# Forcing algebras/Induced torsors/Textabschnitt

As $T_{U}$ is a $V_{U}$-torsor, and as every $V$-torsor is represented by a unique cohomology class, there should be a natural cohomology class coming from the forcing data. To see this, let $R$ be a noetherian ring and $I={\left(f_{1},\ldots ,f_{n}\right)}$ be an ideal. Then on $U=D(I)$ we have the short exact sequence
    $0\longrightarrow \operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)}\longrightarrow {\mathcal {O}}_{U}^{n}\longrightarrow {\mathcal {O}}_{U}\longrightarrow 0.$
An element $f\in R$ defines an element $f\in \Gamma (U,{\mathcal {O}}_{U})$ and hence a cohomology class $\delta (f)\in H^{1}(U,\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)})$. Hence $f$ defines in fact a $\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)}$-torsor over $U$. We will see that this torsor is induced by the forcing algebra given by $f_{1},\ldots ,f_{n}$ and $f$.

## Theorem

Let $R$ denote a noetherian ring, let $I={\left(f_{1},\ldots ,f_{n}\right)}$ denote an ideal and let $f\in R$ be another element. Let $c\in H^{1}(D(I),\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)})$ be the corresponding cohomology class and let
    $B=R[T_{1},\ldots ,T_{n}]/{\left(f_{1}T_{1}+\cdots +f_{n}T_{n}-f\right)}\,$
denote the forcing algebra for these data.
Then the scheme $\operatorname {Spec} {\left(B\right)}{|}_{D(I)}$ together with the natural action of the syzygy bundle on it is isomorphic to the torsor given by $c$.

### Proof

We compute the cohomology class $\delta (f)\in H^{1}(U,\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)})$ and the cohomology class given by the forcing algebra. For the first computation we look at the short exact sequence
    $0\longrightarrow \operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)}\longrightarrow {\mathcal {O}}_{U}^{n}{\stackrel {f_{1},\ldots ,f_{n}}{\longrightarrow }}{\mathcal {O}}_{U}\longrightarrow 0.$
On $D(f_{i})$, the element $f$ is the image of $\left(0,\ldots ,0,\,{\frac {f}{f_{i}}},\,0,\ldots ,0)\right)$ (the non-zero entry is at the $i$th place). The cohomology class is therefore represented by the family of differences
    $\left(0,\ldots ,0,\,{\frac {f}{f_{i}}},\,0,\ldots ,0,\,-{\frac {f}{f_{j}}},\,0,\ldots ,0\right)\in \Gamma (D(f_{i})\cap D(f_{j}),\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)}).$
On the other hand, there are isomorphisms
    $V{|}_{D(f_{i})}\longrightarrow T{|}_{D(f_{i})},\,\left(s_{1},\ldots ,s_{n}\right)\longmapsto \left(s_{1},\ldots ,s_{i-1},\,s_{i}+{\frac {f}{f_{i}}},\,s_{i+1},\ldots ,s_{n}\right).$
The composition of two such isomorphisms on $D(f_{i}f_{j})$ is the identity plus the same section as before.
$\Box$

## Example

Let $(R,{\mathfrak {m}})$ denote a two-dimensional normal local noetherian domain and let $f$ and $g$ be two parameters in $R$. On $U=D({\mathfrak {m}})$ we have the short exact sequence
    $0\longrightarrow {\mathcal {O}}_{U}\cong \operatorname {Syz} {\left(f,g\right)}\longrightarrow {\mathcal {O}}_{U}^{2}{\stackrel {f,g}{\longrightarrow }}{\mathcal {O}}_{U}\longrightarrow 0$
and its corresponding long exact sequence of cohomology,
    $0\longrightarrow R\longrightarrow R^{2}{\stackrel {f,g}{\longrightarrow }}R{\stackrel {\delta }{\longrightarrow }}H^{1}(U,{\mathcal {O}}_{X})\longrightarrow \ldots .$
The connecting homomorphism $\delta$ sends an element $h\in R$ to ${\frac {h}{fg}}$. The torsor given by such a cohomology class $c={\frac {h}{fg}}\in H^{1}{\left(U,{\mathcal {O}}_{X}\right)}$ can be realized by the forcing algebra
    $R[T_{1},T_{2}]/{\left(fT_{1}+gT_{2}-h\right)}.$
Note that different forcing algebras may give the same torsor, because the torsor depends only on the spectrum of the forcing algebra restricted to the punctured spectrum of $R$. For example, the cohomology class ${\frac {1}{fg}}={\frac {fg}{f^{2}g^{2}}}$ defines one torsor, but the two fractions yield the two forcing algebras $R[T_{1},T_{2}]/{\left(fT_{1}+gT_{2}-1\right)}$ and $R[T_{1},T_{2}]/{\left(f^{2}T_{1}+g^{2}T_{2}-fg\right)}$, which are quite different. The fiber over the maximal ideal of the first one is empty, whereas the fiber over the maximal ideal of the second one is a plane.
If $R$ is regular, say $R=K[X,Y]$ (or the localization of this at $(X,Y)$ or the corresponding power series ring) then the first cohomology classes are $K$-linear combinations of ${\frac {1}{x^{i}y^{j}}}$, $i,j\geq 1$.
    $K[X,Y,T_{1},T_{2}]/{\left(X^{i}T_{1}+Y^{j}T_{2}-1\right)}.$