Flatness is all about avoiding an annoying property of the tensor product: it does not preserve submodules.
Fix a commutative ring . We say that an
-module
is flat if the functor
, is exact; it is always right-exact. It is sufficient to show that for any injection
the tensored sequence is exact, i.e.
It is immediate that the direct sum or tensor product of flat modules is flat, and that a direct summand in a flat module is flat. Also, the module itself is clearly flat. So any free module is flat, and thus so is any projective module. In the local case, we’ll see that the converse is true at least if we stick to finitely generated modules over a noetherian ring (more generally, for finitely presented modules over any ring).
Since completion preserves exactness, is a flat
-module where
is an
-adic completion of
for
noetherian. Localization does to, so any localization is flat. So many common examples are flat. But because of the above injectivity characterization, a nonzerodivisor in
must act by a nonzerodivisor on any flat module. Thus for instance,
is not flat over
.
Interpretation via Tor
Recall that the derived functors of the tensor product functor are denoted by . So
. These functors are symmetric in both variables, and given an exact sequence
we have a long exact sequence
For this, cf. books on homological algebra.
One of the basic applications of this is that for a flat module , the tor-functors vanish for
(whatever be
). Indeed, recall that
is computed by taking a projective resolution of
,
tensoring with , and taking the homology. But tensoring with
is exact if we have flatness, so the higher
modules vanish.
Conversely, suppose for all
and
. Then given an exact sequence,
we get exactness of
so the -vanishing (as we saw, we only needed it for
) gives flatness. By taking direct (filtered) limits, which preserve exactness, we can reduce to the case of checking that
for
finitely generated. By using a filtration on
whose quotients are of the form
for
an ideal, we find that:
Proposition 1
is flat iff
for all finitely generated ideals
.
Note that there is an exact sequence and so
is exact, and it suffices to show that
is injective for all .
Theorem 2 If
is noetherian local and
finitely generated and flat, then
is free.
Indeed, let be the maximal ideal and
the residue field. Now
is a finitely generated
-vector space; choose a basis. This induces an isomorphism
which we can lift to a map
which is surjective by Nakayama’s lemma, and which becomes an isomorphism upon being tensored with . Let
be the kernel:
then we have an exact sequence
As a result , which implies
by Nakayama, so
is an isomorphism, q.e.d.
The local criterion of flatness
In Proposition 1 above, the need to check all ideals is potentially burdensome, so the following is nice.
Let be a noetherian local ring with maximal ideal
and residue field
,
be a local finitely generated
-algebra with
for
the maximal ideal of
, and
a finitely generated
-module.
Theorem 3
is flat over
iff
![]()
Necessity is immediate. What we have to prove is sufficiency.
First, I make the following claim. If is an
-module of finite length, then
This is because has by devissage a filtration
whose quotients are of the form
for
prime and (by finite length hypothesis)
. Then the result is true
trivially. We climb up the filtration piece by piece inductively; if
, then the exact sequence
yields
from the long exact sequence of and the hypothesis on
. The claim is proved.
The idea behind this proof is to show that is injective for any
. We will use some diagram chasing and the Krull intersection theorem on the kernel
of this map, to interpolate between it and various quotients by powers of
. First we write some exact sequences.
We have an exact sequence
which we tensor with :
The sequence
is also exact, and tensoring with yields an exact sequence:
because (finite length hypothesis.
Let us draw the following commutative diagram:
where the column and the row are exact. As a result, if an element in goes to zero in
(a fortiori in
) it must come from
for all
. Thus it belongs to
for all
, and the Krull intersection theorem (applied to
, since
) implies it is zero.
The infinitesimal criterion
Theorem 4 Let be a module over a noetherian local ring
. Then
is flat iff
is flat over
for all
.
One direction is easy, because if is a flat
-module and
an
-algebra, then
is flat over
. For the other direction, take the same commutative diagram as before:
The horizontal sequence was always exact. The vertical sequence can be argued to be exact by tensoring the exact sequence
of -modules with
, and using flatness. Thus we get flatness of
as before.
Incidentally, if we combine the local and infinitesimal criteria fo flatness, we get a little more. Moreover, the same reasoning shows that we only need the hypothesis of the theorem for sufficiencly large , or even arbitrarily large.
October 29, 2010 at 11:10 pm
[…] Climbing Mount Bourbaki on local and infinitesimal criteria for flatness, might help you with the exercises below. […]