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I am confused about the connection between vector bundles and locally free sheaves over schemes. In a past lecture we followed Andreas Gathmann: Algebraic Geometry (2002/2003), the definition of locally free vector sheaf and vector bundle from that source are repeated here:

Definition: Let $X$ be a scheme. A sheaf of $O_X$-modules $\mathcal{F}$ is called locally free of rank $r$, if there is an open cover $\{ U_i \}$ of $X$ such that $\mathcal{F} \mid_{U_i} \simeq O_{U_i}^{\oplus r}$ for all $i$.

Definition: A vector bundle of rank $r$ on a scheme $X$ over a field $k$ is a $k$-scheme $F$ and a $k$-morphism $\pi:F \rightarrow X$, together with the additional data consisting of an open covering $\{ U_i \}$ of $X$ and isomorphisms $\Psi_i: \pi^{-1}(U_i) \rightarrow U_i \times \mathbb{A}_k^r$ over $U_i$, such that the automorphism $\Psi_i \circ \Psi_j^{-1}$ of $(U_i \cap U_J) \times \mathbb{A}^r$ is linear in the coordinates of $\mathbb{A}^r$ for all $i$, $j$.

It is argued later, that free rank $r$ sheaves are in 1:1 correspondence with vector bundles of rank $r$. Is this to be understood as: Fix $X$ to be a scheme over $k$. Then vector bundles of rank $r$ on $X$ are in 1:1 correspondence with free sheaves of rank $r$ on $X$.

I am not sure, whether or not I understood correctly, because $k$ is mentioned in the second definition but not in the first.

Also, I do not understand what a $k$-scheme is. (Neither our lecture nor Hartshorne gives a definition, the only definition I found that sounds similar is definition 32.20.1. of the Stacks project. However trying to understand this a get lost backtracking dozens of previous definitions) But it seems that in this case it is enough to think of $F$ as a scheme over $k$. The other $k$-scheme properties will be forced on $F$ by the special local structure. Is that correct?

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  • $\begingroup$ You were right in the comment to my answer. I stand by my statement that a $k$-scheme is a scheme over $k$. $\endgroup$ – Daniel Robert-Nicoud Mar 11 '17 at 23:29
  • $\begingroup$ @DanielRobert-Nicoud But why does the definition of "algebraic $k$-scheme" on the Stacks project look so complicated? See here: stacks.math.columbia.edu/tag/06LF . Why does the structure morphism have to be of finite type? But as far as vector bundles are concerned: If I just take $k$-schemes to be schemes over $k$, then I get the 1:1 correspondence. A vector bundle probably is of finite type in some sense, so may be something better than an ordinary scheme over $k$, but I don't really need to worry about that. Thanks for the reply! $\endgroup$ – user424591 Mar 13 '17 at 15:10
  • $\begingroup$ @user424591 well, $k$-scheme is a scheme over $k$ according to all of the modern sources I'm aqcuainted with. Algebraic $k$-scheme is different from $k$-schemes, I'm not familiar with them, but according to the link in your comment they something looking similar to varieties, since section on algebraic schemes is contained in the chapter on varieties on the website above. This is where the finiteness condition is coming from. $\endgroup$ – Mihail Mar 8 at 23:11

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