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3.3 Nonregular 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
noncompatible objects. Therefore, there is no concept analogously to the
regularity in the ideal case. Nevertheless, the nonregular
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
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Up: 3. The Extension of
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