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Here's a question that I'm wondering about:

Let $A$ be a Euclidean domain with Euclidean function $\delta$, but $A$ is not a field. Is it true that $\delta \left({A^*}\right)$ (where $A^* = A \setminus\{0\}$) has at least $3$ distinct elements ?

I know a result that: If A is a Euclidean domain, then A is a field if and only if $\delta\left(A^*\right) $ has only 1 element, but when apply to the question, it just only proves that $\delta(A^*)$ has at least 2 elements!

Thank you for your help

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The answer is yes.

Assume that $\delta(A^*)=\{d_1,d_2\}$ where $d_1<d_2$. Then every $u \in A^*$ with $\delta(u)=d_1$ is a unit. It follows that for all non-units $b \in A^*$ we have that $A/(b)$ is a field, thus $(b)$ is a maximal ideal. By the same argument $(b^2)$ is a maximal ideal too, which implies $(b)=(b^2)$ i.e. $b=qb^2$ for some $q \in A$. But since $b$ is not a zero divisor, we have $qb=1$ (contradiction).

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