Let $R$ be an Euclidean domain with valuation $\delta$. Then we know that for each $a,b\in R$ with $b\neq0$, there exist $q,r\in R$ such that $a=qb+r$, where either $r=0$ or $\delta(r)<\delta(b)$. Now I need to prove that these elements $q,r$ are unique if and only if $\delta(a+b)\leq\max\{\delta(a),\delta(b)\}$. I can prove the forward direction but failed to prove the other direction. Could someone please help? Thanks.
Asked
Active
Viewed 243 times
2
-
See also this question. – Dietrich Burde Jun 05 '17 at 15:43
-
1It doesn't seem to help. – Janitha357 Jun 05 '17 at 15:55
1 Answers
3
Maybe in your definition of valuation, $\delta(a)\leq \delta(ab)$ for all $a,b\neq 0$. Suppose this is the case. Let $a=q_1b+r_1$ and $a=q_2b+r_2$. Then $(q_1-q_2)b=r_2-r_1$. Assume $q_1\neq q_2$ and $r_1\neq r_2$.
If $r_1=0$ and $r_2\neq 0$, then $\delta(b)\leq \delta((q_1-q_2)b)=\delta(r_2)<\delta(b)$.
If $r_2=0$ and $r_1\neq 0$, then $\delta(b)\leq \delta((q_1-q_2)b)=\delta(-r_1)=\delta(r_1)<\delta(b)$.
If $r_1\neq 0$ and $r_2\neq 0$, then $\delta(b)\leq \delta((q_1-q_2)b)=\delta(r_2-r_1)\leq \max\{\delta(r_1),\delta(r_2)\}<\delta(b)$.
All of them are contradictions.
KC.
- 408