The definition of a (geometric) vector bundle over a scheme $X$ can be rewritten as follows in terms of 'not-so-geometrical algebra' if $X=Spec R$ is affine and if I am not missing something.

A vector bundle of rank $n$ over $R$ is an $R$-algebra $A$ such that

  • for every $p\in Spec R$ there is a isomorphism (belonging to the data) $$ \phi_p:k(p)[X_1,...,X_n]\xrightarrow{\cong} A\otimes_R k(p) $$ where $k(p)$ is the residue field $R_p/m_p$ and

  • there are elements $\{a_i\}_{i\in I}$ of $R$ such that the $D(a_i)=\{ I\in Spec R\mid a\notin I \}$ cover $Spec R$ and for every $i\in I$ there is an $R_a$-algebra isomorphism $$ A\otimes_R R_{a_i}\xrightarrow{\cong}R[X_1,\ldots,X_n]\otimes_RR_{a_i} $$ which induces for every $p\in Spec R$ with $a_i\notin p$ a $k(p)$-linear $k(p)$-algebra isomorphism $$ A\otimes_R k(p)~\xleftarrow{\phi_p}k(p)[X_1,...,X_n]\to k(p)[X_1,...,X_n]\cong R[X_1,...,X_n]\otimes_R k(p). $$

The (isomorphism classes) of such vector bundles over $R$ should correspond to (isomorphism classes) of finitely gererated projective modules over the ring $R$.

How can this correspondence be seen?

  • $\begingroup$ Isn't this just a special case of the correspondence between vector bundles and locally free $\mathcal{O}_X$-modules? In which case, take your vector bundle and look at its sheaf of global sections (which will just be an $R$-module in your affine setting). $\endgroup$ Nov 2, 2010 at 18:47
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    $\begingroup$ Yes, it is a special case. Unfortunately I do not understand the general case and what happens there with the symmetric algebra. This is why I would like to see how this works in the affine case. $\endgroup$
    – roger123
    Nov 2, 2010 at 18:51
  • $\begingroup$ It is a standard commutative algebra fact that a module is locally free of finite rank if and only if it is finitely generated and projective. You can find it in every good textbook (and of course, in Bourbaki). $\endgroup$ Nov 2, 2010 at 21:15
  • $\begingroup$ Ok, I understand that. But why does the above correspond to a locally free module of finite rank? $\endgroup$
    – roger123
    Nov 2, 2010 at 21:57
  • $\begingroup$ This is also discussed in EGA 2 in the section on the relative proj construction. $\endgroup$ Nov 3, 2010 at 6:15

3 Answers 3


Given any $R$-module $M$, there is a scheme which corresponds to the 'total space' of $M$, given by $$ Tot(M):=Spec( Sym_R(M*))$$ where $M*$ is the dual module $Hom_R(M,R)$ and $Sym_RM*$ is the symmetric algebra of $M^*$ over $R$. If $M$ happens to a free rank $n$ $R$-module, then $Sym_RM\simeq R[X_1,...,X_n]$. The scheme $Tot(M)$ has a natural map to $Spec(R)$, which is dual to the obvious inclusion $$R\rightarrow Sym_RM$$

If you start with a locally free, finite rank $R$-module $P$, and then consider its total space $Tot(P)$, the corresponding scheme is a vector bundle by your definition. This follows from considering open sets on which $P$ is free, and considering the restriction of $Tot(P)$ over those open sets. Since restriction to an open set is the same as tensoring over the localization, and localization commutes with forming symmetric algebras, the locally freeness becomes your second condition. The first condition is also straightforward.

Then, observe that every vector bundle by your definition arises this way. To see this, follow Mike's comment. Associate to a vector bundle $V$ its sheaf of sections over $Spec(R)$, which is an $R$-module in a natural way. It will be free over the open cover $D(a_i)$, with constant rank $n$.

Edit: As pointed out by roger, the total space construction should use the dual of $M$. As a side note, this means that it is the same if you replace $M$ with $M^{**}$, and so it is not interesting to apply this construction to non-reflexive modules.

  • $\begingroup$ Thank you, Greg. Don't I have to consider $Sym_R M^*$, the symmetric algebra of the dual of $M$? $\endgroup$
    – roger123
    Nov 8, 2010 at 22:53
  • $\begingroup$ Ah, yes, that is correct. Sorry, I changed it at the last minute from the free case. $\endgroup$ Nov 8, 2010 at 23:34

It may make more sense to look at the isomorphism class of a vector bundle as its corresponding isomorphism class as a locally free sheaf (Hartshorne: Algebraic Geometry - Exercise 5.18)

Then, it becomes clearer to see the connection by looking at the sheaf associated to a projective module $M$ - which is a locally free sheaf, since the the stalk at any $p \in SpecR$ is isomorphic to $M_p \simeq R^n$, and we have an open cover of $X=SpecR$ by the $D(a_i)$.


Even though there are many excellent answers above and included in the comments, let me try to offer an alternative view.

Let's say that $E\to X$ is a trivial geometric vector bundle. As a scheme, $E\simeq X\times \mathbb A^r$ (and I am simplifying a little bit, but the point here is intuition I suppose). Anyway, now if $X={\rm Spec}R$, then what is $E$? Well, what else but ${\rm Spec} R[x_1,\dots,x_r]$? And what is $E_x$ for some $x\in X$? If $x$ corresponds to the prime ideal $\mathfrak p\subseteq R$ then $E_x$ is just $\kappa(\mathfrak p)[x_1,\dots,x_r]$.

OK, now if $E\to X$ is not trivial, but locally trivial, then the only thing that changes is that these should hold over some open cover of $X$. The topology of $X$ has a base by open sets of the form $D(a)$, so we may as well take our open cover consists of such open sets.

In other words we recovered your two conditions. Now go and check (possibly at the suggested references) that you can do the construction backwards.


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