In this post, we shall accomplish the goal stated earlier: we shall show that a formally smooth morphism of noetherian rings (which is essentially of finite type) is flat. We shall even get an equivalence: flatness together with smoothness on the fibers will be both necessary and sufficient to ensure that a given such morphism is formally smooth.
In order to do this, we shall use a refinement of the criterion in the first post for when a quotient of a formally smooth algebra is formally smooth. We shall need a bit of local algebra to do this, but the reward will be a very convenient Jacobian criterion, which will then enable us to prove (using the results from last time on lifting flatness from the fibers) the final characterization of smoothness.
3. The Jacobian criterion
So now we want a characterization of when a morphism is smooth. Let us motivate this with an analogy from standard differential topology. Consider real-valued functions . Now, if
are such that their gradients
form a matrix of rank
, then we can define a manifold near zero which is the common zero set of all the
. We are going to give a relative version of this in the algebraic setting.
Recall that a map of rings is essentially of finite presentation if
is the localization of a finitely presented
-algebra.
Proposition 5 Let
be a local homomorphism of local rings such that
is essentially of finite presentation. Suppose
for some finitely generated ideal
, where
is a prime ideal in the polynomial ring.Then
is generated as a
-module by polynomials
whose Jacobian matrix has maximal rank in
if and only if
is formally smooth over
. In this case,
is even freely generated by the
.
Proof: Indeed, we know that polynomial rings are formally smooth. In particular is formally smooth over
, because localization preserves formal smoothness. (This is straightforward.) Note also that
is a free
-module, because this is true for a polynomial ring and Kähler differentials commute with localization.
So the previous result implies that
is a split injection precisely when is formally smooth over
. Suppose that this holds. Now
is then a summand of the free module
, so it is projective, hence free as
is local. Let
. It follows that the map
is an injection. This map sends . Hence the assertion is clear.
Conversely, suppose that has such generators. Then the map
is a split injection. However, if a map of finitely generated modules over a local ring, with the target free, is such that tensoring with the residue field makes it an injection, then it is a split injection. (We shall prove this below.) Thus is a split injection. In view of the criterion for formal smoothness, we find that
is formally smooth.
4. A minor lemma on modules over local rings
When one has a morphism of finitely generated free modules over a local ring, a well-known criterion (and nice exercise) states that the map is a split injection if and only if it is an injection modulo the residue field.
Lemma 6 If
is a local ring with residue field
,
a finitely generated
-module,
a finitely generated projective
-module, then a map
is a split injection if and only if
is an injection.
Proof: One direction is clear, so it suffices to show that is a split injection if the map on fibers is an injection.
Let be a “free approximation” to
, that is, a free module
together with a map
which is an isomorphism modulo
. By Nakayama’s lemma,
is surjective. Then the map
is such that the
is injective, so
is a split injection (by an elementary criterion). It follows that we can find a splitting
, which when composed with
is a splitting of
.
6. Formal smoothness implies flatness
Now, we can get the characterization promised:
Theorem 9 (EGA IV 17.5.1) Let
be a morphism of local noetherian rings such that
is the localization of a f.p.
-algebra at a prime ideal,
. Then if
is formally smooth,
is a flat
-algebra.
Proof: Let . Then
is a local ring, essentially of finite type over
, and we have morphisms of local rings
Moroever, is a flat
-module, and we are going to apply the fiberwise criterion for regularity to
.
Now we know that is a
-module generated by polynomials
whose Jacobian matrix has maximal rank in
. The claim is that the
are linearly independent in
. This will be the first key step in the proof. In other words, if
is a family of elements of
, not all non-units, we do not have
For if we did, then we could take derivatives and find
for each . This contradicts the gradients of the
being linearly independent in
.
Now we want to show that the form a regular sequence in
. To do this, we shall reduce to the case where
is a field. Indeed, let us make the base-change
where
is the residue field. Then
are formally smooth local rings over a field
. We also know that
is a regular local ring, since it is a localization of a polynomial ring over a field.
Let us denote the maximal ideal of by
; this is just the image of
.
Now the have images in
that are linearly independent in
. It follows that the
form a regular sequence in
, by general facts about regular local rings; indeed, each of the successive quotients
will then be regular. It follows from the fiberwise criterion (
being flat) that the
form a regular sequence in
itself, and that the quotient
is
-flat.
In fact, we can get a general criterion now:
Theorem 10 Let
be a morphism of local noetherian rings such that
is the localization of a f.p.
-algebra at a prime ideal,
. Then
is formally smooth over
if
is
-flat and
is formally smooth over
.
Proof: One direction is immediate from what we have already shown. Now we need to show that if is
-flat, and
is formally smooth over
, then
is itself formally smooth over
.
As before, write the sequence
where is a localization of a polynomial ring at a prime ideal, and in particular is formally smooth over
. We know that
, where
.
To check that is formally smooth over
, we need to show (
being formally smooth) that the conormal sequence
is split exact.
Let be the base changes of
to
; let
be the kernel of
. Note that
by flatness of
. Then we know that the sequence
is split exact, because is a formally smooth
-algebra.
In particular, there are polynomials (which is the localization of a polynomial ring over
) which generate
and whose gradients are linearly independent in
. These lift to polynomials in
which generate (by Nakayama)
and whose gradients are linearly independent in the residue field
. Now by the Jacobian criterion, we are done.
February 15, 2012 at 3:27 pm
Should it not be ${\overline{\mathfrak{q}}/\overline{\mathfrak{q}}^2 = \mathfrak{q}/\mathfrak{q}^2+\mathfrak{m}}$? If yes, why are the $f_i$ still independent?
February 18, 2012 at 1:12 am
I think this is ok, since
is contained in $\mathfrak{q}$.