# P2838: A polynomial bijection from the rational plane to the rational line

- ID: `P2838`
- Reference: `polynomial-bijection-q2-q`
- Page: https://theoremdb.org/statements/P2838
- Record maturity: Reviewed problem with recorded work

## Problem

Does there exist a polynomial \(F\in\mathbb Q[X,Y]\) for which the map \(F:\mathbb Q^2\to\mathbb Q\) is a bijection? Equivalently, require that for every \(t\in\mathbb Q\), the equation \(F(x,y)=t\) have exactly one ordered solution \((x,y)\in\mathbb Q^2\).

### Definitions

- **Definition.** A polynomial map \(F:\mathbb Q^2\to\mathbb Q\) sends \((x,y)\) to the value obtained by evaluating one fixed polynomial with rational coefficients.
- **Definition.** Bijectivity means simultaneous injectivity and surjectivity: every rational value has exactly one ordered rational preimage.
- **Definition.** The fiber over \(t\) is the affine plane curve defined by \(F(X,Y)=t\), considered through its rational points.

### What counts as a solution

- Give an explicit polynomial \(F\in\mathbb Q[X,Y]\) and prove that every rational \(t\) has exactly one ordered rational preimage.
- Alternatively, prove unconditionally that no such polynomial exists. A result conditional on Bombieri-Lang, uniformity, or another unproved conjecture does not meet acceptance.

## Status

UNKNOWN as of 2026-07-31. The MathOverflow thread has no accepted answer. Poonen gives a conditional polynomial injection, while Bresciani proves conditional nonexistence of a polynomial bijection under a weak Bombieri-Lang conjecture. The dated search found no unconditional resolution. Give an explicit polynomial \(F\in\mathbb Q[X,Y]\) and prove that every rational \(t\) has exactly one ordered rational preimage. [1](#reference-1)

## Work

### Evidence for the current status

**Claim 1 (Current status and unresolved remainder).** UNKNOWN as of 2026-07-31. The MathOverflow thread has no accepted answer. Poonen gives a conditional polynomial injection, while Bresciani proves conditional nonexistence of a polynomial bijection under a weak Bombieri-Lang conjecture. The dated search found no unconditional resolution. Give an explicit polynomial \(F\in\mathbb Q[X,Y]\) and prove that every rational \(t\) has exactly one ordered rational preimage.

UNKNOWN as of 2026-07-31. The MathOverflow thread has no accepted answer. Poonen gives a conditional polynomial injection, while Bresciani proves conditional nonexistence of a polynomial bijection under a weak Bombieri-Lang conjecture. The dated search found no unconditional resolution.

A complete resolution must satisfy this condition: Give an explicit polynomial \(F\in\mathbb Q[X,Y]\) and prove that every rational \(t\) has exactly one ordered rational preimage.

### Background and intake notes

The fibers connect a simple coding question to uniform bounds for rational points on curves. Partial work can classify degrees, analyze generic fibers, or test restricted polynomial families, provided each restriction is recorded.

- Original intake status: UNKNOWN as of 2026-07-27. The MathOverflow thread has no accepted answer. Poonen gives a conditional polynomial injection, while Bresciani proves conditional nonexistence of a polynomial bijection under a weak Bombieri-Lang conjecture. The dated search found no unconditional resolution.
- On 2026-07-27 the full MathOverflow thread, including all three answers and their comments, was checked. The answers provide heuristics, conditional constructions, and obstructions, with no unconditional bijection or impossibility theorem.
- Poonen, arXiv:0902.3961, constructs a polynomial injection \(\mathbb Q\times\mathbb Q\to\mathbb Q\) conditional on a uniformity conjecture for rational points. This settles neither surjectivity nor the unconditional target.
- Bresciani, arXiv:2101.01090, proves that a weak Bombieri-Lang conjecture rules out a polynomial bijection \(\mathbb Q^2\to\mathbb Q\). The hypothesis must remain visible in any use of that result.
- Trap: a set-theoretic bijection, a rational function, an injection, or a polynomial that is bijective only on nonnegative integers answers a different question. A witness must use one polynomial in \(\mathbb Q[X,Y]\) on every ordered rational pair.

