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Let $R$ be an integral domain. Let $P$ be a finitely generated $R$-module.

The problem:

If we additionally assume that $R$ is a local ring (that is, $R$ is a local domain), is the following statement true? "$P$ is torsion-free if and only if $P$ is projective."


Some facts I know/encountered and the result of my search... I apologize if this might be messy but I want to ensure everybody is on the same page as me.

  1. If $R$ is just an integral domain then the statement is not necessarily true.

  2. If $R$ is a principal ideal domain (PID) then the statement is true (easily proven using the structure theorem of modules over PIDs).

  3. If $R$ is a local ring and we drop the integral domain assumption then the statement is not necessarily true.

  4. If $R$ is a discrete valuation ring (= local ring + PID) then the statement is true (since it is a module over PID).

  5. Discrete valuation rings are valuation rings but the converse is not necessarily true.

  6. Valuation rings are integral domains and local rings.

  7. A valuation ring is Noetherian if and only if it is a discrete valuation ring or a field. (Thus, the statement above is true for Noetherian valuation rings)

  8. An integral domain is a valuation ring if and only if it is a Bézout ring and a local ring. [Proposition 1.5, Krull]

  9. A local ring is a Bézout domain if and only if it is a valuation ring.

  10. A Bézout domain is a Prüfer domain.

  11. "A finitely generated module M over a Prüfer domain is projective if and only if it is torsion-free." (Converse direction "A finitely generated torsion-free module $M$ over a Prüfer domain $R$ is projective." proven in textbook [Theorem 2.7, Modules over Non-Noetherian Domains by Fuchs & Salce, original result proven by Kaplansky])

The conclusion I have from all the statements above is:

$$\{ \textit{discrete valuation ring} \} \supset \{ \textit{valuation ring} \} \supset \{ \textit{local rings} \} \cap \{ \textit{Bézout rings} \} \supset \{ \textit{local rings} \} \cap \{ \textit{Prüfer domain}\}$$

and any ring in any collection above will make the statement true.


The subquestions I am currently thinking about in hopes to answer the question above:

  1. If $R$ is a local domain, then is $R$ a Prüfer domain? If this is true, then we are done. If it is not true, I am hoping to construct a counter-example...

  2. In case this question is simpler, are there examples of rings that is a local domain but not a valuation ring?

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  • $\begingroup$ Possibly relevant additional source: ams.org/journals/tran/1952-072-02/S0002-9947-1952-0046349-0/… $\endgroup$ Jan 3, 2023 at 3:28
  • $\begingroup$ @MarianoSuárez-Álvarez Thank you. Do you recommend any introductory-level text that discusses this type of ring more in-depth? Maybe it is because I haven't taken any commutative algebra or algebraic geometry courses yet? $\endgroup$ Jan 3, 2023 at 19:58

1 Answer 1

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As Mariano has pointed out in the comments, over a local ring, projective and free are equivalent, so you are asking whether every finitely generated torsion-free module over a local domain is free.

This is not the case. Indeed, first observe that over any commutative ring $A$, an ideal $I$ is free if and only if it is principal and torsion-free. If $I$ is principal and torsion-free, then the singleton consisting of any generator is a basis for $I$. On the other hand, note that any two distinct elements $x, y \in I$ are linearly dependent, because $y \cdot x - x \cdot y = 0$ is a nontrivial $A$-linear relation.

Hence, it suffices to give an example of a local domain $A$ with an ideal $I$ that is not principal, since $I$ will be torsion-free but not free. Mariano already suggested an example of such a ring in the comments: for $F$ a field, consider the ring $A := F[X, Y]_{\langle X, Y \rangle}$. Then $A$ is a local ring with maximal ideal $M := \langle X, Y \rangle$. But $M$ cannot be generated by a single element; there are many proofs of this, but one is to observe that $M/M^{2}$ is an $A/M \cong F$-vector space of dimension $2$, and so $M$ cannot be principal or else $M/M^{2}$ would be a $1$-dimensional $F$-vector space.

I think it's worth noting that this kind of example is quite common. If $A$ is a Noetherian local ring with unique maximal ideal $M$, then by Krull's principal ideal theorem, the minimum number of generators of $M$ as an $A$-module is bounded below by the Krull dimension of $A$. Examples of local domains with Krull dimension greater than one abound, and give rise to an example in the same way highlighted above. The particular example we gave belongs to an important class of rings you will encounter when studying commutative algebra, namely so-called regular local rings. A Noetherian local ring $A$ with unique maximal ideal $M$ is a regular local ring if the minimal number of generators of $M$ as an $A$-module is equal to the Krull dimension of $A$.

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