All rings in this question are integral.

It is known that flat modules are torsion-free. Conversely, torsion-free modules over Prüfer domain (in particular, Dedekind domain) are flat, please see here. My questions are:

  • Is there a general condition under which a torsion free module is flat?
  • What is the simplest example of torsion-free non-flat module?

edit: Since the example in Georges's answer is the coordinate ring of a singular curve, I want to ask the same questions for the coordinate ring $R$ of a smooth algebraic variety:

  • Is there a general condition under which a torsion free module over $R$ is flat?
  • what is the simplest example of torsion-free non-flat module over $R$
  • 1
    $\begingroup$ @liu hang: One of the equivalent definitions wikipedia lists is exactly "Every torsion-free R-modul is flat." Isn't your first question already answered with that? $\endgroup$ Jan 4, 2011 at 11:16
  • $\begingroup$ @Johannes: It is true for Prüfer domain, but not for general commutative rings. $\endgroup$
    – Liu Hang
    Jan 4, 2011 at 12:31
  • 10
    $\begingroup$ Take $R=k[x,y]$ and take $M=(x,y)\subset R$. This is torsion-free, but not flat. $\endgroup$ Jan 4, 2011 at 14:14
  • 1
    $\begingroup$ Keerthi's example is right, and it is certainly the ``simplest'' example of a torsion-free non-flat module over the coordinate ring of a smooth variety. Generally speaking, EVERY coordinate ring of an algebraic variety of dimension > 1 will have torsion free modules which are not flat (the ideal of any closed point is exactly such a module). $\endgroup$ Jan 4, 2011 at 15:02
  • 2
    $\begingroup$ Another source of free-torsion algebras comes from blowing-ups. Take $R=k[x,y]$ and $A=k[x, y/x]\subset k(x,y)$. $\endgroup$
    – Qing Liu
    Jan 4, 2011 at 20:36

4 Answers 4


Dear liu,

1) If $A$ is a domain in which every finitely generated ideal is principal, then a module over $A$ is flat iff it is torsion free (Bourbaki, Comm.Alg.,I,§2, 4, Prop.3). Of course a PID has this property, but the ring $\mathcal O(U)$ of holomorphic functions over a connected open subset $U\subset \mathbb C$ also has it, although it is not a PID.[Ah, I have just checked the definition of Prüfer and it seems that these rings are actually Prüfer, so you knew this. I'll leave this class of examples for the benefit of those who, like me, don't know the concept "Prüfer ring".]

2) A simple example of torsion free non flat module $E$ over a ring $A$ is $A=\mathbb C [t^2, t^3] \subset E=\mathbb C [t]$. This corresponds to the normalization of the cusp $S=Spec (\mathbb C[X,Y]/(Y^2-X^3))$ i.e. to the morphism $f: \mathbb A ^1 \to S \subset \mathbb A ^2$ given by $x=t^2, y=t^3$. Non-flatness is due to the fact that the fiber of $f$ at the origin is a double point on $\mathbb A ^1$ (supported at the origin) whereas the other fibers are simple points. This is a tiny particular case of the constancy of Hilbert polynomials in flat families. Of course there are direct proofs by pure algebra, but I hope you like geometry...

The end of Mumford's red book Here is a vast generalization of 2). Consider a locally noetherian integral scheme. If the scheme is not normal, its normalization is never flat over it.This is essentially proved in Matsumura's Commutative Ring Theory, Corollary to Theorem 23.9. (So if you are arithmetically inclined, $\mathbb Z[2i] \subset \mathbb Z[i]$ is a non flat algebra) . It is very easy to find a (slightly different) statement in the literature: the one and a half last lines of Mumford's Red Book !

Edit: I'm happy to report that I just located non-flatness of normalization in our friend Qing Liu's book Algebraic Geometry and Arithmetic Curves : 4.3.1. Example 3.5, page 136.

  • 11
    $\begingroup$ Just to make your answer complete: Lam proves in Theorem 4.69 (see also Prop. 4.20) in his "Lectures on Modules and Rings" (Springer GTM 189) that a (commutative) domain is a Prüfer ring if and only if every torsion-free module is flat. $\endgroup$ Jan 4, 2011 at 11:26
  • 1
    $\begingroup$ Thank you very much, Theo, for this extremely interesting and precise reference. $\endgroup$ Jan 4, 2011 at 11:34
  • $\begingroup$ Thanks for the intuitive explanation. But since the ring in your example is the coordinate ring a singular curve, I want to ask the same questiones for the coordinate ring $R$ of a smooth algebraic variety: 1)Is there a general condition under which a torsion free module over $R$ is flat? 2)what is the simplest example of torsion-free non-flat module over $R$ $\endgroup$
    – Liu Hang
    Jan 4, 2011 at 13:47
  • 4
    $\begingroup$ Re question 1: If your $R$-module $M$ is finitely generated, then it's flat if and only if it's projective. So if $R = k[X]$ for some smooth algebraic variety $X$ is such that every torsion free $R$-module is flat, then every ideal of $R$ is projective, which is equivalent to $R$ being Dedekind and to $X$ being a smooth curve. Re question 2: Take $R = k[x,y]$ and $M = xR + yR$. Then $M$ isn't flat because if it was, it would have to be projective (being finitely generated), and then $R/M$ would have projective dimension 1. But it's known that this module has projective dimension 2. $\endgroup$
    – user91132
    Jan 4, 2011 at 14:13
  • $\begingroup$ Dear Konstantin, in your equivalence related to question 1, you have to assume M finitely presented. If you replace finitely presented by finitely generated, there are counterexamples to "flat + finitely generated implies projective" . Cf. Example 4.37, page 135 in Lam's book quoted by Theo in his comment above. Of course your nice comment is not affected by my nit-picking, since in your case $R$ is noetherian. $\endgroup$ Jan 4, 2011 at 15:42

