Question

(a) Show that $\mathbf{1}\mathbf{+}\mathit{i}$is not a unit in $\mathrm{\mathbb{Z}}\left[\mathrm{i}\right]$.

(b) Show that 2 is not irreducible in$\mathbf{\mathbb{Z}}\mathbf{\left[}\mathbf{i}\mathbf{\right]}$.

Short answer

1. It is proved that$1+i$ is not a unit in$\mathrm{\mathbb{Z}}\left[\mathrm{i}\right]$ .
2. It is proved that 2 is not irreducible in$\mathrm{\mathbb{Z}}\left[\mathrm{i}\right]$ .

Step-by-step solution

Show that 1+i is not a unit in ℤ[i]

Now,$\mathbb{Z}\left[i\right]$ is a set which is defined as $\mathbb{Z}\left[i\right]=\{a+ib/a,b\in \mathbb{Z}\}$

Now, let $u=1+i$ be the unit of $\mathbb{Z}\left[i\right]$

So, by definition $u^{-1}$exist

$u^{-1}={(1+i)}^{-1}=\frac{1}{2}-\frac{i}{2}\notin \mathbb{Z}\left[i\right]$

Which is contradiction.

Therefore, our assumption is false.

Hence $u=1+i$is not a unit of $\mathbb{Z}\left[i\right]$.

Show that 2 is not an irreducible in ℤ[i]

By using definition of irreducible element, we can say that,A nonzero element $\mathit{p}\mathbf{\in }\mathit{R}$ is said to be irreducible provided that p is not a unit and the only divisor of p are its associates and the units of R.

Suppose $p=2$is irreducible in $\mathbb{Z}\left[i\right]$

Now, $p=2=(1+i)(1-i)$

Here, clearly, 2 is divisible by $(1+i)$and $(1-i)$which is not a unit of $\mathbb{Z}\left[i\right]$.