# function of a unit in a Euclidean domain

Let $R$ be a Euclidean domain and let $u$ be a unit in $R$. If we denote $\delta$ the corresponding function, is it true that $\delta(c)=\delta(uc)$ for every non-zero $c \in R$?

I know that an integral domain $R$ is a Euclidean domain if there is a function $\delta$ from the nonzero elements of $R$ to the non-negative integers with the following properties: (1) If $a$ and $b$ are nonzero elements of $R$, then $\delta(a) \le \delta(b)$. (2) If $a,b \in R$ and $b \ne 0_R$, then there exist $q,r \in R$ such that $a=bq+r$ and either $r=0_R$ or $\delta(r) < \delta(b)$.

Here is what I've attempted:

We know that if $u$ is a unit of $R$, then $\delta(u)=\delta(1_R)$.

Therefore $\delta(c)=\delta(uc)$ for some nonzero $c \in R$, with $c=1_R$ because $\delta(1_R)=\delta(u)=\delta(u \cdot 1_R)$.

I fell like going one way was not enough so I also proved backwards:

If $\delta(c)=\delta(uc)$ for some nonzero $c \in R$, then according to the definition of a Euclidean domain (with $c$ and $uc$ in place of $a$ and $b$, there exist $q,r \in R$ such that

$c=(uc)q+r$ and either $r=0_R$ or $\delta(r) < \delta(uc)$.

If $\delta(r)< \delta(uc)$, then by part (1) of the definition with $\delta(u)=\delta(1_R)$, we have

$\delta(c) \le \delta(c(1_R-uq))=\delta(c-ucq)=\delta(r)<\delta(uc)=\delta(c)$,

so that $\delta(c) < \delta(c)$, which is a contradiction. Hence, we must have $r=0_R$. Thus $c=(uc)q$, which implies that $1_R=uq$. Therefore, $u$ is a unit.

Did I do it correctly? Or did I do too much? Any suggestions will be greatly appreciated. Thanks in advance!

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You have not shown that for every non-zero $c \in R$ that $\delta(c) = \delta(uc)$, just for the very specific case $c = 1_R$. –  Henry Swanson Dec 11 '13 at 6:18
Also, you may want to check property $1$. As it is, it requires $\delta$ to be a constant function. –  Henry Swanson Dec 11 '13 at 6:21

Let $v \in R$ be the inverse of $u$, and let $c \in R$ be non-zero. By the multiplicative property of the norm, we know that $$\delta(c) \le \delta(cu) \le \delta(cuv) = \delta(c)$$ So $\delta(c) = \delta(cu)$ for all non-zero $c \in R$.
For the reverse: it seems good, but make sure that $1_R - uq$ isn't zero, otherwise the $\delta(a) \le \delta(ab)$ doesn't hold. (Although, in that case, $uq = 1$, so it's a unit anyways)