Auflösung eines Ideals/Erste Syzygien/Motivierendes Beispiel für lokal frei/en/Textabschnitt

Let ${}R$ be a commutative ring and let ${}I$ be an ideal generated by finitely many elements ${}I=(f_{1},\ldots ,f_{n})$. The free resolution of the residue class ring ${}R/I$ is the exact complex

$\ldots \longrightarrow R^{n_{2}}\longrightarrow R^{n}{\stackrel {f_{1},\ldots ,f_{n}}{\longrightarrow }}R\longrightarrow R/I\longrightarrow 0.$

This resolution goes (unless ${}I$ has finite projective dimension) on forever, but we can break it up to obtain the exact complex

$0\longrightarrow M=\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)}\longrightarrow R^{n}{\stackrel {f_{1},\ldots ,f_{n}}{\longrightarrow }}R\longrightarrow R/I\longrightarrow 0,$

where the module ${}M$ is just defined to be the kernel of the ${}R$-module-homomorphism

$R^{n}\longrightarrow R,\,(s_{1},\ldots ,s_{n})\longmapsto \sum _{i=1}^{n}s_{i}f_{i}.$

This kernel consists exactly of the syzygies for these elements, hence it is called (the first) syzygy module. This module can be already quite complicated, however, we can make the following observation. Let us fix one ${}i$, say ${}i=1$, and look at the induced sequence over the localization ${}R_{f_{1}}$. As localization is an exact functor, we still get an exact sequence, and since ${}f_{1}\in I$, the ideal ${}I_{f_{1}}$ contains now a unit and therefore we have ${}(R/I)_{f_{1}}=0$, so we can rewrite the induced sequence as

$0\longrightarrow M_{f_{1}}\longrightarrow (R_{f_{1}})^{n}{\stackrel {f_{1},\ldots ,f_{n}}{\longrightarrow }}R_{f_{1}}\longrightarrow 0.$

We claim that we have an ${}R_{f_{1}}$-module isomorphism

$(R_{f_{1}})^{n-1}\longrightarrow M_{f_{1}}\cong (\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)})_{f_{1}}$

by sending the ${}j$-th standard vector ${}e_{j}$ ($j=2,\ldots ,n$) to

$v_{j}=(-{\frac {f_{j}}{f_{1}}},\ldots ,0,1,0,\ldots ,0)$

(the ${}1$ stands at the ${}j$th position). This is obviously well-defined, since ${}f_{1}$ is a unit in ${}R_{f_{1}}$, and evidently the given tuple is a syzygy. If ${}s=(s_{1},\ldots ,s_{n})$ is a syzygy, then ${}\sum _{j=2}^{n}s_{j}e_{j}$ is a preimage, since it is mapped under this homomorphism to

${}\sum _{j=2}^{n}s_{j}v_{j}=(-\sum _{j=2}^{n}s_{j}{\frac {f_{j}}{f_{1}}},s_{2},\ldots ,s_{n})=(s_{1},s_{2},\ldots ,s_{n})\,.$

Hence we have a surjection. The injectivity follows immediately by looking at the components ${}2$ to ${}n$ in the syzygy.

This means that the syzygy module when restricted to the open subset ${}D(f_{1})$ (viewed as an ${}R_{f_{1}}$-module) is free of rank ${}n-1$, and the same holds for all ${}D(f_{j})$. Hence the syzygy module restricted to the open subset

$U=\bigcup _{j=1}^{n}D(f_{j})$

has the property that there exists a covering by open subsets such that the restrictions to these open subsets are free modules. In general, the syzygy module is not free as an ${}R$-module nor as an ${}{\mathcal {O}}_{U}$-module on ${}U$. The above given explicit isomorphism on ${}D(f_{1})$ (such an isomorphism is called a local trivialization of ${}M$ on $D(f_{1})$) uses that ${}f_{1}$ is a unit, hence this can not be extended to give an isomorphism on ${}U$.

On the intersection ${}D(f_{1})\cap D(f_{2})=D(f_{1}f_{2})$ ${}f_{1}$ as well as ${}f_{2}$ are units, hence the above isomorphisms (let's call them ${}\psi _{1}$ on ${}D(f_{1})$ and ${}\psi _{2}$ on ${}D(f_{2})$) induce two different isomorphisms on ${}D(f_{1}f_{2})$ between ${}(R_{f_{1}f_{2}})^{n-1}$ and ${}M_{f_{1}f_{2}}$ We can connect them to get an isomorphism

$\psi _{2}^{-1}\circ \psi _{1}\colon (R_{f_{1}f_{2}})^{n-1}\longrightarrow (R_{f_{1}f_{2}})^{n-1}$

which is given by the (over $R_{f_{1}f_{2}}$) invertible ${}(n-1)\times (n-1)$-matrix

${\begin{pmatrix}-{\frac {f_{2}}{f_{1}}}&-{\frac {f_{3}}{f_{1}}}&-{\frac {f_{4}}{f_{1}}}&\ldots &-{\frac {f_{n}}{f_{1}}}\\0&-1&0&\ldots &0\\0&0&-1&\ddots &\vdots \\\vdots &\vdots &\ddots &\ddots &0\\0&\ldots &\ldots &0&-1\end{pmatrix}}.$