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Nov 21, 2012 at 9:06 vote accept Lennart Meier
Nov 20, 2012 at 15:46 answer added Will Sawin timeline score: 1
Nov 20, 2012 at 15:34 answer added Mahdi Majidi-Zolbanin timeline score: 1
Nov 20, 2012 at 15:08 history edited Lennart Meier CC BY-SA 3.0
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Nov 20, 2012 at 14:05 comment added Jason Starr @Lennart: With your definition, my morphism $X\hookrightarrow Y$ below is "locally free of finite rank". Yet I give a locally free $\mathcal{O}_X$-module $\widetilde{M}|_X$ such that $\Gamma(X,\widetilde{M}|_X)$ is not a locally free module over $\Gamma(X,\mathcal{O}_X)$.
Nov 20, 2012 at 13:49 comment added Lennart Meier I would define a map $f: X \to Y$ to be locally free of finite rank if for every $x\in X$, the local ring $\mathcal{O}_{X,x}$ is free of finite rank over $\mathcal{O}_{Y,f(x)}$.
Nov 20, 2012 at 13:43 history edited Lennart Meier CC BY-SA 3.0
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Nov 20, 2012 at 13:11 comment added Qfwfq What do you mean by "the natural map $X\to\mathrm{spec}\Gamma(\mathcal{O}_X)$ is locally free of finite rank"?
Nov 20, 2012 at 13:06 answer added Jason Starr timeline score: 10
Nov 20, 2012 at 13:04 comment added Fred Rohrer Dear @Lennart, your statement about affine schemes contradicts an example given in EGA I.1.4.4.1 (1970 edition).
Nov 20, 2012 at 12:45 history asked Lennart Meier CC BY-SA 3.0