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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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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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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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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