It may sound like a basic question, but I am finding it hard to prove using the definition of Euclidean domain.
So far I have thought that there should be an element $a \in R$ such that $d(a)$ is minimum. This will imply that for any $x \in R$, $x = pa + r$ where $r=0$ or $d(r) < d(a)$. But as $a$ is chosen such that $d(a)$ is minimum, $r$ must be $0$.
So, $a$ should divide every element of $R$. I am stuck here.
Definition provided in the book: An integral domain R is called a Euclidean domain if for all $a \in R$, $a \neq 0$ there is defined a non -ve integer $d(a)$ s.t.,
- for all $a,b \in R$, $a \neq 0$, $b \neq 0$, $d(a) \leq d(ab)$
- for all $a,b \in R$, $a \neq 0$, $b \neq 0$, $\exists$ $t$ and $r$ in $R$ s.t., $a = tb + r$, where either $r=0$ or $d(r) \lt d(d)$.
Definition of integral domain in the book: "A commutative ring $R$ is called an Integral domain if $ab=0 \implies$ either $a=0$ or $b=0$".
Best Answer
As $a$ divides every element of $R$, it divides itself. So $a = ta$ for some $t \in R$. So for every $b \in R$, $ab = tab \implies a(b-tb)=0 \implies b=tb$. So $t$ is the identity of $R$.