Let $R$ be a Euclidean Domain and $f$ be an Euclidean valuation on $R$. Show that if $a$ and $b$ are associates in $R$, then $f (a) = f (b)$.
What I tried:
I know that $f$ is an Euclidean Valuation so,
$f: R-\{0\} \to \mathbb{N}$.
Let $a$ and $b$ elements of $R$, $b\ne 0$, $a=bq+r$ such that $r=0$ or $f(r)<f(b)$.
Let $a$,$b$ elements of $R-\{0\}$, $f(a)\le f(ab)$.
Since $a$ and $b$ are associates, $a=bu$ and $b=au^{-1}$ where $u$ is a unit.
ED iff PID iff Field, so $a\mid b$ and $b\mid a$.
Here comes nothing. How can I prove it?
Best Answer
If $a=\mu b$, with $\mu \in R^*$ we have $f(a)=f(\mu b)\ge f(b), $ so $f(a)\ge f(b)$. But also $b=a\mu^{-1}$ so $f(b)=f(a\mu^{-1})\ge f(a)$. And now we conclude that $f(a)=f(b)$.