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I have an integral scheme $X$, and two coherent ideal sheaves $\mathcal I$ and $\mathcal J$ on $X$. I know there is a (maybe not closed) point $x$ of $X$ such that $\mathcal I$ and $\mathcal J$ define the same ideal sheaf in the stalk $\mathcal O_{X,x}$. Does it follow that there exists an open set $U$ containing $x$ such that $\mathcal I\vert_U = \mathcal J\vert_U$? I'm having trouble translating this into commutative algebra.

tim
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Because of Hartshorne's (not so great ...) book, the word "coherent" is often not used in its proper meaning. Coherent modules are defined as finitely generated modules whose finitely generated submodules are finitely presented. Thus, coherent modules are finitely presented and finitely presented modules are finitely generated, but over a non-noetherian base all these notions may differ. Over a non-noetherian base, being finitely presented is often the most natural finiteness condition. For ideals, being of finite type often suffices (in fact, if $I \subseteq \mathcal{O}_X$ is of finite type then $\mathcal{O}_X/I$ is of finite presentation). This is also the case here.

Let $X$ be an arbitrary ringed space. Let $I,J \subseteq \mathcal{O}_X$ be ideals of finite type with $I_x=J_x$ for some point $x \in X$. Choose generators $s_1,\dotsc,s_n \in \Gamma(W,I)$ of $I|_W$ for some open neighborhood $W$ of $x$. Then $(s_1)_x,\dotsc,(s_n)_x$ lie in $I_x=J_x$. It follows that there is some open neighborhood $U \subseteq W$ of $x$ such that $s_1|_U,\dotsc,s_n|_U$ lie in $\Gamma(W,J)$. (For each $s_i$ we find one and then we intersect; here we use the finiteness!) Since $s_1|_U,\dotsc,s_n|_U$ generate $I|_U$, we see $I|_U \subseteq J|_U$. Similarly, we find some open neighborhood $V$ of $x$ with $J|_V \subseteq I|_V$. Thus, $I$ and $J$ agree on $U \cap V$.

Community
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Martin Brandenburg
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    I like to remember this as an example of the support of a finitely generated module (here $I/(I\cap J)$) being closed. – Hoot Dec 12 '14 at 19:56
  • @Hoot: This is a very elegant alternative proof. In my proof I basically repeat the general argument that supp(f.g. module) is closed. – Martin Brandenburg Dec 12 '14 at 19:59