Question

In which of these domains is 5 an irreducible element.

1. $\mathbf{\mathbb{Z}}$
2. $\mathbf{\mathbb{Z}}\left[i\right]$
3. $\mathbf{\mathbb{Z}}\left[\sqrt{-2}\right]$
   

Short answer

5 is irreducible in$\mathrm{\mathbb{Z}}$ and$\mathbb{Z}\left[\sqrt{-2}\right]$ .

Step-by-step solution

(a)Step 1: Prove that 5 is irreducible in  ℤ

Since 5 is a prime number, therefore it cannot be factorized further.

Hence, 5 is irreducible in $\mathrm{\mathbb{Z}}$.

(b)Step 2: Show that 5 is reducible in ℤ[i]

Since$5=(2+i)(2-i)$ where $(2+\mathrm{i}),(2-\mathrm{i})\in \mathrm{\mathbb{Z}}\left[\mathrm{i}\right]$

Therefore, 5 is reducible in $\mathrm{\mathbb{Z}}\left[\mathrm{i}\right]$.

(c)Step 3: Prove that 5 is irreducible in  ℤ[-2]

Here, -2 is a square free integer.

$5\in \mathrm{\mathbb{Z}}\left[\sqrt{-2}\right]$and $N\left(5\right)=5$is a prime in $\mathrm{\mathbb{Z}}$.

Thus, by Corollary 10.22, 5 is irreducible in $\mathbb{Z}\left[\sqrt{-2}\right]$.

Conclusion

Thus, it can be concluded that 5 is irreducible in $\mathrm{\mathbb{Z}}$and $\mathbb{Z}\left[\sqrt{-2}\right]$and reducible in$\mathbb{Z}\left[i\right]$ .