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Let $f:Z\to X$ be an immersion of schemes. Let $E$ be a vector bundle on $X$(coherent and locally free of finite type and say constant rank $n$).

Suppose that $O_X$ is not necessarily coherent.
It seems that the $O_Z$-Module $f^*E$ need not to be a vector bundle. Which kind of conditions on $f$ could be added to ensure $f^*E$ to be a vector bundle?

(I know that locally free of finite type means (flat+of finite presentation), so maybe a question related is when does a pull-back of a flat $O_X$-Module flat?)

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    $\begingroup$ I see absolutely no reason to assume that a "vector bundle" should correspond to a coherent sheaf. This would make the notion of vector bundle not functorial. $\endgroup$
    – Angelo
    Commented Sep 20, 2010 at 11:53
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    $\begingroup$ I am confused. In which sense can $O_X$ be not coherent? Are you schemes not noetherian? Or did you mean $f^* O_X$ not coherent? $\endgroup$
    – Bugs Bunny
    Commented Sep 20, 2010 at 13:21
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    $\begingroup$ I am still confused: even on a non-noetherian scheme, how could the structure sheaf fail to be coherent? It should be trivial that it is coherent. $\endgroup$ Commented Sep 20, 2010 at 16:10
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    $\begingroup$ Hartshorne defines coherent sheaves on Noetherian schemes only, in which case the two definitions agree and the structure sheaf is always coherent. (A submodule of a finitely generated module over a Noetherian ring is always finitely generated, so the condition on kernels becomes redundant in the Noetherian case.) $\endgroup$ Commented Sep 20, 2010 at 16:23
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    $\begingroup$ The right definition can be found in EGA for instance. Hartshorne's book is very nice, but this one place where I think he could have been more explicit that he was taking a short cut. $\endgroup$ Commented Sep 20, 2010 at 16:44

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For noetherian schemes the pull-back of coherent (quasicoherent, locally free) sheaf retains that property, cf. II.5.8 in Hartshorne. The notion of coherence becomes subtle in non-noetherian case but the pull-back of locally free sheaf is locally free. However, locally free finite rank may not imply coherent.

If I understand it correctly, the question as posed has little substance. If $O_X$ is not coherent, any locally free $O_X$-module is not coherent either. Hence, $X$ has no "vector bundles"...

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  • $\begingroup$ Also, Hartshorne uses the wrong definition of coherence. $\endgroup$ Commented Sep 20, 2010 at 15:33
  • $\begingroup$ Although Hartshorne does say (after defining coherence) something along the lines of "only for use with Noetherian schemes". Hartshorne's definition does certainly cause substantial confusion though. $\endgroup$ Commented Sep 20, 2010 at 16:52
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    $\begingroup$ Hartshorne is not interested in non-noetherian schemes. His definition is OK in this case. Nothing I said is wrong, btw. $\endgroup$
    – Bugs Bunny
    Commented Sep 21, 2010 at 9:10
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Let $f \colon X \to Y$ be a map of locally ringed spaces, and $\mathcal{E}$ a sheaf of locally free finite rank $\mathcal{O}_Y$-modules. Then $f^*\mathcal{E}$ is a sheaf of locally free finite rank $\mathcal{O}_X$-modules.

(If you agree that a vector bundle is a a sheaf of locally free finite rank $\mathcal{O}$-modules, I assume this is a fairy general answer. I guess it works also for ringed topos, but I will sketch the proof only in the space case.)

Indeed, the property of being "locally free of finite rank" is local, therefore the problem reduces to the case of a free finite rank sheaf, but in this case its inverse image is free because $f^*$ is an additive functor.

For the right notion of coherent sheaf I suggest to look at EGA $0_I$ 5.3 (new edition). Hartshorne's definition works well only for noetherian schemes. Also, for a discussion on the relationship between locally free sheaves (or, more generally sheaves of $\mathcal{O}$-modules) and its associated shemes (that play the role of bundle spaces) look at EGA I, 9.4 (new edition). In the old edition I think it is somewhere in EGA II.

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