14
$\begingroup$

I asked this question on SE a long time ago, but never received an answer:

The Govorov-Lazard Theorem states that a (left) module over an unital ring is flat iff it is a direct limit of finitely generated free (left) modules.

The theorem is contained in many textbooks like Eisenbud (Commutative Algebra) or Rotman (Introduction to Homological Algebra). However, no applications are given there.

Question: Are there interesting applications of the Govorov-Lazard Theorem ?

N.b.: The only application I've seen so far, was in a question on SE, where someone remarked that if $A,B$ are commutative $R$-algebras and $M$ is a flat $A$-module and $N$ a flat $B$-module, then it follows from Govorov-Lazard that $M\otimes_R N$ is a flat $A\otimes_R B$-module. But, of course, this follows more easily from standard properties of the tensor product.

$\endgroup$
7
  • $\begingroup$ YMMV on whether this is interesting, but it helps generalize some relative version of Poincaré-Birkhoff-Witt to Lie algebras which are flat as modules over rings: mathoverflow.net/questions/65675/… $\endgroup$ Apr 17, 2013 at 1:45
  • 1
    $\begingroup$ (Probably it's part of the proof that Poincaré-Birkhoff-Witt itself, not just some relative version, holds for Lie algebras which are flat as modules over the base ring. I have not studied this proof but I think it should be in P. J. Higgins, Baer Invariants and the Birkhoff-Witt Theorem, Journal of Algebra 11, pp. 469-482 (1969).) $\endgroup$ Apr 17, 2013 at 1:48
  • 3
    $\begingroup$ Closely related to the G-L theorem is the equational criterion for flatness, which I really like. It provides a nice counterpoint to the more common category-theoretic or homological points of view on flatness. $\endgroup$
    – JBorger
    Apr 17, 2013 at 9:15
  • $\begingroup$ @James: What is the equational criterion ? $\endgroup$
    – Ralph
    Apr 17, 2013 at 18:35
  • 1
    $\begingroup$ @darij: As far as I can see, L-G is neither used in Higgins paper nor does Higgins treat the flat case. If I'm not missing something this is quite surprising because the flat case follows immediately from Higgins results and L-G. Higgins paper was submitted 1968 while the paper from Govorov was published 1965 and that of Lazard 1969. So it unclear, if Higgins should had known L-G when he was writing his paper. In any case L-G is an interesting application on the flat Birkhoff-Witt. Wouldn't you like to post this as an answer ? $\endgroup$
    – Ralph
    Apr 17, 2013 at 19:11

4 Answers 4

7
$\begingroup$

One application which I find particularly beautiful is the following:

Theorem: If $R$ is countable then any flat left $R$-module $M$ has projective dimension at most 1

This appeared first in

Jensen, 1966, On homological dimensions of rings with countably generated ideals

It has been used more recently by my advisor, Mark Hovey, in his paper "On Freyd's Generating Hypothesis" because the ring of stable homotopy groups of spheres is countable, and many other objects of interest have homotopy which is flat over it.

It would appear that this observation of Jensen is also related to the projectivity criterion of Raynaud and Gruson. First, observe that

Theorem: For any $R$, if $M$ is countably presented and flat then $M$ has projective dimension at most 1

Using this, one gets the projectivity criterion for countably presented flat left $R$-modules, which states that $M$ is projective iff whenever $M$ is the direct limit of finitely generated free left $R$-modules $M_n$ then the inverse system (Hom$_R(M_n,R))_n$ satisfies the Mittag-Leffler condition.

A corollary of the proof is that if $R$ is countable then a left module $M$ is countably presented iff it's countably generated

$\endgroup$
3
  • $\begingroup$ Sure, no problem. Thanks for an interesting question $\endgroup$ May 3, 2013 at 2:08
  • $\begingroup$ The proof in Jensen does not appear to use Govorov-Lazard, and in fact it is quite involved. Is there a simple proof of your first quoted result using G-L? The projectivity criterion you cite seems to be stacks.math.columbia.edu/tag/059X, which indeed follows from G-L, but I don't see an obvious modification of the proof that would work for your quoted result $\endgroup$ Oct 27, 2020 at 17:00
  • $\begingroup$ I'm sorry, but this answer is from seven and a half years ago, and is not related to my current research at all. So, I don't have the time or expertise to answer the question you just asked. Good luck! $\endgroup$ Oct 27, 2020 at 19:01
10
$\begingroup$

Interesting is always in the eye of the beholder but this theorem plays a crucial role in the proof of the proper base change theorem in SGA4 Exposé XVII (by P.Deligne). One wishes to prove an isomorphism between complexes and the proof is pure EGA/SGA reduction style: first reduce to complexes concentrated in a single degree, then to modules over $\mathbb{Z}/n\mathbb Z$, then to flat modules and then (by the theorem of Govorov-Lazard, which I knew solely as a theorem of Lazard) to free modules by the commutativity of $H^{i}(X_{s},-)$ with direct limits. For free modules, the result is obvious. The moral of the story to me as always been that whenever a functor commutes with direct limits, then it is enough to consider free modules if one wishes to prove something about flat modules.

