Punctured spectrums of local rings

Question

Let $A$ be a local ring with the unique maximal ideal $\mathfrak{m}$. The punctured spectrum of $A$ is the open subset $\text{Spec}(A)\setminus \{\mathfrak{m}\}$. I have seen many papers (for instance Horrocks' papers) studying vector bundles over algebraic varieties (in particular, projective spaces) by putting them over a punctured spectrum of a local ring.

However, I am wondering why punctured spectrum is better than varieties in this satiation. I feel that geometric pictures of varieties are clearer than that of a punctured spectrum. More essential, are the categories of coherent sheaves over a variety and its punctured spectrum equivalent? How much information about coherent sheave (in particular, vector bundles) can be recovered from punctured spectrums? For second question, I am thinking examples that reflect some relations between those two gadgets.

Answer 1

Here is a general principle. If a problem is formulated over a general (say noetherian) base, by localization arguments one can often reduce to the case of a local base. Arguing by induction on the dimension of the base, we can exploit the fact that the punctured spectrum, while often not affine, is of smaller dimension! Then we can try to "extend" the solution over the punctured spectrum to a solution over the local spectrum (by taking into account local cohomology or other tools). I think this brilliant idea is due to Grothendieck (truly exploiting the concept of a scheme).

For example, observe that the punctured spectrum of a 2-dimensional regular local ring is a (typically non-affine!) Dedekind scheme, over which flatness is a hands-on concept. Much deeper, the proof of Zariski's Main Theorem in EGA IV$_3$ goes via dimension induction on the base using the above trick with a punctured spectrum having smaller dimension.

For the study of coherent sheaves, the categories on the spectrum and punctured spectrum are not at all equivalent, though the situation for vector bundles is more accessible; the problem is governed by local cohomology and depth considerations (so the situation is more tractable over a regular base). This is studied in depth (no pun intended) in the middle of SGA2.

Answer 2

This is not directly responsive to your question, but it does show that if you want to study vector bundles, you can't always replace the punctured spectrum with the full spectrum:

Let $A$ be a regular local ring of dimension 2, and $U$ its punctured spectrum. Then every vector bundle over $Spec(A)\times {\bf A}^1=Spec(A[X])$ is trivial, but there are certainly vector bundles over $U\times{\bf A}^1$ that are non-trivial.

To construct one, let the maximial ideal of $A$ be generated by elements $a,b$, let ${\cal O}$ be the structure sheaf on $U\times{\bf A}^1$ and let ${\cal F}$ be the kernel of the map ${\cal O}^3\rightarrow{\cal O}$ given by $(a,b,X)$.

It's not too hard to check that ${\cal F}$ can't be trivial. If it were, then (extending to the full spectrum and observing that global sections don't change) we'd have a sequence of $A[x]$ modules

$$0\rightarrow A[X]^2\rightarrow A[X]^3\rightarrow A[X] \rightarrow k \rightarrow 0$$

which gives an $A[X]$-projective resolution of $k$ of length 2, contradicting the fact that, over $A[X]$, $k$ has projective dimension 3.

I am almost sure I learned this from Chuck Weibel,and almost equally sure that the argument is due to Dick Swan.

Answer 3

I think we can adapt the argument from my other answer to give an answer that is directly responsive to your question:

Let $A$ be the local ring of the origin in 3-space over your favorite field $k$. Let $X,Y,Z$ be the coordinates on 3-space. Map $A^3\rightarrow A$ by $(X,Y,Z)$. Let $M$ be the kernel. Restricting $M$ to the punctured spectrum, we get a vector bundle. But this vector bundle can't be free, or we'd have a projective resolution of $k$ by length 2 over $A$, whereas the projective dimension of $k$ is 3.

Therefore the punctured spectrum supports a non-trivial vector bundle, wherease the full spectrum doesn't.