A Euclidean domain $D$ has a function $F$ from nonzero elements of $D$ to nonnegative integers that makes division with remainder possible in $D$: Let $D$ have elements $t$ and $s$, and $t$ is nonzero. Then $D$ has elements $q$ and $r$ for which $s = tq + r$, and if the remainder $r$ is nonzero too, then $F(t) > F(r)$. If $r$ happens to be zero, then $t$ divides $s$ because $s=tq$. If $t$ happens to divides $s$, then is $r$ zero? If $t$ divides $s$, then $s = ut$ for an element $u$ in $D$. Then $r=t(u-q)$ is an element of the ideal generated by $t$, so the ideal generated by $r$ is contained in the ideal $(t)$.
Do any of these mean $r=0$?
We have $t$ divides $r$, which is unusual for nonzero remainders, but, unlike with the Euclidean domain of the integers $\mathbb Z$, I don't see the precise contradiction for a general Euclidean domain $D$. Bernard poses a good question. Usually, the $F$ that allows us to deduce $r=0$.
Something that might help: Can we deduce the existence of a function $G$ that satisfies $G(p) \le G(pv)$ and $G(p)=0$ if and only if $p=0$?
Thanks, and have a happy new year!