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 ${f_1, \dots, f_p \in C^{\infty}(\mathbb{R}^n)}$. Now, if ${f_1, f_2, \dots, f_p}$ are such that their gradients ${\nabla f_i}$ form a matrix of rank ${p}$, then we can define a manifold near zero which is the common zero set of all the ${f_i}$. We are going to give a relative version of this in the algebraic setting.

Recall that a map of rings ${A \rightarrow B}$ is essentially of finite presentation if ${B}$ is the localization of a finitely presented ${A}$-algebra.

Proposition 5 Let ${(A, \mathfrak{m}) \rightarrow (B, \mathfrak{n})}$ be a local homomorphism of local rings such that ${B}$ is essentially of finite presentation. Suppose ${B = (A[X_1, \dots, X_n])_{\mathfrak{q}}/I}$ for some finitely generated ideal ${I \subset A[X_1, \dots, X_n]_{\mathfrak{q}}}$, where ${\mathfrak{q}}$ is a prime ideal in the polynomial ring.Then ${I/I^2}$ is generated as a ${B}$-module by polynomials ${f_1, \dots, f_k \in A[X_1, \dots, X_n]}$ whose Jacobian matrix has maximal rank in ${B/\mathfrak{n}}$ if and only if ${B}$ is formally smooth over ${A}$. In this case, ${I/I^2}$ is even freely generated by the ${f_i}$.

Proof: Indeed, we know that polynomial rings are formally smooth. In particular ${D = A[X_1, \dots, X_n]_{\mathfrak{q}}}$ is formally smooth over ${A}$, because localization preserves formal smoothness. (This is straightforward.) Note also that ${\Omega_{D/A}}$ is a free ${D}$-module, because this is true for a polynomial ring and Kähler differentials commute with localization.

So the previous result implies that

$\displaystyle I/I^2 \rightarrow \Omega_{D/A} \otimes_D B$

is a split injection precisely when ${B}$ is formally smooth over ${A}$. Suppose that this holds. Now ${I/I^2}$ is then a summand of the free module ${\Omega_{D/A} \otimes_D B}$, so it is projective, hence free as ${B}$ is local. Let ${K = B/\mathfrak{n}}$. It follows that the map

$\displaystyle I/I^2 \otimes K \rightarrow \Omega_{D/A} \otimes_D K = K^n$

is an injection. This map sends ${f \mapsto df}$. Hence the assertion is clear.

Conversely, suppose that ${I/I^2}$ has such generators. Then the map

$\displaystyle I/I^2 \otimes K \rightarrow K^n, \quad f\mapsto df$

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 ${I/I^2 \rightarrow \Omega_{D/A} \otimes_D B}$ is a split injection. In view of the criterion for formal smoothness, we find that ${B}$ 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 ${(A, \mathfrak{m})}$ is a local ring with residue field ${k}$, ${M}$ a finitely generated ${A}$-module, ${N}$ a finitely generated projective ${A}$-module, then a map

$\displaystyle \phi: M \rightarrow N$

is a split injection if and only if

$\displaystyle M \otimes k \rightarrow N \otimes k$

is an injection.

Proof: One direction is clear, so it suffices to show that ${M \rightarrow N}$ is a split injection if the map on fibers is an injection.

Let ${L}$ be a “free approximation” to ${M}$, that is, a free module ${L}$ together with a map ${L \rightarrow M}$ which is an isomorphism modulo ${k}$. By Nakayama’s lemma, ${L \rightarrow M}$ is surjective. Then the map ${L \rightarrow M \rightarrow N}$ is such that the ${L \otimes k \rightarrow N \otimes k}$ is injective, so ${L \rightarrow N}$ is a split injection (by an elementary criterion). It follows that we can find a splitting ${N \rightarrow L}$, which when composed with ${L \rightarrow M}$ is a splitting of ${M \rightarrow N}$.

6. Formal smoothness implies flatness

Now, we can get the characterization promised:

Theorem 9 (EGA IV 17.5.1) Let ${(A, \mathfrak{m}) \rightarrow (B, \mathfrak{n})}$ be a morphism of local noetherian rings such that ${B}$ is the localization of a f.p. ${A}$-algebra at a prime ideal, ${B = (A[X_1, \dots, X_n])_{\mathfrak{q}}/I}$. Then if ${A \rightarrow B}$ is formally smooth, ${B}$ is a flat ${A}$-algebra.

Proof: Let ${C = (A[X_1, \dots, X_n])_{\mathfrak{q}}}$. Then ${C}$ is a local ring, essentially of finite type over ${A}$, and we have morphisms of local rings

$\displaystyle (A, \mathfrak{m}) \rightarrow (C, \mathfrak{q}) \twoheadrightarrow (B, \mathfrak{n}).$

Moroever, ${C}$ is a flat ${A}$-module, and we are going to apply the fiberwise criterion for regularity to ${C}$.

