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Assume that there is no polynomial $f(x,y)\in{\mathbb Q}[x,y]{}$ such that $f\colon {\mathbb Q}\times{\mathbb Q}\to {\mathbb Q}$ is a bjiection. Does this imply that there is no polynomial $f(x,y,z)\in{\mathbb Q}[x,y,z]{}$ such that $f:{\mathbb Q}\times{\mathbb Q}\times{\mathbb Q}\to {\mathbb Q}$ is a bijection ?

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    What do you think about it? Show please your attempts. – Michael Rozenberg Apr 22 '20 at 04:38
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    Why "Again" in the title? Was this question already asked before? – Wojowu Apr 22 '20 at 11:33
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    @Wowoju https://mathoverflow.net/q/21003/17064 (OP should have linked this) – Gro-Tsen Apr 22 '20 at 12:12
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    Trivial, but worth noting that the converse holds: $\mathbb{Q}^2\leftrightarrow \mathbb{Q} \implies \mathbb{Q}^2\times\mathbb{Q}\leftrightarrow \mathbb{Q}\times\mathbb{Q}\leftrightarrow\mathbb{Q}$. – Yaakov Baruch Apr 22 '20 at 13:50
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    @MichaelRozenberg. What is the presumption? Maybe OP has no fruitful attempts or promising ideas to show - that wouldn't invalidate the question. A trivial answer would, but is there one? – Yaakov Baruch Apr 22 '20 at 14:16
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    I doubt there is a simple "formal" argument for this. In various categories (e.g. groups) it is possible to have an object $A$ isomorphic to $A^3$ but not isomorphic to $A^2$. – Wojowu Apr 22 '20 at 15:34
  • @Wojowu: Is there a natural category whose isomorphisms are the polynomial bijections? – R. van Dobben de Bruyn Apr 22 '20 at 19:06
  • @R.vanDobbendeBruyn I doubt it (and I didn't mean to imply that). If there is, it definitely cannot have the obvious concretization, because the inverse of $\mathbb R\to\mathbb R,x\mapsto x^3$ is not a polynomial bijection. – Wojowu Apr 22 '20 at 19:54
  • Maybe you can localise the category of polynomial maps at the maps that induce bijections... – R. van Dobben de Bruyn Apr 22 '20 at 21:36
  • Can anyone extend Bresciani's conditional results on $\mathbb{Q}\times \mathbb{Q}\to \mathbb{Q}$ to answer this question? https://arxiv.org/abs/2101.01090 –  Feb 20 '21 at 14:46
  • @MattF. If you assume the geometric Lang conjecture in addition to the weak Bombieri-Lang conjecture, then my result rules out the case in which the generic fiber is of general type. It remains then to address the cases in which the fiber has lower Kodaira dimension, but that's probably easy. – Giulio Bresciani Mar 01 '21 at 17:15
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    Actually, if one uses more generally the methods of my paper rather than the result as stated, the geometric Lang conjecture can be avoided. – Giulio Bresciani Mar 02 '21 at 12:02

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