Mathoverflow 507128
There exists a proper ideal I in a (commutative) total ring R of fractions that is an invertible module. If I ⊊ R is such an example, I must have infinite order in the Picard group, and R must not be Noetherian (otherwise it must be semi-local and therefore have trivial Picard group).
▸Motivation
There exists a proper ideal I in a (commutative) total ring R of fractions that is an invertible module. If I ⊊ R is such an example, I must have infinite order in the Picard group, and R must not be Noetherian (otherwise it must be semi-local and therefore have trivial Picard group).
Adapted from formal-conjectures, Mathoverflow/507128.lean. Catalogued at https://mathoverflow.net/questions/507128/embeddability-order-on-picard-groups.
▸Lean API
import Mathlib.RingTheory.PicardGroup
namespace Conjectura.MO0001
/-- There exists a proper ideal `I` in a (commutative) total ring `R` of fractions that is an invertible module. If `I ⊊ R` is such an example, `I` must have infinite order in the Picard group, and `R` must not be Noetherian (otherwise it must be semi-local and therefore have trivial Picard group). -/
def goal : Prop :=
∃ (R : Type) (_ : CommRing R) (_ : IsFractionRing R R) (I : Ideal R),
I ≠ ⊤ ∧ Module.Invertible R I
end Conjectura.MO0001▸Definition4
- CommRingCommRing
class
- IsFractionRingIsFractionRing
abbrev
- IdealIdeal
structure
- InvertibleModule.Invertible
class
▸Related work2
- Mathoverflow 507128—
The catalogue entry, with references and status.
- formal-conjecturesThe Formal Conjectures Authors (Google DeepMind) · 2025
Source of the Lean formalization adapted here.
▸For your AI
I am proving a theorem in Lean 4 and submitting it to Conjectura.
## Problem MO0001 — Mathoverflow 507128
There exists a proper ideal `I` in a (commutative) total ring `R` of fractions that is an invertible module. If `I ⊊ R` is such an example, `I` must have infinite order in the Picard group, and `R` must not be Noetherian (otherwise it must be semi-local and therefore have trivial Picard group).
## Environment (fixed — do not assume anything newer)
- Lean toolchain: `leanprover/lean4:v4.33.0-rc1`
- Mathlib: `v4.33.0-rc1`
If a lemma you want does not exist in that Mathlib, prove it inline instead of
importing something newer.
## The exact statement I must prove
```lean
theorem solution : Conjectura.MO0001.goal := by
sorry
```
## The file I submit
```lean
import Conjectura.Problems.MO0001.Statement
namespace Submission
theorem solution : Conjectura.MO0001.goal := by
sorry
end Submission
```
## The Lean definitions of every term in this problem
These are the actual definitions your proof will be checked against. Do not
substitute your own version of any of them.
### CommRing
class
```lean
CommRing
```
Defined in Mathlib.
### IsFractionRing
abbrev
```lean
IsFractionRing
```
Defined in Mathlib.
### Ideal
structure
```lean
Ideal
```
Defined in Mathlib.
### Invertible
class
```lean
Module.Invertible
```
Defined in Mathlib.
## Rules — submissions violating these are rejected automatically
1. **Do not change the name or type of `solution`.** It must satisfy the
statement above exactly.
2. **Do not redefine or shadow anything from the problem's Statement module.**
Declaring your own `goal`, or redefining a name it depends on, produces a
proof of a *different* statement and is rejected. This is the single most
common rejection.
3. **No `sorry`** anywhere, including in helper lemmas. It surfaces as the
axiom `sorryAx` and is detected transitively through imports.
4. **No `native_decide`** — it surfaces as `Lean.ofReduceBool` and is not
accepted, because it trusts compiled code rather than the kernel.
5. Only these axioms are permitted: `propext`, `Classical.choice`,
`Quot.sound`.
6. Follow Mathlib style: hypotheses left of the colon, explicit types,
`snake_case` theorem names, `UpperCamelCase` types.
## What I want from you
Here is my argument in informal mathematics:
> [PASTE YOUR PROOF SKETCH HERE]
Turn it into Lean 4 that compiles under the environment above and satisfies the
statement exactly. Where you are unsure a lemma exists in this Mathlib version,
say so explicitly rather than guessing a name.Submissions are not open yet
Conjectura is in beta. You can read every statement, every definition and the Lean behind them, and download the exact files the checker uses — but proofs are not being accepted yet.
The reason is a deliberate order of operations. Accepting a proof means running a stranger’s code and standing behind a verdict, and no statement here yet carries a researcher’s name. A machine-checked answer to a question nobody has vouched for is worth very little, so the vouching comes first.
The English write-ups are also switched off during the beta. Nothing on this page is generated by a model.
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