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## 3.3 Non-regular Extensions

Assume to be an arbitrary element in . Let be again a free resolution of and denote by a resolution of :

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

PROOF: The situation is the following:

where or can be zero for or if the resolutions have different length. The mapping is given. We construct the mappings by induction on : For there are mappings such that
 (1)

This condition means that

maps into the kernel of

Now, we start with the indices and . The image of is the ideal . Thus, a standard generator of is mapped into by . We chose a representative of as an element of and define the map by . Then as required.

Assume to be constructed. Let be a standard generator of . From we deduce . Thus, we can chose again a representative of in and we define by . The equation (1) follows as in the beginning of the induction.

Proposition 3.7   The diagram constructed form the homomorphism by reversing the sign of every second map :

constitutes a double complex the total complex of which is a resolution of .

PROOF: The first assertion follows from the construction in Lemma 3.6. The second follows again from the spectral sequence converging to the homology of the total complex. For a fixed the homology of is concentrated in the zeroth term and

The map in the middle is the multiplication by and, consequently, a monomorphism. Thus, this complex is a resolution of and the second statement follows.
REMARK:For an -regular generator the upper resolution could be chosen as identical copy of the lower and, hence, the double complex 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 and .

Definition 3.8   The total complex associated to is denoted by and is called the -th subresolution of in .

Now, asssume that is replaced by an -module . We define again the -th extension ideal by . Here, and are non-compatible objects. Therefore, there is no concept analogously to the -regularity in the ideal case. Nevertheless, the non-regular extension of a resolution applies, too.

Theorem 1   Let be a module (or an ideal) in (or resp.). A resolution of M could be computed from a sequence of resolutions of the extension ideals for .

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 of can contain absolute entries for certain components.
EXAMPLE: Let us consider . We start with . is regular over and a resolution of is given by the usual Koszul complex:

is a zero divisor in and we obtain . Again the Koszul complex is a resolution of .

Now sends to and has a representation by in the lower . A representation for is given by . Thus, we define by and .

by definition of the Koszul complex. We conclude which is just the image of under in the lower Koszul complex. Hence, is defined by .

We see that the total complex of the double complex contains absolute entries, i.e., it is not minimal. Indeed, this property reflects the fact that the syzygy of is not a minimal generator of

Next: 4. The Algorithm Up: 3. The Extension of Previous: 3.2 Regular Extensions
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