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 ${f_i}$. (more…)

Ultimately, we are headed towards a characterization of formal smoothness for reasonable morphisms (e.g. the types one encounters in classical algebraic geometry): we want to show that they are precisely the flat morphisms whose fibers are smooth varieties. This will be a much more usable criterion in practice (formal smoothness is given by a somewhat abstract lifting property, but checking that a concrete variety is smooth is much easier).  This is the intuition between smoothness: one should think of a flat map is a “continuously varying” family of fibers, and one wishes the fibers to be regular. This corresponds to the fact from differential topology that a submersion has submanifolds as its fibers.

It is actually far from obvious that a formally smooth (and finitely presented) morphism is even flat. Ultimately, the idea of the proof is going to be write the ring as a quotient of a localization of a polynomial ring. The advantage is that this auxiliary ring will be clearly flat, and it will also have fibers that are regular local rings.  In a regular local ring, we have a large supply of regular sequences, and the point is that we will be able to lift the regularity of these sequences from the fiber to the full ring.

Thus we shall use the following piece of local algebra.

Theorem Let ${(A, \mathfrak{m}) \rightarrow (B, \mathfrak{n})}$ be a local homomorphism of local noetherian rings. Let ${M}$ be a finitely generated ${B}$-module, which is flat over ${A}$.

Let ${f \in B}$. Then the following are equivalent:

1. ${M/fM}$ is flat over ${A}$ and ${f: M \rightarrow M}$ is injective.
2. ${f: M \otimes k \rightarrow M \otimes k}$ is injective where ${k = A/\mathfrak{m}}$.

This is a useful criterion of checking when an element is ${M}$-regular by checking on the fiber. That is, what really matters is that we can deduce the first statement from the second. (more…)

Flatness is all about avoiding an annoying property of the tensor product: it does not preserve submodules.

Fix a commutative ring ${R}$. We say that an ${R}$-module ${M}$ is flat if the functor ${Mod(R) \rightarrow Mod(R), N \rightarrow M \otimes_R N}$, is exact; it is always right-exact. It is sufficient to show that for any injection

$\displaystyle 0 \rightarrow N' \rightarrow N$

the tensored sequence is exact, i.e.

$\displaystyle 0 \rightarrow N' \otimes M \rightarrow N \otimes M.$

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 ${R}$ 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, ${\hat{R}}$ is a flat ${R}$-module where ${\hat{R}}$ is an ${I}$-adic completion of ${R}$ for ${R}$ 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 $R$ must act by a nonzerodivisor on any flat module.  Thus for instance, $\mathbb{Z}/2\mathbb{Z}$ is not flat over $\mathbb{Z}$(more…)

The previous post got somewhat detailed and long, so today’s will be somewhat lighter. I’ll use completions to illustrate a well-known categorical trick using finite presentations.

The finite presentation trick

Our goal here is:

Theorem 1  Let ${A}$ be a Noetherian ring, and ${I}$ an ideal. If we take all completions with respect to the ${I}$-adic topology,

$\displaystyle \hat{M} = \hat{A} \otimes_A M$

for any f.g. ${A}$-module ${M}$.   (more…)