For an euclidean domain $R$, there is $r\in R \setminus R^×$ with surjective projection $p:R^×\cup \{0\} \rightarrow R/Rr$ [universal side divisors]

Question

I think I should pick $r$ such that the value of $f(r)$ (where $f$ is an euclidean function) is minimal, but I’m not sure what to do next.

Answer 1

I think your approach is the right one. If $r$ is a (non-zero) non-unit element with $f(r)$ minimal among non-units, then for any unit $u$, we can use the division algorithm to write $u=ar+b$ with $f(b)<f(r)$ or $b=0$. If $b=0$, however, then since $u=ar$ is a unit and $1 = u^{-1}u = u^{-1}ar = (u^{-1}a)r$, it follows that $r$ is also a unit, which contradicts our choice of $r$. So $b\neq 0$ is a nonzero element with $f(b)<f(r)$. Since $f(r)$ was minimal among non-units, it follows that $b$ is a unit.

Now $b = u - ar$, so the projection $p(b)$ of $b$ onto the quotient $R/(r)$ is simply $u$. Since $u$ was picked as an arbitrary unit, this shows that $p$ is surjective.