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 5Let 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 6If is a local ring with residue field , a finitely generated -module, a finitely generated projective -module, then a mapis 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 10Let 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 $\mathfrak{q}$.