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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4$\begingroup$ What do you think about it? Show please your attempts. $\endgroup$– Michael RozenbergCommented Apr 22, 2020 at 4:38
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2$\begingroup$ @Wowoju mathoverflow.net/q/21003/17064 (OP should have linked this) $\endgroup$– Gro-TsenCommented Apr 22, 2020 at 12:12
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2$\begingroup$ Trivial, but worth noting that the converse holds: $\mathbb{Q}^2\leftrightarrow \mathbb{Q} \implies \mathbb{Q}^2\times\mathbb{Q}\leftrightarrow \mathbb{Q}\times\mathbb{Q}\leftrightarrow\mathbb{Q}$. $\endgroup$– Yaakov BaruchCommented Apr 22, 2020 at 13:50
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3$\begingroup$ @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? $\endgroup$– Yaakov BaruchCommented Apr 22, 2020 at 14:16
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3$\begingroup$ 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 $A^2$. $\endgroup$– WojowuCommented Apr 22, 2020 at 15:34
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