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 $R$?
Best Answer
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.