Let $w$ be an algebraic element over $\mathbb{C}[x]$, with minimal polynomial $f(t)=c_mt^m+\cdots+c_1t+c_0$, $c_i \in \mathbb{C}[x]$.

Is it possible to characterize (in terms of the $c_j$'s) all algebraic elements $w$, such that $R=\mathbb{C}[x][w]$ has no prime elements?

Please see this related question, in which the special case $k[x^2][x^3]$ is dealt with (notice that in this special case, $m=2$ with $c_2=1$, namely, $w$ is integral).

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    $\begingroup$ This is far from being a full characterisation: In the integral case, if the curve given by $f$ is smooth, then there are no prime elements whenever the curve has positive genus. Because a prime element would mean that the corresponding point on the curve is linear equivalent to the point at infinity in the projective closure. This can only happen for rational curves by a well known lemma. $\endgroup$
    – MooS
    May 11 '17 at 18:14
  • $\begingroup$ Thank you very much! Additional partial answers (from you or other people) are also welcome. $\endgroup$
    – user237522
    May 11 '17 at 18:55
  • $\begingroup$ @MooS This is not quite correct, several things can go wrong. In particular, there could be several points at infinity and then things don't work so easily. $\endgroup$
    – Mohan
    May 11 '17 at 21:57
  • $\begingroup$ @Mohan, please, do you have any ideas what happens if $f$ is not monic, namely, $w$ is not integral? Which case (integral or not integral) is easier to deal with, in your opinion? $\endgroup$
    – user237522
    May 11 '17 at 22:09
  • $\begingroup$ I would suspect the integral case is simpler, but I do not think there can be any simple criterion without dealing with what happens at infinity. $\endgroup$
    – Mohan
    May 12 '17 at 0:00

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