Now we know that ${I/I^2}$ is a ${B}$-module generated by polynomials ${f_1, \dots, f_n \in A[X_1, \dots, X_n]}$ whose Jacobian matrix has maximal rank in ${B/\mathfrak{n}}$. The claim is that the ${f_i}$ are linearly independent in ${\mathfrak{q}/\mathfrak{q}^2}$. This will be the first key step in the proof. In other words, if ${\left\{u_i\right\}}$ is a family of elements of ${C}$, not all non-units, we do not have

$\displaystyle \sum u_i f_i \in \mathfrak{q}^2.$

For if we did, then we could take derivatives and find

$\displaystyle \sum u_i \partial_j f_i \in \mathfrak{q}$

for each ${j}$. This contradicts the gradients of the ${f_i}$ being linearly independent in ${B/\mathfrak{n} = C/\mathfrak{q}}$.

Now we want to show that the ${\left\{f_i\right\}}$ form a regular sequence in ${C}$. To do this, we shall reduce to the case where ${A}$ is a field. Indeed, let us make the base-change ${A \rightarrow k, B \rightarrow \overline{B} = B \otimes k, C \rightarrow \overline{C}=C \otimes k}$ where ${k = A/\mathfrak{m}}$ is the residue field. Then ${\overline{B},\overline{C}}$ are formally smooth local rings over a field ${k}$. We also know that ${\overline{C}}$ is a regular local ring, since it is a localization of a polynomial ring over a field.

Let us denote the maximal ideal of ${\overline{C}}$ by ${\overline{\mathfrak{q}}}$; this is just the image of ${\mathfrak{q}}$.

Now the ${\left\{f_i\right\}}$ have images in ${\overline{C}}$ that are linearly independent in ${\overline{\mathfrak{q}}/\overline{\mathfrak{q}}^2 = \mathfrak{q}/\mathfrak{q}^2}$. It follows that the ${\left\{f_i\right\}}$ form a regular sequence in ${\overline{C}}$, by general facts about regular local rings; indeed, each of the successive quotients ${\overline{C}/(f_1, \dots, f_k)}$ will then be regular. It follows from the fiberwise criterion (${C}$ being flat) that the ${\left\{f_i\right\}}$ form a regular sequence in ${C}$ itself, and that the quotient ${C/(f_i) = B}$ is ${A}$-flat.

In fact, we can get a general criterion now:

Theorem 10 Let ${(A, \mathfrak{m}) \rightarrow (B, \mathfrak{n})}$ be a morphism of local noetherian rings such that ${B}$ is the localization of a f.p. ${A}$-algebra at a prime ideal, ${B = (A[X_1, \dots, X_n])_{\mathfrak{q}}/I}$. Then ${B}$ is formally smooth over ${A}$ if ${B}$ is ${A}$-flat and ${B/\mathfrak{m}B}$ is formally smooth over ${A/\mathfrak{m}}$.

Proof: One direction is immediate from what we have already shown. Now we need to show that if ${B}$ is ${A}$-flat, and ${B/\mathfrak{m}B}$ is formally smooth over ${A/\mathfrak{m}}$, then ${B}$ is itself formally smooth over ${A}$.

As before, write the sequence

$\displaystyle (A, \mathfrak{m}) \rightarrow (C, \mathfrak{q}) \twoheadrightarrow (B,\mathfrak{n})$

where ${C}$ is a localization of a polynomial ring at a prime ideal, and in particular is formally smooth over ${A}$. We know that ${B = C/I}$, where ${I \subset \mathfrak{q}}$.

To check that ${B}$ is formally smooth over ${A}$, we need to show (${C}$ being formally smooth) that the conormal sequence

$\displaystyle I/I^2 \rightarrow \Omega_{C/A} / I \Omega_{C/A} \rightarrow \Omega_{C/B} \rightarrow 0.$

is split exact.

Let ${\overline{A}, \overline{C}, \overline{B}}$ be the base changes of ${A, B, C}$ to ${k = A/\mathfrak{m}}$; let ${\overline{I}}$ be the kernel of ${\overline{C} \twoheadrightarrow \overline{B}}$. Note that ${\overline{I} = I/\mathfrak{m}I}$ by flatness of ${B}$. Then we know that the sequence

$\displaystyle \overline{I}/\overline{I}^2 \rightarrow \Omega_{\overline{C}/k} / \overline{I} \Omega_{\overline{C}/k} \rightarrow \Omega_{\overline{C}/\overline{B}} \rightarrow 0$

is split exact, because ${\overline{C}}$ is a formally smooth ${k}$-algebra.

In particular, there are polynomials ${\overline{P}_1, \dots, \overline{P}_r \in \overline{C}}$ (which is the localization of a polynomial ring over ${k}$) which generate ${\overline{I}/\overline{I}^2}$ and whose gradients are linearly independent in ${\overline{C}/\mathfrak{\overline{q}}}$. These lift to polynomials in ${I}$ which generate (by Nakayama) ${I/I^2}$ and whose gradients are linearly independent in the residue field ${C/\mathfrak{q} = \overline{C}/\overline{\mathfrak{q}}}$. Now by the Jacobian criterion, we are done.