:44
• 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? Dec 19, 2016 at 11:06
• 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 $x = x_0$. Dec 19, 2016 at 11:16

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.