Question

Show that the set $I=\left\{\left.\left(k,0\right)\right|k\in \mathrm{\mathbb{Z}}\right\}$is an ideal in the ring$\mathrm{\mathbb{Z}}\times \mathrm{\mathbb{Z}}$$\mathrm{\mathbb{Z}}\times \mathrm{\mathbb{Z}}$.

Short answer

It is proved$I$ is an ideal.

Step-by-step solution

Use Theorem 6.1

Using Theorem 6.1,

Consider $\left(0,0\right)\in I$; thus,$I$is non-empty.

Now, consider $\left(k_1,0\right),\left(k_2,0\right)\in \mathrm{\mathbb{Z}}\times \mathrm{\mathbb{Z}}$. Then,

$\left(k_1,0\right)-\left(k_2,0\right)=\left(k_1-k_2,0\right)\in I$

As $k_1-k_2\in \mathrm{\mathbb{Z}}$, condition (i) of Theorem 6.1 is satisfied.

Determine set I=k,0k∈ℤ as an ideal in the ring ℤ×ℤ

Now, 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 $I$is an ideal.