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Let $f(x),g(x)$ be polynomials in $\mathbb{Q}[x]$. If $\mathrm{deg}(f)\geq2$ and $f$ irreducible, is the composition $f(g(x))$ always reduced (has no repeated irreducible factors)?

(If we do not ask $\mathrm{deg}(f)\geq2$ we can take $f(x)=x-1, g(x)=x^2+1$; if we do not ask $f$ be reducible, we can take $f(x)=(x-1)x$ and $g=x^2+1$.)

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  • $\begingroup$ @M.G. Thanks the example! And apologize that I should modify the question where "$f\circ g$ irreducible" replaced by ''$f\circ g$ reduced'' (no repeated factors). $\endgroup$
    – user39380
    Oct 23, 2021 at 12:38
  • $\begingroup$ No problem! I will now delete my comment as it is no longer applicable. $\endgroup$
    – M.G.
    Oct 23, 2021 at 15:26
  • $\begingroup$ @M.G. Please feel free to leave your comment there, as it is a very nice reference! (You referred to this link: math.stackexchange.com/questions/1909625/…) $\endgroup$
    – user39380
    Oct 23, 2021 at 15:27
  • $\begingroup$ Sorry, I already deleted it. But your comment contains the link, so it's all good :-) $\endgroup$
    – M.G.
    Oct 23, 2021 at 15:32

1 Answer 1

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It may have repeated irreducible factor. Take $f(x)=x^2+1$ and $g(x)=x+f(x) h(x)$ so that $g'(i)=0$. Then $f^2$ divides $f(g) $.

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    $\begingroup$ For instance $h(x)=x/2$, so $g(x)=(x^3+3x)/2$, yields $f\circ g(x)=\frac14(x^2+1)^2(x^2+4)$. $\endgroup$
    – YCor
    Oct 23, 2021 at 13:16
  • $\begingroup$ @Fedor Thanks for this very nice example! Would you explain a bit about the idea? I could verify the equivalence $f^2|f(g)$ with $f|1+2hx$ with $g'(i)=0$ in this specific form by expanding, but not sure how this came up? (I try to generalize $\mathbb{Q}$ to $\mathbb{C}(t)$, not sure if there is a similar example.) $\endgroup$
    – user39380
    Oct 23, 2021 at 13:41
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    $\begingroup$ The idea is that a double root of $f(g)$ corresponds to $f(g)=(f(g))'=f'(g)g'=0$, the first multiple can not be 0 but the second multiple can $\endgroup$ Oct 23, 2021 at 15:15

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