# Is every locally free module of rank $1$ over a commutative ring concretely invertible?

Since this subject is full of misunderstandings (see here, here, here, and here) let us fix a precise terminology.
Let $$A$$ be a commutative ring and $$P$$ an $$A$$-module.

I) We'll say that $$P$$ is a locally free module of rank one or is invertible if $$P$$ is finitely generated, projective and of rank one in the sense that for every prime ideal $$\mathfrak p$$ of $$A$$ the localized $$A_\mathfrak p$$- module $$P_\mathfrak p$$ (which is free by projectiveness) is of dimension $$1$$.
These modules correspond bijectively, by a well known result of Serre in FAC, to locally free sheaves $$\tilde P$$ of rank $$1$$ on $$\operatorname {Spec}A$$, also known as invertible sheaves. This is one motivation for the above terminology.
Another justification for the terminology "invertible" is that these modules are exactly those for which the canonical evaluation map $$P^*\otimes_AP\to A$$ is an isomorphism.

II) If $$B\supset A$$ is an overring of $$A$$ and $$P\subset B$$ is an $$A$$-module, we'll say that it is concretely invertible with respect to $$B$$ if $$P.(A:P)_B=A$$.
[As is standard $$(A:P)_B$$ denotes the set of elements $$b\in B$$ such that $$bP\subset A$$]
Lam proves in his Lectures on Modules and Rings, that these concretely invertible modules are invertible. What about the converse?

Question:
Is an invertible $$A$$-module $$P$$ isomorphic as an $$A$$-module to a concretely invertible module $$P'\subset B$$ with respect to a suitable overring $$B\supset A$$?

Remarks
a) Denote by $$\operatorname {Quot} A$$ the total quotient ring of $$A$$ obtained by inverting the regular (=not zero-divisors) of $$A$$, so that $$A\hookrightarrow \operatorname {Quot} A$$ is injective.
Then a submodule $$P\subset \operatorname {Quot}A$$ is invertible if and only if it is concretely invertible.
b) The answer is "yes" if $$A$$ is an integral domain: we can take $$P'$$ sitting inside $$B=\operatorname {Frac}A$$.
c) The answer is "yes" if $$A$$ is semi-local, since then $$P$$ is free of rank $$1$$: see here.
d) The answer is "yes" if $$A$$ is noetherian: in Eisenbud's Commutative Algebra, page 253, Theorem 11.6 b., it is proven that every invertible module $$P$$ over a noetherian ring $$A$$ is isomorphic to a concretely invertible submodule $$P'\subset \operatorname {Quot} A$$ of its total quotient ring.
e) Whatever the answer to the question is, it is definitely not true that we can always find the required $$P'$$ inside the total quotient ring $$B=\operatorname {Quot} A$$.
Lam gives a counter-example in his Lectures on Modules and Rings, Example (2.22)(A), page 37.

• Since the result holds over noetherian bases, the obvious idea would be to make a noetherian approximation (write $A$ as a colimit of its f.g. subrings) and use that $P$ is f.p. Did you try that? – Martin Brandenburg Dec 24 '19 at 20:23
• @Martin Brandenburg. No, I didn't. – Georges Elencwajg Dec 24 '19 at 23:41

The answer is yes. Recall that given an invertible $$A$$-module $$P$$ and $$n \in \mathbf{Z}$$ there is an invertible $$A$$-module $$P^{\otimes n}$$ such that $$P^{\otimes 0} = A$$, $$P^{\otimes 1} = P$$, and $$P^{\otimes n} \otimes_A P^{\otimes m} = P^{\otimes n + m}$$. Set $$B = \bigoplus_{n \in \mathbf{Z}} P^{\otimes n}$$; this is a commutative $$\mathbf{Z}$$-graded $$A$$-algebra (details omitted). Then $$P \subset B$$ in degree $$1$$ and $$(P : A)_B = P^{\otimes -1}$$ sitting in degree $$-1$$ and we have $$P \cdot P^{\otimes -1} = A$$.

