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Let $p(t),q(t)\in\Bbb R[t]$. Do there always exist nonconstant $f(t),g(t)\in\Bbb R[t]$ such that $f(p(t))=g(q(t))$?

I started by looking at a tricky example $$x=t^5+2t^3+3t, y=t^6+t^4+t^2$$where there is really nothing to cancel out by comparing the RHS. I have really no idea for this specific example, and I doubt the question is true.

Rewrite the question in general: Let $k$ be a field and let $p,q\in k[x]$ be nonconstant. Is it always true that the subrings $k[p]$ and $k[q]$ generated by $p$ and $q$ respectively intersect nontrivally?

user108580
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    I think what you actually want is a polynomial $F \in \mathbb{R}[x,y]$ such that $F(p(t), q(t)) = 0$. This is known as implicitization; here are some relevant posts: 1, 2. I used SageMath to compute $F$ for the example you gave: $x^{6} - 14 x^{4} y + 6 x^{4} - 7 x^{2} y^{3} + 43 x^{2} y^{2} - 69 x^{2} y + 9 x^{2} - y^{5} + 12 y^{4} - 54 y^{3} + 108 y^{2} - 81 y$. – Viktor Vaughn Jul 06 '24 at 06:21
  • @ViktorVaughn Right. That makes better sense. Thank you. – user108580 Jul 08 '24 at 02:45

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I think the answer is no. Take $p = t^2 + t$, $q = t^2$, and suppose we have $f$ and $g$ as stated. Writing $f$ as a polynomial in $p$, it has the form $a_n p^n + a_{n - 1} p^{n - 1} + ... + a_0$ for some $n > 0$.

$p^n = t^n(t + 1)^n = t^n(t^n + nt^{n - 1} + ... + 1)$, so the coefficient of $t^{2n - 1}$ in $p^n$ is $n$. The coefficient of $t^{2n - 1}$ in $p^{n - 1}$ and all smaller powers of $p$ is evidently zero. So the coefficient of $t^{2n - 1}$ in $f(p(t))$ is $n a_n$, which is nonzero. But in $g(q(t))$, the coefficients of all odd powers of $t$ are zero. So this is impossible.

This argument appears to work for all fields $k$ (EDIT: of characteristic zero, and for $\Bbb{R}$ in particular).

Hew Wolff
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