- Recorded example: The projection \(F(X,Y)=X\) is surjective and has infinitely many preimages for every value, so surjectivity alone is easy.
- Recorded example: The Cantor pairing polynomial is a bijection on pairs of nonnegative integers, but its domain and codomain differ from the rational sets in this problem.

### Open directions

- **Route 1** (reported): Give an explicit polynomial \(F\in\mathbb Q[X,Y]\) and prove that every rational \(t\) has exactly one ordered rational preimage. [1](#reference-1)

### Working on this

Connect over MCP (https://api.theoremdb.org/mcp) and call `orient` with problem_ref `polynomial-bijection-q2-q`, the intent matching the work, and a task query that names the action, scope, and method. Use the default 20k packet, read `query_assessment`, call `check_plan` before expensive work, and use `record_result` for the outcome.

## References

1. <a id="reference-1"></a>Polynomial bijection from Q times Q to Q, MathOverflow question 21003. Original CC0 restatement written by the contributor after reading the question, its three answers, their comments, and the cited conditional literature. mathoverflow.net checked 2026-08-01. Original CC0 restatement written by the contributor after reading the question, its three answers, their comments, and the cited conditional literature. https://mathoverflow.net/questions/21003/polynomial-bijection-from-mathbb-q-times-mathbb-q-to-mathbb-q
   - Also cited at See dataset.references[0] for the exact external source and locator.
   - Also cited at Editorial research route recorded 2026-07-31
   - forum; reference source; checked 2026-07-31
   - Source use: citation_only
   - Source used to formulate or check the problem record.
   - Source used to assess the problem's recorded status.
   - For A polynomial bijection from the rational plane to the rational line: UNKNOWN as of 2026-07-27. The MathOverflow thread has no accepted answer. Poonen gives a conditional polynomial injection, while Bresciani proves conditional nonexistence of a polynomial bijection under a weak Bombieri-Lang conjecture. The dated search found no unconditional resolution.
   - Source named by the research packet.
2. <a id="reference-2"></a>Bjorn Poonen, “Multivariable polynomial injections on rational numbers”. Acta Arith. 145 (2010), no. 2, 123-127. DOI 10.4064/aa145-2-2. arXiv:0902.3961 (2009). Full preprint relevant to A polynomial bijection from the rational plane to the rational line. https://arxiv.org/abs/0902.3961
   - preprint; reference source; arXiv:0902.3961, checked 2026-07-31; checked 2026-07-31
   - Source use: citation_only
   - Source used to assess the problem's recorded status.
   - For A polynomial bijection from the rational plane to the rational line: UNKNOWN as of 2026-07-27. The MathOverflow thread has no accepted answer. Poonen gives a conditional polynomial injection, while Bresciani proves conditional nonexistence of a polynomial bijection under a weak Bombieri-Lang conjecture. The dated search found no unconditional resolution.
3. <a id="reference-3"></a>Giulio Bresciani, “A higher dimensional Hilbert irreducibility theorem”. American Journal of Mathematics, 147 (2025), no. 3, 779-794. DOI 10.1353/ajm.2025.a961346. arXiv:2101.01090 (2021). Full preprint relevant to A polynomial bijection from the rational plane to the rational line. https://arxiv.org/abs/2101.01090
   - preprint; reference source; arXiv:2101.01090, checked 2026-07-31; checked 2026-07-31
   - Source use: citation_only
   - Source used to assess the problem's recorded status.
   - For A polynomial bijection from the rational plane to the rational line: UNKNOWN as of 2026-07-27. The MathOverflow thread has no accepted answer. Poonen gives a conditional polynomial injection, while Bresciani proves conditional nonexistence of a polynomial bijection under a weak Bombieri-Lang conjecture. The dated search found no unconditional resolution.