All irreducible elements in Gaussian integers

Question

I have shown that for $q \in \mathbb Z[i]$, if $N(q)=p$ ($p$ prime), or $N(q)=p^2$ ($p$ prime, $p \equiv 3 \pmod 4)$, then $q$ is irreducible ($(N(q)$ denotes the norm of $q$). But how can I prove that these are the only irreducible elements in $\mathbb Z[i]$?

Answer 1

Let $\pi\in\mathbb Z[i]$ irreducible. As you know $N(\pi)\ne 1$, so $N(\pi)$ is a product of primes in $\mathbb Z_{>0}$.
If $N(\pi)$ is prime, you are done.
Suppose $N(\pi)$ is not prime.
If there is a prime $p\equiv 3\pmod 4$ such that $p\mid N(\pi)=\pi\bar{\pi}$ then $p\mid\pi$, so $\pi$ and $p$ are associates in $\mathbb Z[i]$ and therefore $N(\pi)=N(p)=p^2$. (Here I've used that any prime $p\equiv 3\pmod 4$ is prime in $\mathbb Z[i]$.)
Otherwise, $N(\pi)$ is a product of primes $p\equiv1\pmod 4$. Let $p$ be such a prime. We also know that $p=z\bar z$ where $z\in\mathbb Z[i]$ is a prime element. Then $z\mid \pi\bar{\pi}$ and therefore $\pi$ (or $\bar{\pi}$) and $z$ are associates, so $N(\pi)=N(z)=p$, a contradiction.

Answer 2

If $q \in \mathbb Z[i]$ is irreducible, it generates a maximal ideal, since we work in an euclidean domain. So $\mathbb Z[i]/(q)$ is a finite field, which is - as an abelian group - generated by at most two elements, so it is isomorphic to $\mathbb F_p$ or $\mathbb F_{p^2}$ for some prime number $p$. This shows $N(q)=p$ or $N(q)=p^2$.

In the latter case we have to show $p \equiv 3 \mod 4$. To that account, we note that we have $p \in (q)$, hence $q | p$ and thus $q = p$ (up to a unit) due to $N(q)=p^2=N(p)$. So $p$ is prime in $\mathbb Z[i]$ and I think you already know that this implies $p \equiv 3 \pmod 4$.

Answer 3

Here is a more pedestrian step-by-step answer. I will only assume the following facts:

1. An Euclidian domain is a principal domain.

2. Principal domains are unique factorization domains.

3. Every irreducible element is prime in factorization domains.

With this in mind let us begin our quest:

In an euclidian domain, $\alpha$ is invertible if and only if $N(\alpha)=N(1)$

$(\Rightarrow )$ $N(1)\leq N(1\cdot \alpha)$ and $N(\alpha)\leq N(\alpha\alpha^{-1})=N(1)$

$(\Leftarrow)$ $1=\alpha q+r$ and if $r\not=0$, $N(r)<N(\alpha)$ and we must have $N(1)\leq N(r)<N(\alpha)$, which is absurd. $\square$

If $u$ is a unit and $i$ is irreducible, $ui$ is irreducible.

$ N(1)<N(i)\leq N(ui)$ so $ui$ is non invertible. If $ui=ab$ with both $a$ and $b$ non invertible, $i=(u^{-1}a) b$. A similar computation shows $(u^{-1}a)$ is non invertible. Contradiction with the fact $i$ is irreducible. $\square$

From now on, let us take $D=\mathbb{Z}[i]$ where $N(\alpha_1 \alpha_2)=N(\alpha_1)N(\alpha_2)$.

Let $p$ be a prime with $p \equiv 3 \mod{4}$. $p$ is irreducible in $\mathbb{Z}[i]$.

It can be proven that if $p|a^2+b^2$, we must have $p|a$ and $p|b$.

If $p$ is reducible $p^2=N(p)=N(ab)=(a^2_1+a^2_2)(b^2_1+b^2_2)$ where both of them are non invertible. This means, $1<a^2_1+a^2_2=p$. If $p\equiv 3 \mod(4)$, $p|a_1$ and $p|a_2$ so $p^2| a^2_1+a^2_2=p$. This is absurd. So one of them should have been invertible and $p$ is irreducible. $\square$

Let $p$ be a prime with $p \equiv 1 \mod{4}$. $p$ is reducible in $\mathbb{Z}[i]$.

Another number theory digression teaches us that for some integers $p=a^2+b^2$. Therefore $p=(a+bi)(a-bi)$ and it is not ireducible because $N(a\pm bi)=p\not=1$.$\square$

If $N(\alpha)=p$, we must have $\alpha$ is irreducible.

Indeed, if $ab=\alpha$ we must have $N(a)N(b)=p$ and because $N(k)$ is a positive integer, we must have one of them being one, say $N(a)=1=N(1)$, so $a$ is invertible and $\alpha$ is irreducible. $\square$

If $N(\alpha)=\prod_{i\geq 1} p_i$ then $\alpha$ is irreducible if and only if $N(\alpha)=2, \: p_i \text{or}\: p_j^2$ where $p_j \equiv 3 \mod(4)$ and $p_i \equiv 1 \mod(4)$.

$(\Rightarrow)$ Suppouse $\alpha$ is irreducible, we know it is prime.

As $\alpha \overline{\alpha}=\prod_{i>1} p_i $, we must have $\alpha |\prod_{i>1} p_i= ((1+i)(1-i))^k \prod_i p_i \prod_j (a_j-ib_j)(a_j+ib_j)$. Here we have writen $2=(1+i)(1-i)$ as the product of irreducibles and $ p_j \equiv 1 \mod( 4)$ as $p_j=a_j^2+b_j^2=(a_j-ib_j)(a_j+ib_j)$ (product of irreducibles). Let $f_i$ denote a general irreducible.

Because $\alpha$ is prime $\alpha| f_i$ and $\alpha =u f_i $. Because we are in a unique factorization domain $u $ must be invertible. Hence $N(\alpha)=N(u)N(f_i)=N(f_i)=2, p_i^2 \text{or} p_j$.

$(\Leftarrow)$ If $N(\alpha)=p$ we are done. Otherwise $N(\alpha)=p_j^2$. Because $p_j^2=\alpha \overline{\alpha}$ and $p_j$ is irreducible (hence prime), $p_j|\alpha$ or $p_j| \overline{\alpha}\Rightarrow p_j=\overline{p_j}| \alpha $. Hence $p_j|\alpha$ and $p_jk=\alpha$. If $k$ were non invertible, $N(k)>1$ which is absurd. Thus, $k$ is invertible and $\alpha$ is related to an irreducible and must be irreducible itself. $\square$