Are there any $f,g \in \mathbb{Q}[x]$ such that for every root of unity $\zeta$, and every $a,b \in \mathbb{Q}(\zeta)$, $f(a) \neq g(b)?$
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7
I guess you won't be satisfied with the answer $f=0$ and $g=1$. :)
But the answer is yes even if you assume that $f$ and $g$ are nonconstant. For example, consider $f(x)=2x^3$ and $g(x)=(x^3-2)^3$. If $f(a)=g(b)$ for some $a,b \in \mathbb{Q}(\zeta)$, then one finds that $2$ is a cube in $\mathbb{Q}(\zeta)$, which is a contradiction.
answered Nov 13, 2014 at 23:08
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1$\begingroup$ You beat me to it. I was about to post "A silly observation is you can take $f(x)=1$ and $g(x)=2$. So presumably you want $f$ and $g$ non-constant," and then follow that with the observation that the OP is looking at points on curves defined over $\mathbb{Q}^{\text{ab}}$, for which it shouldn't be hard to produce examples with no points (as you just did). $\endgroup$ Commented Nov 13, 2014 at 23:10
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4$\begingroup$ So doesn't it mean that the answer is yes? $\endgroup$– ACLCommented Nov 13, 2014 at 23:13
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1$\begingroup$ And what if $\mathbb{Q}^{\text{sol}}$ is taken instead of $\mathbb{Q}^{\text{ab}}$? $\endgroup$– PabloCommented Nov 13, 2014 at 23:26
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$\begingroup$ @ACL: Oops, yes. I'll fix this. $\endgroup$ Commented Nov 14, 2014 at 2:48
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2$\begingroup$ @Pablo: I don't know how to do it for $\mathbb{Q}^{\operatorname{sol}}$ instead of $\mathbb{Q}^{\operatorname{ab}}$. The trick behind my example above was to choose $f$ and $g$ so that $f(x)=g(y)$ has no geometrically irreducible component over $\mathbb{Q}^{\operatorname{ab}}$, but that idea doesn't seem to be implementable over $\mathbb{Q}^{\operatorname{sol}}$. Whether every geometrically irreducible curve over $\mathbb{Q}^{\operatorname{sol}}$ has a rational point is an open question, as far as I know. $\endgroup$ Commented Nov 14, 2014 at 3:02