Question

(b) Show that the set $\mathit{T}\mathbf{=}\left\{\left.\left(\mathit{k},\mathit{k}\right)\right|\mathit{k}\in \mathbb{Z}\right\}$ is not an ideal in $\mathbf{\mathbb{Z}}\mathbf{\times }\mathbf{\mathbb{Z}}$.

Short answer

Tis not ideal, it is proved.

Step-by-step solution

Determine that T={k,kk∈ℤ} is not an ideal in ℤ×ℤ

From (a) part, consider that $\left(a,b\right)\in \mathrm{\mathbb{Z}}\times \mathrm{\mathbb{Z}}$; then,

$\left(k_1,0\right)\left(a,b\right)=\left(k_{1}a,0b\right)=\left(k_{1}a,0\right)\in I$.

As $k_{1}a\in \mathrm{\mathbb{Z}}$, $\left(a,b\right)\left(k_1,0\right)=\left(a{k}_1,0\right)\in I$.

Also, $a{k}_1\in \mathrm{\mathbb{Z}}$. Thus, condition (ii) of Theorem 6.1 is satisfied.

Thus, by Theorem 6.1, it is concluded is an ideal.

Determine that T={k,kk∈ℤ} is not an ideal in ℤ×ℤ

Consider$\left(1,1\right)\in T$and $\left(1,0\right)\in \mathrm{\mathbb{Z}}\times \mathrm{\mathbb{Z}}$; then,

$\left(1,1\right)\left(1,0\right)=\left(1·1,1·0\right)=\left(1,0\right)\notin T$.

This means that it is not true for all $r\in \mathrm{\mathbb{Z}}\times \mathrm{\mathbb{Z}}$and $a\in T$.

Also, $ar\in T$, and$ra\in T$ . Therefore, T is not ideal.