In proving the following statement:
Every Euclidean domain is a Principal Ideal Domain.
We are trying to show that every non-zero ideal in Euclidean Domain is a Principal Ideal. So, let $A(\neq 0)$ be any non-zero ideal of D. Let $b(\neq 0)$ be any element in A s.t $d(b)$ is minimum where $d$ is a euclidean valuation i.e $\exists$ no element $c\in A$ s.t $d(c)\lt d(b)$.
My question is:
Can $d(b)=0$ or it has to be positive always i.e $\gt 0$?
My attempt:
It can be $0$ and I am reading a book which says if $a(\neq 0)\in D$ where $D$ is a euclidean domain s.t $d(a)=0$ then a is a unit.(I need a help proving this statement though!). So keeping this statment in mind if $d(b)=0$, then $b$ will be unit and hence it will generate the whole Domain, which nowhere violates the conditions of it being a principal ideal domain and hence this case is permissible. Am I correct? Please throw some light. Thank you.
First, let's prove that if $d(a)=0$ for some nonzero $a$, then $a$ is a unit.
Consider the elements $1$ and $a $. Then, there exists $q $ and $r $ such that $1=aq+r $ with $r=0$ or $d (r)\lt d (a)=0$. But $d (r)\lt 0$ is not possible and hence $r=0$. So $1=aq $ and hence $a $ is a unit.
What you wrote is correct; $d(b)$ can be zero. In that case, you have $d(1)=0$.