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Is it very obvious that on a Euclidean Domain, two elements $x$ and $y$ have the same Euclidean norm $\nu(x) = \nu(y)$ then they are associates?

Can someone give me a proof of this?

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Might it be possible to give a characterization of all Euclidean domains (which are not fields) for which equality of norms implies elements are associates? Assuming the norm is multiplicative (as I like to...), this implies an injection of monoids $(R \setminus \{0\})/R^{\times} \rightarrow \mathbb{Z}^+$, which seems quite restrictive. – Pete L. Clark May 22 '12 at 18:50

4 Answers 4

up vote 6 down vote accepted

No. Consider the Euclidean domain $\mathbb R[x]$, which has Euclidean norm $\nu(f)=\mathrm{deg}(f)$. Then for example $x^2+1$ and $x^2+x+1$ have the same norm but are not associates, as the units are nonzero elements of $\mathbb R$.

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Ah ok, so I guess that's only true in the Gaussian integers? – user26069 May 22 '12 at 13:30
@morphism, not even there. See my answer. – lhf May 22 '12 at 13:31
Oh yeah, that's odd because my lecture notes use this when proving: ''If $x \in \mathbb{Z}[i]$ and $\nu (x)$ is prime in $\mathbb{Z}$ then $x$ is Gaussian prime.'' And we have that if $\nu (x) = p$ and if $\nu(y) | \nu(x)$ then $\nu (y)$ is $1$ or $p$, and if $\nu(y) = p$, 'then $x$ is associate to $y$'... – user26069 May 22 '12 at 13:34
@morphism Just because it holds for primes doesn't mean it holds for all natural numbers. But the statement "If $x\in \mathbb Z[i]$ and $\nu(x)$ is prime in $\mathbb Z$ then $x$ is Gaussian prime.'' is actually false, for example, $5$ is not a Gaussian prime. – Alex Becker May 22 '12 at 13:37
@AlexBecker, $N(5)=25$. – lhf May 22 '12 at 13:44

In $\mathbb Z[i]$, we have $N(1+7i)=50=N(5+5i)$ but $1+7i$ and $5+5i$ are not associates.

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A simpler counterexample is $1 + 2i, 1 - 2i$. – Qiaochu Yuan May 22 '12 at 17:04

While this is not true, what you seem to want is true (based on the comments): an element $x$ of a Euclidean domain is a unit if and only if $\nu(x)=\nu(1)$.

To see this, note that $\nu(1)\leq\nu(z)$ for all $z\neq 0$ (since $\nu(1)\leq\nu(1z)$). If $x$ is a unit, then $\nu(x)\leq\nu(xx^{-1})=\nu(1)$ gives the equality.

Conversely, if $\nu(x)=\nu(1)$, divide $1$ by $x$ to get $1 = xy+r$ with $r=0$ or $\nu(r)\lt(x)=\nu(1)$. But $r\neq 0$ implies $\nu(r)\geq \nu(1)$, so we conclude that $r=0$ so $x$ is a unit.

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Hint $\ $ If it were true, then in a norm-euclidean quadratic field, ramification would be rampant

$$\rm\:N(w) = ww' = N(w')\:\Rightarrow\:(w) = (w')\:\Rightarrow\:(p) = (p,w)(p,w') = (p,w)^2 $$

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