Romanian National Olympiad 1999 - Grade 12 - Problem 4

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Romanian National Olympiad 1999 - Grade 12 - Problem 4

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integral domain

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Source: Romanian National Olympiad 1999 - Grade 12 - Problem 4

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#1 h Jan 30, 2025, 4:01 PM

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Let $A$ be an integral domain and $A[X]$ be its associated ring of polynomials. For every integer $n \ge 2$ we define the map $\varphi_n : A[X] \to A[X],$ $\varphi_n(f)=f^n$ and we assume that the set $M= \Big\{ n \in \mathbb{Z}_{\ge 2} : \varphi_n \mathrm{~is~an~endomorphism~of~the~ring~} A[X] \Big\}$ is nonempty.

Prove that there exists a unique prime number $p$ such that $M=\{p,p^2,p^3, \ldots\}.$

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#2 h Jan 30, 2025, 4:16 PM

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Analyze the cases \operatorname{char}(A)=0 and \operatorname{char}(A)=p where p is a prime. (
/wiki/Integral_domain#Characteristic_and_homomorphisms )

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#5 h Jan 31, 2025, 2:17 AM

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Let $n \in M.$ Then
$x^n+1=\phi_n(x)+\phi_n(1)=\phi_n(x+1)=(x+1)^n=x^n+1+\sum_{k=1}^{n-1}\binom{n}{k}x^k$ and so
$\binom{n}{k}1_A=0, \ \ \ \ \ \ \ \ \\ (*)$ for all $1 \le k \le n-1.$ In particular, $M \ne \emptyset$ implies that the characteristic of $A$ is nonzero. Let $p$ be the characteristic of $A.$ Now, $(*)$ gives
$p \mid d_n:=\gcd \left \{\binom{n}{1}, \cdots \binom{n}{n-1} \right \}.$ But it is a well-known fact (see here for a proof) that $d_n > 1$ if and only if $n=q^m$ for some integer $m \ge 1$ and some prime number $q,$ in which case $d_n=q.$ So $p \mid q$ hence $q=p$ and $n=p^m.$

The converse is trivial, that is, it's clear that if $n=p^m$ for some integer $m \ge 1,$ then $n \in M.$

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1668 posts

#7 h Jan 31, 2025, 4:22 AM

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In fact, the result given in your problem holds for any infinite domain not just polynomial rings over domains. Let $R$ be an infinite domain. Let $n \ge 2$ be an integer and suppose that the map $f: R \to R$ defined by $f(a)=a^n, \ a \in R,$ is a ring homomorphism. Then $2^n=f(2)=2f(1)=2$ and so $(2^n-2)1_R=0$ implying that the characteristic of $R$ is a prime number, say $p.$ So $R$ contains a copy of $\mathbb{Z}_p.$ Now consider the polynomial $g(x):=(1+x)^n-1-x^n=\sum_{k=1}^{n-1}\binom{n}{k}x^k \in \mathbb{Z}_p[x].$ Then for any $a \in R,$
$g(a)=(1+a)^n-1-a^n=f(1+a)-f(1)-f(a)=0$ and so every element of $R$ is a root of $g.$ Thus $g=0$ because $R$ is not finite. So $p \mid \binom{n}{k}$ for all positive integers $k \le n-1$ and hence, by the link in my previous post, $n=p^m$ for some positive integer $m.$

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#9 h Mar 29, 2025, 7:26 AM

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Actually $A$ being a domain is not needed.

Claim. If the characteristic of $A$ is a prime $p$ , then $M$ is in the desired form.
Proof. All powers of $p$ are in $M$ due to freshman's dream.

If $n\in M$ , let $q$ be the largest power of $p$ dividing $n$ so $\binom{n}{q}$ is relatively prime with $p$ . Considering $(1+x)^n=1+x^n$ , we get $n=q$ , as desired. $\square$

For $n\in M$ , from $(1+x)^n=1+x^n$ we deduce that the characteristic of $A$ is nonzero, say $m$ . Taking reduction of $A[x]$ mod $p$ for a prime factor $p\mid m$ yields that all elements in $M$ are a power of $p$ . Hence $m$ has only one prime factor so it is a power of $p$ . Suppose $p^\alpha\in M$ for some positive integer $\alpha$ . Since $\binom{p^\alpha}{p^{\alpha-1}}$ has $p$ -adic valuation equal to $1$ , it is nonzero if $m\neq p$ . This is a contradiction by considering $(1+x)^{p^\alpha}=1+x^{p^\alpha}$ . Thus $m=p$ and the Claim finishes. $\square$

This post has been edited 2 times. Last edited by KevinYang2.71, Mar 29, 2025, 7:28 AM

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