The Zariski cancellation problem
If A[x] is a polynomial ring, must A be one?
▸Motivation
Zariski, 1949. True for n ≤ 2 over ℂ (Fujita, Miyanishi, Sugie).
Neena Gupta showed in 2014 that it fails in positive characteristic for n ≥ 3, which won her the Ramanujan Prize. Characteristic zero remains open, and Gupta's counterexamples use Frobenius, so they say nothing about it.
Oscar Zariski, 1949. Positive characteristic settled negatively by Neena Gupta, 2014.
▸Lean API
import Mathlib.RingTheory.Polynomial.Basic
import Mathlib.Data.Complex.Basic
import Mathlib.Algebra.MvPolynomial.Basic
namespace Conjectura.AL002
/-- If `A[x]` is a polynomial ring over a field in `n+1` variables, must `A` itself be
a polynomial ring in `n`? True for `n ≤ 2` over ℂ; Gupta showed it fails in positive
characteristic for `n ≥ 3`. Characteristic zero is open. -/
def goal : Prop :=
∀ (A : Type) [CommRing A] [Algebra ℂ A] (n : ℕ),
Nonempty (Polynomial A ≃ₐ[ℂ] MvPolynomial (Fin (n + 1)) ℂ) →
Nonempty (A ≃ₐ[ℂ] MvPolynomial (Fin n) ℂ)
end Conjectura.AL002▸Definition5
- CommRingCommRing
class
- AlgebraAlgebra
class
- NonemptyNonempty
abbrev
- PolynomialPolynomial
structure
- MvPolynomialMvPolynomial
abbrev
▸Related work1
- On the cancellation problem for the affine space A³ in characteristic pNeena Gupta · 2014
Disproves the positive-characteristic case; characteristic zero untouched.
▸For your AI
I am proving a theorem in Lean 4 and submitting it to Conjectura.
## Problem AL002 — The Zariski cancellation problem
If A[x] is a polynomial ring, must A be one?
## 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 : ∀ (A : Type) [CommRing A] [Algebra ℂ A] (n : ℕ),
Nonempty (Polynomial A ≃ₐ[ℂ] MvPolynomial (Fin (n + 1)) ℂ) →
Nonempty (A ≃ₐ[ℂ] MvPolynomial (Fin n) ℂ) := by
sorry
```
## The file I submit
```lean
import Conjectura.Problems.AL002.Statement
namespace Submission
theorem solution : Conjectura.AL002.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.
### Algebra
class
```lean
Algebra
```
Defined in Mathlib.
### Nonempty
abbrev
```lean
Nonempty
```
Defined in Mathlib.
### Polynomial
structure
```lean
Polynomial
```
Defined in Mathlib.
### MvPolynomial
abbrev
```lean
MvPolynomial
```
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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