Remark. The spectrum of $$B$$ is the $$\mathbf{G}_m$$-torsor over $$\text{Spec}(A)$$ corresponding to $$P$$.

• What an amazingly crisp, elegant and brilliant solution: I can't begin to tell you how grateful I am for this definitive answer, dear darx! – Georges Elencwajg Dec 26 '19 at 22:41
• I don't think the solution is great, because the ring $B$ depends on $P$. It would be much better if there were a single ring that worked for all invertible $R$-modules $P$ (which is what I thought was being asked). – Jesse Elliott Dec 27 '19 at 1:00

This is a partial answer of a more general problem.

If $$A$$ is a commutative ring with few zerodivisors (which holds iff $$\text{Quot}(A)$$ is semilocal), then the answer is yes: in that case, an invertible $$A$$-module is equivalently a concretely invertible $$A$$-submodule of $$\text{Quot}(A)$$. This result includes the cases you mention where $$A$$ is a domain or $$A$$ is semilocal or $$A$$ is Noetherian. For an outline of the proof, see Exercises 14 and 15 of Section 2.5 of my new book, "Rings, Modules, and Closure Operations." (Sorry, this isn't meant as a shameless plug :) I suspect that in general one needs to use the complete ring of quotients $$Q(R)$$. In other words, I'd conjecture that, for a general commutative ring $$A$$, an $$A$$-module is invertible iff it is a concretely invertible $$A$$-submodule of $$Q(A)$$. (This might even generalize to noncommutative rings.)

EDIT (to clear up some confusion):

In the answer above, $$Q(R)$$ denotes the complete ring of quotients, also known as the maximal ring of quotients, which in general contains but can be much bigger than $$\text{Quot}(R)$$. See planetmath.org/completeringofquotients. Unlike the total quotient ring, the maximal ring of quotients is defined even for noncommutative rings (although then one must distinguish between the maximal left ring of quotients and the maximal right ring of quotients).

ANOTHER EDIT: I thought the original question was to find an overring $$B$$ of $$A$$ that every invertible $$A$$-module is isomorphic to some concretely invertible $$A$$-submodule of $$B$$. This is what the total quotient ring $$B = \text{Quot}(A)$$ achieves in the case where $$A$$ is a ring with few divisors, and I suspect that this is what the complete ring of quotients $$B = Q(A)$$ achieves for any commutative ring $$A$$. The answer that was accepted, however, doesn't require that a single overring $$B$$ work for every $$P$$. The problem of determining a singe $$B$$ that works for every $$P$$ is still open.

• You speak of rings with “few zero divisors” as being equivalent to $\operatorname{Quot}(A)$ being semilocal: is this a definition of having “few zero divisors”? If so, what is the logic behind this terminology? And if not, what is the definition? – Gro-Tsen Jan 1 at 13:46
• Def: A ring $A$ has few zerodivisors if the the set of all zerodivisors of $A$ is a finite union of primes. From prime avoidance and the fact that the set of all zerodivisors of any ring is the union of the primes that are maximal with respect to not containing a zerodivisor, it follows that a ring has few zerodivisors if and only if its total quotient ring is semilocal. Being able to represent the set of zerodivisors as a finite union of primes is a finiteness condition on the set of zerodivisors, hence the terminology. – Jesse Elliott Jan 1 at 14:12
• I think the condition that the total quotient ring be semilocal is more intuitive than the set of zerodivisors being a finite union of primes, so that's why I gave the former definition instead of the latter (which is the standard definition given in references). – Jesse Elliott Jan 1 at 14:20
• Note also that a ring is said to be Marot if every ideal containing a non-zerodivisor is generated by non-zerodivisors. Marot rings are therefore rings with "lots" of non-zerodivisors. It is known that every ring with few zerodivsors is Marot. Intuitively, this means that a ring with few zerodivsors has lots of non-zerodivsors. This gives some further justification for the terminology. – Jesse Elliott Jan 2 at 1:05