$\endgroup$
2
  • $\begingroup$ Thanks for the example. I think the principle you formulated in the last sentence is good to keep in mind. $\endgroup$
    – Ralph
    Apr 18, 2013 at 10:29
  • 1
    $\begingroup$ The special case of Lazard's theorem over $\mathbf{Z}/n\mathbf{Z}$ can be proved much more easily than the general case (we can assume $n$ is a prime power to be over an artin local ring, and then lift a basis over the residue field and use the vanishing of a power of the maximal ideal, etc. -- this appears very early on in Matsumura's book "Commutative Ring Theory", for example), so it could be debatable to say that Lazard's theorem plays a "crucial role" in the proof of the proper base change theorem. But the final sentence is not debatable. :) $\endgroup$
    – user30180
    Apr 19, 2013 at 10:09
3
$\begingroup$

An interesting example arises in the consideration of the $n$th symmetric power of a flat scheme morphism (such as for "directly" constructing the Hilbert scheme of $n$ points on a curve and relating it to the Picard scheme, building on one of Weil's original approaches to constructing the Jacobian of a smooth curve). More specifically, if $A$ is a flat $R$-algebra and $S_n$ denotes the $n$th symmetric group then the subalgebra $(A^{\otimes n})^{S_n}$ is $R$-flat and its formation commutes with any base change on $R$. (Thus, more globally, if $X$ is a flat projective scheme over a ring $R$ then the projective $R$-scheme ${\rm{Sym}}^n(X) := (X^n)/S_n$ is $R$-flat and its formation commutes with any base change on $R$.) A key point is that we do not impose any "lazy" hypothesis concerning the size of $S_n$ being a unit in $R$.

To see such properties for the subalgebra of symmetric tensors, one forgets about the algebra structure and aims to show more generally that if $M$ is any flat $R$-module then $(M^{\otimes n})^{S_n}$ is $R$-flat and its formation commutes (via the evident map) with any base change on $R$. This module problem is compatible with direct limits in $M$, so by the Lazard theorem we are reduced to the case when $M$ is finite free, which in turn is clear by inspection! See pp. 252-254 in "Neron Models" for a discussion (with references) for the application to Hilbert schemes of curves. In a similar spirit, if $M$ is a flat $A$-module then its symmetric and exterior powers are $A$-flat (as we see by using Lazard's theorem to pass to the direct limit on the elementary case of finite free modules).

Another place where the symmetric power of algebras arises is in the construction of the relative Verscheibung morphisms ${\rm{Ver}}_{G/S}: G^{(p)} \rightarrow G$ for any flat commutative group scheme $G \rightarrow S$ over an ${\mathbf{F}}_p$-scheme $S$, compatible with any base change on $S$. For this, one uses the $p$th symmetric power of suitable affine opens in the underlying $S$-scheme. See 4.2-4.3 in Exp. VII$_{\rm{A}}$ in SGA3 for further details (Lazard's theorem arises at the end of 4.2).

For yet another application, if $A \rightarrow B$ is a flat map of rings and $I$ is an ideal of $A$ equipped with a divided power structure $\gamma$ then the divided power structure extends (visibly uniquely) to $IB$. The proof involves reducing the existence (or rather, well-definedness) problem to one in which $B$ intervenes only through its underlying $A$-module structure, so one can use Lazard's theorem to reduce the newly formulated problem to the case of finite free modules, where the necessary computations are easy.

$\endgroup$
1
  • $\begingroup$ Thanks for the examples. Why have you made your answer community wiki ? $\endgroup$
    – Ralph
    Apr 18, 2013 at 10:22
3
$\begingroup$

I cannot offer any deep theorem, but I stumbled upon the following basic question: Let $C$ be a AB5 abelian $R$-linear tensor category. There is a notion of flat object in $C$. Since directed colimits and coproducts of monos are mono, one can see that flat objects are closed under directed colimits as well as coproducts. Now, there is a cocontinous tensor functor $\mathsf{Mod}(R) \to C, M \mapsto M \otimes_R 1$. Does it preserve flat objects? In fact, this follows immediately from Lazard's Theorem.

Let's try to prove this directly: If $X \to Y$ is a monomorphism in $C$, we have to prove that $M \otimes_R X \to M \otimes_R Y$ is a monomorphism in $C$. But how can we use flatness when $X,Y$ are no $R$-modules? Besides, even if $X,Y$ are $R$-modules, this is not clear at all, since $\mathsf{Mod}(R) \to C$ doesn't preserve monos (it only preserves colimits, in particular epis). I don't think that there is a direct proof.

Remark that Lazard's Theorem transforms a statement of the form "for all modules ... is exact" into a statement of the form "there exists a direct system of modules such that ... is an isomorphism". The latter is easier to check and invariant under many functors, whereas the first would require the functor to be essentially surjective or something like that in order to lift the test module.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.