We have seen that the fibers of the spectrum of a forcing algebra are
(empty or)
affine spaces. However, this is not only fiberwise true, but more general: If we localize the forcing algebra at
we get
-
![{\displaystyle {\left(R[T_{1},\ldots ,T_{n}]/{\left(f_{1}T_{1}+\cdots +f_{n}T_{n}-f\right)}\right)}_{f_{i}}\cong R_{f_{i}}[T_{1},\ldots ,T_{i-1},T_{i+1},\ldots ,T_{n}]\,,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e543d798afe63140348d5f0332f0c929d78060a4)
since we can write
-
![{\displaystyle {}T_{i}=-\sum _{j\neq i}{\frac {f_{j}}{f_{i}}}T_{j}+{\frac {f}{f_{i}}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2e26ce7052a34c71b34bb1c42426fe7a8e4afe3)
So over every
the spectrum of the forcing algebra is an
-dimensional affine space over the base. So locally, restricted to
, we have isomorphisms
-
![{\displaystyle {}T{|}_{D(f_{i})}\cong D(f_{i})\times {{\mathbb {A} }_{}^{n-1}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/004104e44a0c2221a09a42b3e04e05ed6da7a61e)
On the intersections
we get two identifications with affine space, and the transition morphisms are linear if
,
but only affine-linear in general
(because of the translation with
).
So the forcing algebra has locally the form
and its spectrum
has locally the form
. This description holds on the union
.
Moreover, in the homogeneous case
(
)
the transition mappings are linear. Hence
, where
is the spectrum of a homogeneous forcing algebra, is a geometric vector bundle according to the following definition.