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3.3 Non-regular Extensions

Assume $f_i$ to be an arbitrary element in $R\setminus I_{i-1}$. Let $K_{i-1}$ be again a free resolution of $R/I_{i-1}$ and denote by $L_{i}$ a resolution of $R/J_i$:

\begin{displaymath}L_i: 0 \rightarrow R^{m_l} \rightarrow \ldots \rightarrow R^{m_2} \rightarrow R^{m_1} \rightarrow R^1 \rightarrow 0.\end{displaymath}

Lemma 3.6   The multiplication by $f_i$ on the $0$-level induces a homomorphism of complexes:

\begin{displaymath}F_{\_}:L_i \rightarrow K_{i-1}.\end{displaymath}

PROOF: The situation is the following:

& & & d'_{l-1} & & & & d'_1& &...
& & & d_{l-1} & & & & d_1 & & d_0 & & & \\

where $R^{m_s}$ or $R^{n_t}$ can be zero for $s>s_0$ or $t>t_0$ if the resolutions have different length. The mapping $F_0:=f_i$ is given. We construct the mappings $F_j$ by induction on $i$: For $i=1,\ldots,l-1$ there are mappings $F_j$ such that
\end{displaymath} (1)

This condition means that

\begin{displaymath}F_j\oplus d'_j: R^{m_j}\longrightarrow R_{n_j}\oplus R_{m_{j-1}}\end{displaymath}

maps into the kernel of

\begin{displaymath}d_{j-1}-F_{j-1}: R_{n_j}\oplus R_{m_{j-1}}\longrightarrow R_{n_{j-1}}.\end{displaymath}

Now, we start with the indices $1$ and $0$. The image of $d'_0$ is the ideal $J_i=(I_{i-1}:f_i)=(g_1,\ldots,g_l)\subset R^1$. Thus, a standard generator $e_s$ of $R^{m_1}$ is mapped into $I_{i-1}$ by $f_id'_0$. We chose a representative $E_s\in R^{i-1}$ of $f_id'_0(e_s)=f_ig_s$ as an element of $I_{i-1}$ and define the map $F_1$ by $F_1(e_s)=E_s$. Then $(d_0F_1-F_0d'_1)e_s=f_ig_s-f_ig_s=0$ as required.

Assume $F_j$ to be constructed. Let $e_s$ be a standard generator of $R^{m_{j+1}}$. From $F_{j-1}d'_jd'_{j+1}(e_s)=0$ we deduce $F_jd'_{j+1}(e_s)\in ker(d_{j-1})$. Thus, we can chose again a representative $E_s$ of $F_id'_{j+1}(e_s)$ in $R_{n_{j+1}}$ and we define $F_{j+1}$ by $F_{j+1}(e_s)=E_s$. The equation (1) follows as in the beginning of the induction.$\Box$

Proposition 3.7   The diagram $\hat{K_i}$ constructed form the homomorphism $F_{\_}$ by reversing the sign of every second map $F_j$:

& & & d'_{l-1} & & & & d'_1&...
& & & d_{l-1} & & & & d_1 & & d_0 & & & \\
\end{array} \end{displaymath}

constitutes a double complex the total complex of which is a resolution of $R/I_i$.

PROOF: The first assertion follows from the construction in Lemma 3.6. The second follows again from the spectral sequence $E^{p,q}_1=H^q(\hat{K_i}^{*,p})$ converging to the homology of the total complex. For a fixed $p$ the homology of $\hat{K_i}^{*,p}$ is concentrated in the zeroth term and

\begin{displaymath}H^0(\hat{K_i}^{*,\_}): 0\rightarrow R/J_i\rightarrow R/I_{i-1}\rightarrow 0.\end{displaymath}

The map in the middle is the multiplication by $f_i$ and, consequently, a monomorphism. Thus, this complex is a resolution of $R/I_i$ and the second statement follows.$\Box$
REMARK:For an $I_{i-1}$-regular generator $f_i$ the upper resolution $L_i$ could be chosen as identical copy of the lower $K_{i-1}$ and, hence, the double complex $\hat{K_i}$ is the natural generalization of the Koszul complex.

Note, that the whole construction works over arbitrary rings: The only assumption is the existence of the free resolutions $L_i$ and $K_{i-1}$.

Definition 3.8   The total complex associated to $\hat{K_i}$ is denoted by $K_i$ and is called the $i$-th subresolution of $I$ in $R$.

Now, asssume that $I$ is replaced by an $R$-module $M=(m_1,\ldots,m_n)\subset R^p$. We define again the $i$-th extension ideal by $J_i:=(M_{i-1}:m_i)\subset R$. Here, $J_i$ and $M_{i-1}$ are non-compatible objects. Therefore, there is no concept analogously to the $I_{i-1}$-regularity in the ideal case. Nevertheless, the non-regular extension of a resolution applies, too.

Theorem 1   Let $M=(m_1,\ldots,m_n)$ be a module (or an ideal) in $R^p$ (or $R$ resp.). A resolution of M could be computed from a sequence of resolutions of the extension ideals $J_i=(M_{i-1}:m_i)$ for $i=1,\ldots,n$.

REMARK:The resolution constructed via Proposition 3.7 is not minimal in general: Going back to the proof of Lemma 3.6 we see that a chosen representation $E_s$ of $F_id'_{j+1}(e_s)$ can contain absolute entries for certain components.
EXAMPLE: Let us consider $I=(x^2,y^2,xy)\subset k[x,y]$. We start with $I_2=(x^2,y^2)$. $y^2$ is regular over $(x^2)$ and a resolution of $R/I_2$ is given by the usual Koszul complex:

\begin{displaymath}K_2:0\rightarrow R^1\rightarrow R^2\rightarrow R^1\rightarrow 0.\end{displaymath}

$xy$ is a zero divisor in $R/(x^2,y^2)$ and we obtain $J_3=(x,y)$. Again the Koszul complex is a resolution of $R/J_3$.

\begin{displaymath}L_3:0\rightarrow R^1\rightarrow R^2\rightarrow R^1\rightarrow 0.\end{displaymath}

Now $d'_0$ sends $e_1$ to $x\in R$ and $xyd'_0(e_1)=x^2y$ has a representation by $ye_1$ in the lower $R^2$. A representation for $xyd'_0(e_2)=xy^2$ is given by $xe_2$. Thus, we define $F_1$ by $F_1(e_1)=ye_2$ and $F_1(e_2)=xe_2$.

$d'_1(e_1)=ye_1-xe_2$ by definition of the Koszul complex. We conclude $F_1d'_1(e_1)=y^2e_1-x^2e_2$ which is just the image of $e_1$ under $d_1$ in the lower Koszul complex. Hence, $F_2$ is defined by $F_2(e_1)=e_1$.

We see that the total complex $K_3$ of the double complex $\hat{K}_3$ contains absolute entries, i.e., it is not minimal. Indeed, this property reflects the fact that the syzygy $y^2e_1-x^2e_2$ of $(x^2,y^2)$ is not a minimal generator of


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