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The definition of euclidean ring in my textbook is 1) $N(ab) \geq N(a)$ 2) Euclidean algorithm works. However there is a note that the first condition is not necessary. There is a hint that we can choose a following norm: $N'(a) = \min_{b\sim a} N(b)$. I do not understand why it works. We need to check that euclidean algorithm still works with this norm.

I tried to do the following. If $a \sim b$, then $a = bx, x \in K^*$. Let $N'(a)$ be $N(ax)$. Let us divide $ax$ by $bx$ instead of $b$. We get $ax = bxq + rx$, $N(rx) \leq N(bx)$. Here I am stuck and have no idea how to continue.

Xam
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sooobus
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    There is a discussion in https://en.wikipedia.org/wiki/Euclidean_domain. – lhf Oct 04 '17 at 11:40
  • Also here: https://math.stackexchange.com/q/1156384/133781 and here: http://www.math.uconn.edu/~kconrad/blurbs/ringtheory/euclideanrk.pdf – Xam Oct 04 '17 at 19:57

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Let there be a function $N:R\setminus\{0\}\to \mathbb N_0$ such that for all $a,b\in R$, $b\neq 0$ there exist $q,r\in R$ such that $$a = qb+r,\ r = 0\ \text{or}\ N(r)< N(b)$$

and define $N'\colon R\setminus\{0\}\to \mathbb N_0$ with $N(a) = \min_{x\in R\setminus\{0\}}N(ax).$

First note that this function is well defined because $\mathbb N_0$ is well ordered. Furthermore, let $a,b\in R\setminus \{0\}$ and let $x\in R\setminus\{0\}$ such that $N'(ab) = N(abx)$. Then, $$N'(a) \leq N(abx) = N'(ab).$$

Now let us see what you did. You found $x$ such that $N'(a) = N(ax)$. Why? And how? First of all, $a$ might be $0$ and $N(ax)$ not even defined. And even if $a\neq 0$ and $N(ax)$ is defined, how does $N'(a)$ help you prove that $N'(r)<N'(b)$? There is nothing to connect those.

Instead, let us do more reasonable thing, let $x\in R\setminus\{0\}$ be such that $N'(b) = N(bx)$ and let $q$ and $r'$ in $R$ such that $$ax = q(bx) + r',\ r' = 0\ \text{or}\ N(r') < N(bx).$$

If $r' = 0$, we are done, since $ax = q(bx)$ implies $a = qb$ (we are in integral domain). If not, define $r = a - qb$. Obviously, $a = qb + r$ and $r' = rx$. Finally,

$$N'(r)\leq N(rx) = N(r') < N(bx) = N'(b).$$

Ennar
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