Let $A$ be an Euclidean domain associated to a function $d:A-\{0\}\rightarrow \mathbb{N}$ and $a\in A$, $a\ne0$. Prove that if $d(a) = 1$, then $a$ is inversible or irreducible.
I've tried using the fact that $\exists q,r : x = qa + r, \forall x \in A$ satisfying $1 = d(a) > d(r) \Rightarrow d(r) = 0$ or $r=0$, but I'm not sure where to go from there.
Best Answer
If we do Euclidean division between $1$ and $a$ we get $1=aq+r$ where $d(r)<d(a)=1$ or $r=0$.
If $r=0$ we have $1=aq\implies$ $a$ is invertible.
If $d(r)=0$ then we can demonstrate $a$ must be irreducible: Suppose $a=hg$. We know that $d(h)\le d(gh)=d(a)=1$, if $d(h)=0\implies h$ is invertibile, if $d(h)=1$ we do Euclidean Division between $h$ and $a$: $h=ab+c$ with $d(c)=0$ or $c=0$.
If $c=0\implies h=ab=hgb\implies h(1-gb)=0\implies gb=1$ so $g$ invertible.
If $d(c)=0$ then $c$ is invertible and $h=ab+c=hgb+c\iff h(1-gb)=c\iff hc^{-1}(1-gb )=1\implies h$ invertible.
In every situation at least one between $h$ and $g$ is invertible, so $a$ is irreducible.