Number theory proof regarding norms

Question

How would you prove that if $x$ is a prime in $ℤ[i] \Longleftrightarrow$ $N(x)$ is a prime in $ℤ$

$N(x)$ represents the norm of $x$.

Answer 1

We just prove $\Rightarrow$, since $\Leftarrow$ is trivial. Since $\mathbb{Z}[i]$ is an Euclidean domain, irreducible elements and prime elements are the same thing. Assume that $p=a+bi$, with $ab\neq 0$, is a prime in $\mathbb{Z}[i]$, while its norm splits in $\mathbb{Z}$, $a^2+b^2 = d_1\cdot d_2$, with $d_1,d_2>1$. Without loss of generality we can assume $a,b\in\mathbb{N}_0$ and $\gcd(a,b)=1$. Moreover, we can assume that $a$ and $b$ are not both odd, since otherwise: $$ p = a+bi = \left(\frac{a-b}{2}+\frac{a+b}{2}i\right)\cdot(1-i) $$ cannot be a prime in $\mathbb{Z}[i]$. Every integer prime $q$ that divides $a^2+b^2$ is of the form $4k+1$, since $a^2+b^2$ is odd and $$ a^2+b^2\equiv 0\pmod{q} $$ implies that $-1$ is a quadratic residue $\pmod{q}$. So every prime divisor of $d_1$ and $d_2$ has the same property and can be represented as a sum of two squares. Since the set of integers that equal the sum of two squares is a semigroup, due to the fact that the Lagrange identity: $$ (c^2+d^2)(e^2+f^2) = (cd-fe)^2 + (cf+de)^2 $$ follows from $N(z\cdot w)=N(z)\cdot N(w)$ in $\mathbb{Z}[i]$, we have that both $d_1$ and $d_2$ can be represented as a "primitive" sum of two squares, say: $$ d_1 = u_1^2+v_1^2,\quad d_2=u_2^2+v_2^2,$$ with $u_1,v_1,u_2,v_2\neq 0,\gcd(u_1,v_1)=\gcd(u_2,v_2)=1$. However, this implies that $d_1$ splits as $(u_1+iv_1)(u_1-iv_1)$ and $d_2$ splits as $(u_2+iv_2)(u_2-iv_2)$ over $\mathbb{Z}[i]$, hence $p$ splits too, since: $$ N(p) = N((u_1+i v_1)(u_2+i v_2)) $$ implies $p= U\cdot (u_1+i v_1)(u_2+i v_2)$ with $U\in\{\pm 1,\pm i\}$, contradiction.