Irreducible element in Gauss Ring

Units and Prime Norm

Since $\mathbb{Z}[i]$ is a Euclidean domain with norm $N(a+bi)=a^2+b^2$, $\alpha$ is a unit in $\mathbb{Z}[i]$ iff $N(\alpha )=\pm 1$. Therefore, the units in $\mathbb{Z}[i]$ are $\{\pm 1,\pm i\}$.

If $\beta =x+yi$ and $x^2+y^2$ is a prime, then $\beta$ is irreducible. Otherwise $\beta =(a+bi)(c+di)$ with non-unit elements $a+bi$, $c+di$, it follows $(a^2+b^2)(c^2+d^2)$ is a prime, which is impossible.

Irreducible Integers

If $p$ is a prime in $\mathbb{Z}$ and it is reducible in $\mathbb{Z}[i]$, then $p=(a+bi)(c+di)$ and so $p^2=(a^2+b^2)(c^2+d^2)$. It follows that $a^2+b^2=c^2+d^2=p$ as they are non-units. Therefore, $p$ is a sum of two squares of integers. Conversely, if $p=a^2+b^2$, then $p=(a+bi)(a-bi)$ and so $p$ is reducible.

Recall that when $p\equiv 1(\mathrm{mod}4)$, $p$ can be written as $p=a^2+b^2$ and $2=1^2+1^2$. Hence, $p\equiv 1(\mathrm{mod}4)$ and $p=2$ are reducible. On the other hand, if $p=a^2+b^2$, then $p\equiv 0,1,\text{mbox}or2(\mathrm{mod}4)$. Thus $p\equiv 3(\mathrm{mod}4)$ is irreducible.

Irreducible $a+bi$

Assume that irreducible $a+bi$ has norm $p_1\cdots p_k$ where $p_j$ are primes, then $a+bi$ must divide one of $p_j$ because $a+bi$ is irreducible and is prime. Hence $p_j\equiv 1(\mathrm{mod}4)$ and the factorization of $p_j$ is $(m+ni)(m-ni)$, that is, $a+bi=m\pm ni$.

Up to associates, all the irreducible elements in $\mathbb{Z}[i]$ are as follows:

• The element $1+i$ (of norm 2).
• The primes $p\in \mathbb{Z}$ congruent to 3 modulo 4 (of norm $p^2$ ).
• The distinct irreducible factors $a+bi$ and $a-bi$ (each of norm p) of $p=a^2+b^2$ where $p\in \mathbb{Z}$ is congruent to 1 modulo 4 .