An integral domain $R$ is called nearly euclidean if there is a function $$N:R \setminus \{0\} \to \mathbb{Z}^{>0} $$
satisfying the following property: for every $a,b \in R$ with $a \ne 0$, we have $a \mid b,$ or there exist $x,y \in R$ such that $N(ax-by) < N(a).$ Setting $y = 1$ (I think this is a typo and means $x = 1$) changes the definition from a nearly euclidean to a euclidean domain.
Prove that every nearly euclidean domain is a principal ideal domain.
Let $R$ be a nearly euclidean domain and let $\{0\} \ne U \subset R$ be an ideal (If $U = \{0\},$ we are done since $\{0\} = (0)).$ Consider the set $\mathcal{S} = \{N(a):a \in U, a \ne 0 \}.$ Then $\mathcal{S} \subset \mathbb{Z}^{>0}$ and is non-empty since $U \ne \{0\}.$ Say $\mathcal{S}$ has a least element $d$ and set $N(a) = d$ for $a \in U.$ Suppose $b \in U.$ By the nearly euclidean property, $ax = r + by$ for some $x,y \in R$ with $r = 0$ or $N(r) < N(a).$ Since $a,b \in U,$ it follows that $ax,by \in U$ so that $ax – by = r \in U.$ If $r \ne 0,$ then $N(r) < N(a),$ but since we $N(a)$ is equal to the smallest element in $\mathcal{S},$ it must be the case that $a \mid b.$ Hence $b = aq$ for some $q \in R$ so that $b \in (a).$ Thus $U \subset (a)$ and since $(a) \subset U$ by definition of an ideal, we conclude that $U = (a). \text{ } \Box$
I'm trying the prove that every nearly euclidean domain is a PID, but here I get stuck. In the usual proof (proving that a euclidean domain is a PID), we assume $b \in U$ and forcing $r = 0$ yields $b = aq + r \Longleftrightarrow aq + 0 = aq,$ demonstrating that $b \in (a) = \{ar: r \in R\}.$ This shows that $U \subset (a),$ and since $(a) \subset U$ by definition, $U = (a).$ Here it is a bit more confusing since in this particular case $by$ is a multiple of $a$ instead of $b$ being a multiple of $a$ by itself.
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But this is weaker than the nearly Euclidean property. Either $a|b$ or there are $x,y\in R$ with $ax-by\neq 0$ and $N(ax-by)<N(a)$. The latter is impossible, so $a|b$.