What are the factors of (2+i) in Z[i]?

Question

The complex number $2+i$ factors as $i\cdot (1-2i)$ and $(-i)\cdot (1-2i)$. But those factorizations seem trivial. Are there any other ways to factor 2+i within the Z[i]?

Answer 1

The element $2 + i$ is a prime element in $\Bbb Z[i]$, i.e. any factorisation $2 + i = a \cdot b$ will imply that one of $a$ and $b$ is a unit in $\Bbb Z[i]$ (i.e. $\pm1, \pm i$).

To see this, simply take (squared) norm of the identity $2 + i = a \cdot b$. We get $5 = N(a)N(b)$. Since $5$ is a prime number in $\Bbb Z$, this implies that one of $N(a)$ and $N(b)$ is equal to $1$, hence one of $a$ and $b$ is a unit.

Answer 2

The ring $\mathbb{Z}[i]$ is a principal ideal domain. In particular it is a unique factorisation domain. Thus all elements of $\mathbb{Z}[i]$ can be factored uniquely up to multiplication by a unit. To see that these are the only factorisations notice that the unit group of $\mathbb{Z}[i]$ is generated by $i$ and $2 + i$ is irreducible.