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Please point out any mistakes and flaws in my arguments. Thanks!

Hint: $\forall a\in D\setminus{0}$, $\mu(1) \leq \mu(a)$ (Proof. $\mu(1) \leq \mu(1a) = \mu(a)$)

Proof.

($\implies$) Suppose $\mu(1) = \mu(u)$ s.t. $u$ is a non-unit in $D$. $u = 0$ is not possible since $\mu(0)$ is undefined, so $u \neq 0$. Then $\exists q,r \in D$ s.t. $1 = uq + r$. If $r=0$, then $u$ is a unit, a contradiction. If $\mu(r) < \mu(u)$, then $\mu(r)< \mu(1)$, contradicting the hint. Thus $u$ is a unit.

($\impliedby$) Let $u$ be a unit in $D$. From the hint, $\mu(1) \leq \mu(u)$. Also $\mu(u) \leq \mu(uu^{-1}) = \mu(1)$. Thus $\mu(1) = \mu(u)$.

Arturo Magidin
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    It’s okay, but the forward direction is needlessly complicated by making it a proof by contradiction. You know $u\neq 0$; Write $1=uq+r$, with either $r=0$ or $\mu(r)\lt\mu(u)=\mu(1)$. Since the latter cannot hold, $r=0$, so $u$ is a unit. – Arturo Magidin Apr 02 '21 at 12:56
  • It's a special case $,I=(1),$ of the simple proof by Euclidean descent that ideals in Euclidean domains are principal, generated by any element of minimal Euclidean value, i.e. Euclidean domains are PIDs. You can easily restructure the proof to get this more general result, and much more conceptual insight (cf. my answer in the linked dupe), – Bill Dubuque Apr 02 '21 at 13:23

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