Any field is euclidean domain

Asked Aug 24 '18 at 14:19

Active Aug 24 '18 at 14:19

Viewed 1,606 times

In many sources I have seen that in order to prove that any field is euclidean domain we need to take function $d:F-\{0\}\mapsto \mathbb{Z}_{\geq 0}$ and mostly $d$ is taken to be the $d(x)=1$ for any nonzero $x\in F$.

But if I define the function $d$ in different way, namely $d(x)=0$ for any nonzero $x\in F$ then everything still be fine.

1) $d(xy)\geq d(y)$ holds for any nonzero $x,y\in F$.

2) If $a,b\in F$ where $b\neq 0$ then $\exists q,r\in R$ such that $a=bq+r$. In our case we can take $r=0, q=b^{-1}a$.

P.S. In general we can take $d(x)=k$ where $k$ is any nonnegative integer.

Am I right? Would be very grateful for help.

asked Aug 24 '18 at 14:19

RFZ

17,648
11
51
144

We generally don't think of fields as Euclidean domains; a field only has two ideals, $F$ and ${0}$. – Angina Seng Aug 24 '18 at 14:29
@LordSharktheUnknown, Thanks for reply! But I am aksing if I define the function $d$ in the following is it correct or not? – RFZ Aug 24 '18 at 14:32
1

@RFZ Yes, correct, see the duplicate. The answer is the same as you wrote under 2). – Dietrich Burde Aug 24 '18 at 14:42

Any field is euclidean domain

0 Answers0