This is exactly the sort of basic question that I wish were asked more often.

The short answer is that a "typical" ideal in an integral domain $R$ is torsionfree but not flat.

First observe that if $R$ is a domain, then any ideal $I$ of $R$ is a submodule of the torsionfree $R$-module $R$ and is thus torsionfree.

Next, as Theo Buehler comments in Georges' answer, there is a precise criterion on $R$ for every torsionfree $R$-module to be flat: namely, that $R$ is a Prufer domain, i.e., every finitely generated ideal of $R$ is invertible.

For simplicity, let me restrict attention to Noetherian domains $R$.. (Later on I may come back and add more in the general case.) Let $\mathfrak{p}$ be a prime ideal of height greater than one. (So such ideals exist iff $R$ itself has Krull dimension greater than one, e.g. $k[x,y]$.) As above, $\mathfrak{p}$ is a torsionfree $R$-module. If $\mathfrak{p}$ were flat, then since it is finitely generated it would be projective, or equivalently locally principal. But localizing at $\mathfrak{p}$ and applying Krull's Principal Ideal Theorem we get a contradiction.

Similarly, if $R$ is a one-dimensional Noetherian domain which is not a Dedekind domain, then it admits a non-invertible ideal, which is therefore a torsionfree nonflat module. For instance if $R = \mathbb{Z}[\sqrt{-3}]$, then $\mathfrak{p} = \langle 1 + \sqrt{-3}, 1 - \sqrt{-3} \rangle$ is not locally principal. See e.g. Section 3 of these notes for more details.


Here is an answer to the extra question regarding regular rings:

A finitely generated module is flat if and only if it is locally free, so for finitely generated modules your question translates to:

Q1 When is a torsion-free module/sheaf locally free?


Q2 What is a simple example of a torsion-free (say coherent) sheaf on a smooth algebraic variety that is not locally free?

So, first of all, the locus where a torsion-free coherent sheaf on a smooth algebraic variety fails to be locally free is at least of codimension $2$, in particular we have the following partial answer to Q1:

A 1.1 Any torsion-free coherent sheaf on a smooth algebraic curve is locally free.

This is not true in higher dimensions and you can find a counter example on any smooth algebraic variety of dimension at least $2$:

A 2 Let $X$ be a smooth algebraic variety of dimension at least $2$ (e.g., $\mathbb A^2$) and let $\mathcal F=\mathfrak m_x\subset \mathcal O_X$ the (maximal) ideal of a closed point $x\in X$. Then $\mathcal F$ is not flat. (This should qualify as the simplest example possible with the requirement of $X$ being smooth given A1 above).

Proof of A 2 This ideal is isomorphic to $\mathcal O_X$ on the open set $X\setminus\{x\}$, but cannot be generated with less number of elements than the dimension of the local ring $\dim \mathcal O_{X,x}\geq 2$, so it cannot be locally free.

On a surface you have to do a little better to get local freeness: The locus where a reflexive sheaf is not locally free is at least of codimension $3$, so

A 1.2 Any reflexive sheaf on a smooth algebraic surface is locally free.

Being reflexive is equivalent to torsion-free and $S_2$. See more on $S_2$ in this answer.


If the question is only for commutative domains, then yes, torsion-free=flat iff it is Pruefer (see e.g. Fuchs/Salce, Thm. VI.9.10, actually due to Warfield), as was mentioned earlier. If not only domains are in question, then it can be said that Warfield showed that it is true whenever the ring's localizations at maximal ideals are valuation rings (let's call them Pruefer rings). I haven't checked if those are the only commutative rings for which this is true. If one leaves the realm of commutative rings, there is a much bigger class, so-called RD-rings, for which it is true. These are the rings for which Relative Divisibility is the same as purity (just as over the integers). An example is the first Weyl algebra over any field of char 0 (this is even a domain, though not commutative, so possibly not of interest to the poser of the question).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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