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.
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Yes, that's the idea to find nonunits $p\neq 0$ such that mod $p$ every element of $R$ is congruent to a unit (or $0),,$ i.e. $p$ is a universal side divisor, e.g. $\pm2,\pm3\in\Bbb Z,,$ and degree one polynomials over a field. They prove handy for proving some rings to be noneuclidean, e.g. see this answer. – Bill Dubuque Nov 27 '22 at 20:31
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More generally we can extend this idea to construct the (Motzkin) minimal Euclidean algorithm - see Lemmermeyer's survey The Euclidean Algorithm in Algebraic Number Fields. The universal side divisors are the new elements introduced in the 2nd level of the construction (see the Motzkin set $E_2$ in section $2.3$ of said survey) – Bill Dubuque Nov 27 '22 at 20:31
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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.
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@BillDubuque Thanks for the references. I recall a problem from an algebra exam that guided us through using this technique to show that $\mathbb Z[(1+\sqrt{-19})/2]$ is not a Euclidean domain, but we never did put a name to the concept. – Kenanski Bowspleefi Nov 27 '22 at 22:44
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Yes, that's a common exercise - one version of which I linked above in my first comment. – Bill Dubuque Nov 28 '22 at 03:09