Question

Prove or disprove: Let R be a Euclidean domain; Then$\mathit{I}\mathbf{=}\mathbf{\{}\mathbf{a}\mathbf{\in }\mathbf{R}\mathbf{/}\mathbf{\delta }\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{>}\mathbf{\delta }\mathbf{\left(}{\mathbf{1}}_{\mathbf{R}}\mathbf{\right)}\mathbf{\}}$ is an ideal in R.

Short answer

It is proved that $I=\{a\in R/\delta \left(a\right)>\delta \left(1_R\right)\}$is an ideal in .

Step-by-step solution

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 arole="math" localid="1654663234356" $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{R}$ 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$\mathit{\delta }\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{\le }\mathit{\delta }\mathbf{\left(}\mathbf{b}\mathbf{\right)}$.

Definition of Ideal

Let R be a ring and S is a subring of R is said to be ideal If

1. S is subgroup of R.
2. If $\mathrm{r}\in \mathrm{R}$and $s\in S$, then $rs\in S$and $sr\in S$

Now, we have$I=\{a\in R/\delta \left(a\right)>\delta \left(1_R\right)\}$

Now, take $b\in R$ and $a\in I$

Now, Therefore,

$\begin{array}{l}\Rightarrow \delta \left(ab\right)>\delta \left(1_R\right)\\ \Rightarrow \delta \left(a\right)\le \delta \left(ab\right)>\delta \left(1_R\right)[\because a,b\in R,R\text{isEuclideandomain}]\\ \Rightarrow \delta \left(a\right)>\delta \left(1_R\right)\end{array}$

Also,

$\begin{array}{l}\Rightarrow \delta \left(ba\right)>\delta \left(1_R\right)\\ \Rightarrow \delta \left(ab\right)>\delta \left(1_R\right)[\because R\text{isacommutativering}]\\ \Rightarrow \delta \left(a\right)\le \delta \left(ab\right)>\delta \left(1_R\right)[\because a,b\in R,R\text{isEuclideandomain}]\\ \Rightarrow \delta \left(a\right)>\delta \left(1_R\right)\end{array}$

Therefore,

$ab\in I$ and $ba\in I$

Therefore$I$ is an ideal of $R$