Let $R$ be a Euclidean domain with degree function $\varphi$ and $R$ is not a field. Prove the following:
(1) Let $a \ne 0$ and $b \ne 0$ be two elements in $R$. Suppose that $a\mid b$ and $b \nmid a$. Prove that $\varphi (a) < \varphi(b)$
(2) Suppose that $\varphi(ab)= \varphi(a) \varphi(b)$ for all $a,b \in R\setminus\{0\}$. Prove that an element $u \in R$ is a unit if and only if $\varphi (u)=1$
By my book's definition, a Euclidean domain is a Euclidean ring which is an integral domain, which satisfies the following:
a) If a,b∈R and ab≠0, then $\varphi$(a)≤$\varphi$(ab)
b) If a,b∈R and b≠0, then there exists q,r∈R such that a=qb+r with r=0, or r≠0 and φ(r)<φ(b)
By my book as wel, if R is an Euclidean domain, then R is also an integral domain and a principal ideal domain.
It seems the following.
(1) Since $a,b\in R$ and $b\ne 0$, then there exist $q,r\in R$ such that $a=qb+r$ with $r=0$, or $r\ne 0$ and $\varphi(r)<\varphi(b)$. Since $b \nmid a$, $r\ne 0$. Hence $\varphi(r)<\varphi(b)$. Since $a\mid b$ and $a\mid qb$, $a\mid r$ too. Therefore there exists an element $c\in R$ such that $r=ac$. Since $ac=r\ne 0$, we have $\varphi(a)\le \varphi(r)<\varphi(b)$.
(2) The necessity is already proven by egreg, so we prove the sufficiency. Let $\varphi(u)=1$. There exist $q,r\in R$ such that $1=qu+r$ with $r=0$, or $r\ne 0$ and $\varphi(r)<\varphi(u)$. Assume that $r\ne 0$. Then $\varphi(r)=0$. Since $r=1r\ne 0$, $\varphi(1)\le \varphi(r)=0$, so $\varphi(1)=0$. As showed egreg, in this case $R$ is a field, a contradiction. So $r=0$. Then $u$ is a unit.