When iterates of polynomial is a prime power of a polynomial
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Let $mathbb{K}$ be a field, $f^{(n)}(x)=(g(x))^p$ where $f,g in mathbb{K}[x]$; $p$ is prime. Can we conclude that $f(x)=(h(x))^p$ for some $hinmathbb{K}[x]$?
($f^{(n)}(x)$ denotes an iteration of $f$.) The question seems to be interesting even for cases $mathbb{K}=mathbb{R}$ or $mathbb{K}=mathbb{Q}$.
polynomials field-theory
asked Nov 26 at 10:26
Alex
1
add a comment |
Let $mathbb{K}$ be a field, $f^{(n)}(x)=(g(x))^p$ where $f,g in mathbb{K}[x]$; $p$ is prime. Can we conclude that $f(x)=(h(x))^p$ for some $hinmathbb{K}[x]$?
($f^{(n)}(x)$ denotes an iteration of $f$.) The question seems to be interesting even for cases $mathbb{K}=mathbb{R}$ or $mathbb{K}=mathbb{Q}$.
polynomials field-theory
asked Nov 26 at 10:26
Alex
1
This is false for $n = p$ and $f(x) = x$. Perhaps you had in mind $n$ and $p$ coprimes?
– Rchn
Nov 26 at 11:22
$f^{(n) } $ denotes iteration of $f$
– Alex
Nov 26 at 12:28
add a comment |
Let $mathbb{K}$ be a field, $f^{(n)}(x)=(g(x))^p$ where $f,g in mathbb{K}[x]$; $p$ is prime. Can we conclude that $f(x)=(h(x))^p$ for some $hinmathbb{K}[x]$?
($f^{(n)}(x)$ denotes an iteration of $f$.) The question seems to be interesting even for cases $mathbb{K}=mathbb{R}$ or $mathbb{K}=mathbb{Q}$.
polynomials field-theory
asked Nov 26 at 10:26
Alex
1
Let $mathbb{K}$ be a field, $f^{(n)}(x)=(g(x))^p$ where $f,g in mathbb{K}[x]$; $p$ is prime. Can we conclude that $f(x)=(h(x))^p$ for some $hinmathbb{K}[x]$?
($f^{(n)}(x)$ denotes an iteration of $f$.) The question seems to be interesting even for cases $mathbb{K}=mathbb{R}$ or $mathbb{K}=mathbb{Q}$.
polynomials field-theory
polynomials field-theory
asked Nov 26 at 10:26
Alex
1
asked Nov 26 at 10:26
Alex
1
asked Nov 26 at 10:26
Alex
1
asked Nov 26 at 10:26
Alex
1
asked Nov 26 at 10:26
Alex
1
1
This is false for $n = p$ and $f(x) = x$. Perhaps you had in mind $n$ and $p$ coprimes?
– Rchn
Nov 26 at 11:22
$f^{(n) } $ denotes iteration of $f$
– Alex
Nov 26 at 12:28
add a comment |
This is false for $n = p$ and $f(x) = x$. Perhaps you had in mind $n$ and $p$ coprimes?
– Rchn
Nov 26 at 11:22
$f^{(n) } $ denotes iteration of $f$
– Alex
Nov 26 at 12:28
This is false for $n = p$ and $f(x) = x$. Perhaps you had in mind $n$ and $p$ coprimes?
– Rchn
Nov 26 at 11:22
This is false for $n = p$ and $f(x) = x$. Perhaps you had in mind $n$ and $p$ coprimes?
– Rchn
Nov 26 at 11:22
$f^{(n) } $ denotes iteration of $f$
– Alex
Nov 26 at 12:28
$f^{(n) } $ denotes iteration of $f$
– Alex
Nov 26 at 12:28
add a comment |
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This is false for $n = p$ and $f(x) = x$. Perhaps you had in mind $n$ and $p$ coprimes?
– Rchn
Nov 26 at 11:22
$f^{(n) } $ denotes iteration of $f$
– Alex
Nov 26 at 12:28