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Let $P$ be a polynomial in $R[X]$. Then show that

$P(X) -X$ divides $P(P(X))-X$

Brian Fitzpatrick
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Marko Karbevski
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3 Answers3

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Let $I$ be the ideal of $R[X]$ generated by $P(X)-X.$ You want to show that $P(P(X)) - X$ maps to zero under the canonical map $R[X]\to R[X]/I.$ This is true since in the quotient ring, $P(X)=X$ so $P(P(X) ) - X = P(X)- X=0.$

Ragib Zaman
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  • Is this also a valid solution:

    Let $X_i$ be a solution of $P(X)-X=0$

    Then $P(P(X_i))-X_i = P(P(X_i)-X_i+X_i)-X_i=P(X_i)-X_i=0$ ?

     – Marko Karbevski Mar 13 '14 at 07:26
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    @MarkoKarbevski Simply knowing that all the roots of $f(X)$ are also roots of $g(X)$ does not imply that $f(X)$ divides $g(X).$ Take for example $R=\mathbb{R}, f(X) = (X^2+1)(X-1)$ and $g(X)=X-1.$ The only root of $f$ is $X=1,$ which is also a root of $g,$ but $f$ does not divide $g.$ Your solution is valid if you also know that $f$ factors into a product of a constant and linear factors (so $f=a (X-r_1)(X-r_2)\cdots (X-r_n)$ ) and that the leading coefficient of $f$ divides the leading coefficient of $g.$ This is the case when $R$ is an algebraically closed field for example. – Ragib Zaman Mar 13 '14 at 07:40
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If the degree of $P$ is less or equal $0$ then the result is clear.

Otherwise, let $P(X)=\sum_{k=0}^n a_k X^k,\; n\ge1$.

$$P(P(X))-X=P(P(X))-P(X)+P(X)-X=\sum_{k=0}^n a_k\left((P(X))^k-X^k\right)+P(X)-X\\=\sum_{k=1}^n a_k\left((P(X))^k-X^k\right)+P(X)-X$$ But for $1\le k\le n$, $$(P(X))^k-X^k=(P(X)-X)\left((P(X))^{k-1}+(P(X))^{k-2}X+\cdots+P(X)X^{k-2}+X^{k-1}\right)$$ is divisible by $P(X)-X$ so it's the same for $P(P(X))-X$.

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    This is essentially the Ruffini rule, $P(a)-P(b)=(a-b)Q(a,b)$, where $Q$ is some other polynomial obtained by polynomial division, for example using the Horner scheme. Now set $a=P(X)$ and $b=X$ to get the same result but hiding all the calculation by citing some theorem. – Lutz Lehmann Mar 13 '14 at 19:57
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${\rm mod}\,\ p(x)\!-\!x\!:\,\ \color{#c00}{p(x)\equiv x}\,\Rightarrow\, p(\color{#c00}{p(x)})\equiv p(\color{#c00}x)\equiv x\ $ by the Polynomial Congruence Rule.

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Bill Dubuque
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