Let $x = (a,b) \in \mathbb{Q}^2$ and let $p(x,t) = t^2-at+b$. Does there exist an involution $\tau$ of $\mathbb{Q}^2$ such that for all $\tau(x) \neq x$, $x \in \mathbb{Q}^2$ one of the polynomials $p(x,t)$ or $p(\tau(x),t)$ is irreducible in $\mathbb{Q}[t]$. This is not a homework question and I don't know if this question might be considered research level. It is just a question out of curiosity.
1 Answer
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Of course, for example $\tau(x)\equiv x$. Well, if you want fixed point free involution, the answer is still yes since there are countably many irreducible polynomials and countably many reducible polynomials, you may match them. Maybe, you need something more about $\tau$, say, require it to be linear?
answered Jan 22, 2017 at 7:37
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$\begingroup$ Nice argument, with the countability! $\endgroup$– user6671Commented Jan 22, 2017 at 8:10
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1$\begingroup$ @stackExchangeUser if you really need to make the answer into a concrete construction, you need a bijection $(a,b)\leftrightarrow(a',b')$ between rational pairs $(a,b)$ s.th. $a^2-4b$ is a square and rational pairs $(a',b')$ s.th. $a'^2-4b'$ is not a square. There should be an explicit such bijection, playing with prime factorization, although not particularly nice or simple. A warm-up should be: finding an involution of $\mathbb{Q }$, $q\leftrightarrow q'$ such that $q$ is a square iff $q'$ is not. $\endgroup$ Commented Jan 22, 2017 at 8:23