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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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    If you write down the definition of Euclidean domain, which proof boilerplate does that definition suggest immediately?2011-11-13
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    I think you want that norm to be $p^{1-n}$.2011-11-13
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    @HenningMakholm I'm not quite sure which proof it is meant to suggest, but the first condition that: if a|b then N(a)$\le$N(b) is very easy to prove. It's the second condition that is tricky to prove...i.e. I need to find q and r such that a=bq+r and N(r)$\le$N(b) for a,b$\in R, b\neq0$2011-11-13
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    @Catherine: you can always take $q=0$ or $r=0$ depending on $N(a)< N(b)$ or not.2011-11-13
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    @QiL: thanks, I have noticed that, but I think I need to give a specific value of q for when r=0. (obviously if q=0 then r=a so that's not a problem...)2011-11-13
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    @Catherine: if $N(a)\ge N(b)$, show that $b$ divides $a$ in $R$.2011-11-13
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    @QiL:yes, my attempt so far is like this: Suppose N(a)=m,i.e.the first m-1 terms are 0 and $a_m\neq0$. Similarly, let N(b)=k, $k\le m$. We want to find $q\in R$ such that, a=bq. Given the property $a_{k+1}\equiv a_k(modp^k)$ we note that $p^{k-1}|b_m$ and similarly, $p^{m-1}|a_m$.Then I can write $a_m=p^{m-1}\hat{a_m}$ and $b_m=p^{k-1}\hat{b_m}$.Note that now $\hat{b_m}$ is co-prime with p, so is invertible! By rearranging $p^{m-1}\hat{a_m}=p^{k-1}\hat{b_m}*q$, we get $q_m=p^{m-k}\hat{a_m}\hat{b}^{-1}_m$. Now I'm having trouble proving that $q_{n+1}=q_n$ mod $p^n$ to show that $q\in R$2011-11-14
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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 $R$.2011-11-14
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    @QiL: thank you very much, I got it!! if $N(a)=p^{n-1}$ then $a_n=0 $ mod $p^{n-1}$ similarly, $a_{n+1}, a_{n+2}...$ all divide $p^{n-1}\Rightarrow$ a isn't invertible as every component is not co-prime with $p^k$ (k integers $\ge{n}$). However if we were to take out $p^{n-1}$ i.e. $a=p^{n-1} \hat{a}$. The remaining terms are co-prime with every $p^k$ (since $p^{n-1}\nmid a\Rightarrow p^{k}\nmid a$ as k>n). Therefore every term is invertible in its respective $(\mathbb{Z}/p^k\mathbb{Z})\Rightarrow\hat{a}$ is invertible $\Rightarrow$ a is a unit so every N(a)=$p^{n-1}$ is $p^{n-1}*$ unit2011-11-15
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    @QiL:sorry, I run out of edit time, I learned LaTex only yesterday.The last bit is $\hat{a}$ is a unit. Conversely, we note if a= unit*$p^{n-1}$, $a=(a_1,a_2...)$ where each component is co-prime with respective $p^k$ (so that it is invertible) multiplying by $p^{n-1}$ makes each term up to n-1 a multiple of $p^k$ thus =0 e.g. $a_1*p^{n-1}$ is divisible by p thus $a_1*p^{n-1}=0$ when we reach $a_n*p^{n-1}$ neither $a_n$ or $p^{n-1}$ are divisible by $p^n$ so not equal to 0! I can't thank you enough! If you post an answer I will select it as best:)2011-11-15
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    I am happy to see that you finally succeed to solve the problem !2011-11-15
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    @QiL: I can't thank you enough! I really really wanted to solve this question, I even broke down at one point over it! Maybe if you post anything as an answer, I can select it as best and give you some more points?2011-11-16

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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 :)2011-11-16