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I'm having trouble with the Claim: In a Euclidean domain $R$, every element with minimum norm is a unit.

The proofs I have seen say, e.g., $1 = q a + r$, where $a$ has minimum norm $N(a) = m$. Then, $N(r) < N(a)$, by definition of the Euclidean algorithm. Since $N(a)$ is minimal, $N(r) = 0$ implies $r = 0$. ($N(a) > 0$ for $a \ne 0$.)

I'm having difficulty understanding that "one" $1$ can always be written as, $1 = qa + r$? If $d$ has minimum norm (so $r = 0$), and $a = q d$, then $d | a$. Similarly, if $d = q^{'} d^{'}$, then $d^{'} | d$, and vice-versa. So, $d$ and $d^{'}$ differ by a unit. It's the decomposition of "one" $1$ that I'm having trouble grasping.

Cliff
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  • In the proof, it shouldn't be "Since $N(a)$ is minimal, $N(r)=0$". The proposition $N(r)<N(a)$ is not satisfied by any $r$, since $N(a)$ is minimum. The Euclidean division says "r=0" or $N(r)<N(a)$". So, the proof should say "Since $N(r)<N(a)$ cannot happen, then $r=0$". – NDB Jul 12 '23 at 20:26
  • And the ability to write $1=qa+r$ is not special of $1$. The definition of Euclidean domain is that that can be done for every element $x$, in particular $1$. – NDB Jul 12 '23 at 20:28
  • As here: in a Euclidean domain every element $,a!\ne! 0,$ with $,\rm\color{#c00}{minimal},$ Euclidean value is a unit (invertible), else $,a\nmid b,$ for some $,b,$ so $,b\div a,$ leaves nonzero remainder smaller than $,a.\ \ $ – Bill Dubuque Jul 12 '23 at 20:29
  • As remarked in the prior link, this is a special case (ideal = $(1))$ of the general proof that ideals are principal in Euclidean domains (generated by any minimal element). It is instructive to view it this way. – Bill Dubuque Jul 12 '23 at 20:36
  • Well, yes. Sorry if I was a little loose with my language. Since N(a) = m is the minimal value for the norm N, then it must be that r = 0. If N(r) = s > 0, and N(r) < N(a), then the value s would be the minimum value for the norm N. This is contrary to the fact that m is the minimum value for the norm N. – Cliff Jul 13 '23 at 16:44
  • I think the reason I ha e difficulty with the equation, $1 = qa$, is because I'm thinking of the integers, $\mathbb{Z}$. 2. Otherwise, there seems to be some "ring of fractions" lurking here in the background. 3. I like Keith Conrad's proof of this claim, given as Corollary 3.2, in https://kconrad.math.uconn.edu/blurbs/ringtheory/euclideanrk.pdf
    • – Cliff Jul 15 '23 at 14:17