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Let $ p \in \mathbb{Z}$ be a prime, and define

$$R:=\lbrace a=(a_1,a_2,a_3,\ldots) | a_k \in(\mathbb{Z}/p^k\mathbb{Z})\text{ and }a_{k+1}\equiv a_k \pmod {p^k}\text{ for all }k \in \mathbb{N} \rbrace$$

I have proved that R is a ring, with multiplication and addition defined component wise. The Norm for a= ($a_1,a_2,\ldots)\in R$ with $a\neq0$ is defined as $N(a)=p^{n-1}$ where $n$ is the smallest value of $k$ such that $a_k\neq0$ i.e. $a_n$ is the first non-zero term in the sequence $a$.

I need to prove that, if R is endowed with the map $N: R^*\to\mathbb{N}$, then R is an Euclidean domain.

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I think you want that norm to be $p^{1-n}$. –  Gerry Myerson Nov 13 '11 at 2:53
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@Catherine: you can always take $q=0$ or $r=0$ depending on $N(a)< N(b)$ or not. –  user18119 Nov 13 '11 at 15:55
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@Catherine: if $N(a)\ge N(b)$, show that $b$ divides $a$ in $R$. –  user18119 Nov 13 '11 at 21:56
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@Catherine: sorry, it is better to show directly that $N(a)=p^{n-1}$ if and only if $a$ is $p^{n-1}$ times a unit of $R$. –  user18119 Nov 14 '11 at 23:17
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I am happy to see that you finally succeed to solve the problem ! –  user18119 Nov 15 '11 at 23:45

1 Answer 1

up vote 2 down vote accepted

As required by the OP, here is the hint for solve the problem: show that $N(a)=p^{n−1}$ if and only if $a$ is $p^{n−1}$ times a unit of $R$.

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thank you again! It seems like the right answer, a fellow student also proved it that way :) –  Catherine Nov 16 '11 at 20:00

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