Question

Let R be a Euclidean domain such that $\mathit{\delta }\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{\le }\mathit{m}\mathit{a}\mathit{x}\mathbf{\{}\mathbf{\delta }\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{,}\mathbf{\delta }\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{\}}$ for all nonzero $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{R}$. Prove that q and r in the definition of Euclidean domain are unique.

Short answer

It is Proved that$q$and $r$in the definition of Euclidean domain are unique.

Step-by-step solution

Use the Definition of Euclidean domain

An integral domain R is said to be Euclidean domain if there is a function $\mathit{\delta }$from the non-zero elements of R to non-negative integers with these properties:

1. If $\mathbf{a}\mathbf{}\mathbf{and}\mathbf{}\mathbf{b}$ are non-negative elements of R, then $\mathit{\delta }\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{\le }\mathit{\delta }\mathbf{\left(}\mathbf{a}\mathbf{b}\mathbf{\right)}$.
2. If localid="1654669416494" $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{R}$and $\mathit{b}\mathbf{\ne }{\mathbf{0}}_{\mathbf{R}}$, then there exist $\mathit{q}\mathbf{,}\mathit{r}\mathbf{\in }\mathit{R}$such that $\mathit{a}\mathbf{=}\mathit{b}\mathit{q}\mathbf{+}\mathit{r}$and either $\mathit{r}\mathbf{=}{\mathbf{0}}_{\mathbf{R}}$or $\mathit{\delta }\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{\le }\mathit{\delta }\mathbf{\left(}\mathbf{b}\mathbf{\right)}$

Suppose that for $a,b\in R$with $b\ne 0$there are $q_1,q_2,r_1,r_2$such that $a=b{q}_1+r_1=b{q}_2+r_2$either $r_1=0$or $\delta \left(r_1\right)\le \delta \left(b\right)$and either $r_2=0$ or $\delta \left(r_2\right)\le \delta \left(b\right)$

We then have that

$(q_1-q_{2}b)+(r_1-r_2)=0$

Or equivalent that

$(q_1-q_{2}b)=(r_2-r_1)$

Note that if$r_1=r_2$ then the fact that$R$is an integral domain implies that $q_1=q_2$.

Use the method of contradiction

Suppose that $r_1\ne r_2$, then we would also have that $q_1\ne q_2$from the above equation, and so $q_1-q_2\ne 0$

Note first that for any $r\in R$with $r\ne 0$the condition (i) for an Euclidean domain implies that $\delta \left(r\right)\le \delta (-r)$and $\delta (-r)\le \delta (-(-r))=\delta \left(r\right)$, and

so that $\delta \left(r\right)=\delta (-r)$

Suppose first that $r_1=0\text{and}{r}_2\ne 0$, then

$\delta \left(b\right)\le \delta \left((q_1-q_2)b\right)=\delta \left(r_2\right)<\delta \left(b\right),$A contradiction.

Suppose Now that ,$r_1\ne 0\text{and}{r}_2=0$then

$\delta \left(b\right)\le \delta \left((q_1-q_2)b\right)=\delta \left(r_1\right)<\delta \left(b\right),$ A contradiction.

Suppose finally that $r_1\ne 0\text{and}{r}_2\ne 0$then

$\delta \left(b\right)\le \delta \left((q_1-q_2)b\right)=\delta (r_2-r_1)\le \max \{\delta \left(r_1\right),\delta \left(r_2\right)\}<\delta \left(b\right),$A contradiction.

Therefore, $r_1=r_2$and $q_1=q_2$

Hence$q$ and$r$ are unique.