# The necessary and sufficient condition for a unit element in Euclidean Domain

I am trying to prove that in Euclidean domain D with Euclidean function d, u in D is a unit if and only if d(u)=d(1).

Suppose u is a unit, then there exist v in D such that uv=1, this implies u\1 so d(u)<=d(1), but obviously 1 divides u so d(1)<=d(u). Hence, d(u)=d(1).

Conversely, suppose d(u)=d(1), since u is not zero, there exist q and r in D such that 1=uq+r with r=0 or d(r)< d(u).

If r=0 then u is a unit. Else d(r)< d(u) =d(1), this implies d(r)< d(1). I stop here, because I failed to argue that r must be zero.

Can anyone help me? Thanks.

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Does your $d$ satisfy $d(xy)\le d(x) d(y)$? In other words, what is your definition of Euclidean domain? See en.wikipedia.org/wiki/Euclidean_domain#Definition for a discussion. – lhf Oct 3 '11 at 14:21
My Euclidean domain is arbitrary, so there is no need to define the function d. I cannot say d satisfies d(xy)<=d(x)d(y). – Hassan Muhammad Oct 3 '11 at 14:31
How do you conclude that $d(u)\le d(1)$ from $u\mid 1$? I suggest you spell out your definition of Euclidean domain in the question. – lhf Oct 3 '11 at 14:34
Sorry, I meant $d(x) \le d(xy)$. – lhf Oct 3 '11 at 14:43
Saying d(x)<=d(xy) is the same as saying d(x)<=d(y) whenever x divides y. What you suggest about spelling the definition is good, but if D is a Euclidean domain, then the definition follows trivially. – Hassan Muhammad Oct 3 '11 at 15:47

From the text of your question I assume that your $d$ function satisfies $d(x)\le d(z)$ if $x$ divides $z$. (BTW, this is the same as saying that $d(x)\le d(xy)$.)
Since $1$ divides every element, we have $d(1)\le d(x)$ for all $x$. If $u$ is a unit, then $u$ divides $1$ and so $d(u)\le d(1)$. This implies $d(u)=d(1)$.
Conversely, as you have remarked, $1=uq+r$ with $r=0$ or $d(r)< d(u)$. But if $d(u)=d(1)$, then if $r\ne0$ you'd get $d(r)<d(1)$, which contradicts $d(1)\le d(x)$ for all $x$. Hence $r=0$ and $u$ is a unit.