Gluing vector bundles on irreducible components

Question

Here is a question that I thought it was obvious but more I thought about it less obvious it became. Given vector bundles on irreducible components of a scheme that match up on intersections, is it possible to glue it to a vector bundle on the whole scheme? A simple example is like this. Given a projective $R[s,t]/(s)$ module $P_1$ and a projective $R[s,t]/(t)$ module $P_2$ that are isomorphic at $t=s=0$. How to construct a projective module on $R[s,t]/(st)$ as a gluing?

Answer 1

Let $P_0 = P_1 \otimes_{R[t]} R = P_2 \otimes_{R[s]} R$. Then define $P$ from the exact sequence $$ 0 \to P \to P_1 \oplus P_2 \to P_0 \to 0, $$ where $P_i$ are considered as $R[s,t]/(st)$-modules via the natural morphisms $R[s,t]/(st) \to R[s,t]/(s)$, $R[s,t]/(st) \to R[s,t]/(t)$ and $R[s,t]/(st) \to R[s,t]/(s,t)$. Then $P$ is projective.

Indeed, the claim is local, so we may assume that $P_1$ and $P_2$ (and hence $P_0$ as well) are free. Then the defining sequence of $P$ is isomorphic to the direct sum of several copies of the standard exact sequence $$ 0 \to R[s,t]/(st) \to R[s,t]/(s) \oplus R[s,t]/(t) \to R[s,t]/(s,t) \to 0, $$ hence $P$ is a direct sum of free modules.

The case of a more general gluing is completely analogous.