Last edited by hbghlyj at 2023-2-15 17:40:00设$θ=\frac{1}{2}(1+\sqrt{-19})$, $ℤ\left[θ\right]=\Set{m+nθ:m,n∈ℤ}$,
证明：不存在函数 $d :ℤ\left[θ\right]\setminus\{0\}→ \Bbb N$，满足：
- 若 $a$ 整除 $b$，则 $d ( a ) ≤ d ( b )$。
- 若 $a$ 不整除 $b$，则存在 $q , r ∈ ℤ\left[θ\right]$ 使得 $a = b q + r$，而且 $d ( r ) < d ( b )$。
来自Algebraic structures I, Spring 2011
证明We choose $m \in R$ such that $d(m)$ is as small as possible, subject to $m$ not being 0 or a unit. First we divide 2 by $m$, and get a quotient and remainder:
$$
2=m q+r \text { with } d(r)<d(m) \text { or } r=0 .
$$
This means $r=0,1$, or $-1$. If $r=0$, then $m \mid 2$, which means $m=\pm 2$, since 2 is irreducible, and $m$ is not a unit. Similarly, if $r=-1$, then $m=\pm 3$. The case $r=1$ cannot happen, for if it did, then $m \mid 1$ so $m$ is a unit.
Next we divide $\theta$ by $m$ in the same way, getting
$$
\theta=m q^{\prime}+r^{\prime} \text { with } d\left(r^{\prime}\right)<d(m) \text { or } r^{\prime}=0 .
$$
Again we have $r^{\prime}=0,1$, or $-1$. So one of $\theta, \theta+1$, or $\theta-1$ is divisible by $m$. But $m=\pm 2$ or $\pm 3$, and it is easy to see that none of these quotients is in $R$. This contradiction tells us that $R$ is not a Euclidean domain for any Euclidean function. | 
