What is the prime factorization $6+12i \in \Bbb Z[i]$?
I'm reading an abstract algebra sheet where they had the following problem, but I cannot find any examples on irreducible elements of $\Bbb Z[i]$. If I understand the problem correclty I'm being asked to express $6+12i$ as a product of irreducible elements of $\Bbb Z[i]$?
To start, it's clear that we can factor out a $6$, which is $2\times 3$. Thus:
$$6+12i=2\times 3\times \left(1+2i\right)$$
Of these three factors, $3$ is irreducible in $\mathbb{Z}[i]$, because it cannot be written as a sum of two squares. Also, $\left(1+2i\right)$ is irreducible, because its norm is $5$, which is prime in $\mathbb{Z}$.
On the other hand, since $2$ can be written as a sum of two squares ($2=1+1$), we can factor it in $\mathbb{Z}[i]$: $2=\left(1+i\right)\left(1-i\right)$.
Thus, a full factorization looks like:
$$6+12i = 3\times\left(1+i\right)\times\left(1-i\right)\times\left(1+2i\right)$$
Each of the four factors on the right is irreducble