I'm not sure that this is generally true, but Harthorne p73 seems to suggest it. If it is true could someone give me a hint for the proof?
If $f$ an isomorphism of ringed spaces, is $f$ necessarily an isomorphism of locally ringed spaces?
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4Of course this is true, provided that the stalks are local rings. Because the isomorphism induces isomorphisms of stalks. And an isomorphism of local rings must carry the maximal ideal into the maximal ideal. – 2012-10-15
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1But the induced isomorphism isn't to the same stalk as in the definition of locally ringed space. In particular you get an induced isomorphism to $f_{*}(\mathcal{O}_X)_{f(P)}$ but we need that the homomorphism to $(\mathcal{O}_X)_P$ obtained by composing this isomorphism with the restriction map is a local homomorphism. I don't quite see how to do this! – 2012-10-15
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1In the paragraph you're referring to, Hartshorne is assuming that $f$ is a morphism of locally ringed spaces. – 2012-10-16
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1@EdwardHughes:I am sorry but I don't quite see what you mean. See if you agree with this: $f_{*,x}:\mathcal O_{Y,f(P)} \to f_*(\mathcal O_X)_{f(P)}$ is an isomorphism. – 2012-10-16