Is it true that $a+bi$ is prime in $\mathbb{Z}[i]$ if and only if $a^2+b^2$ is prime in $\mathbb{Z}$? [closed]

Is it true that $a+bi$ is prime in $\mathbb{Z}[i]$ if and only if $a^2+b^2$ is prime in $\mathbb{Z}$?

How can I prove this? Can anybody help me please?

Solution 1:

This is not true when $b=0$ and $a$ is an ordinary prime of the form $4k+3$. And for the same reason it is not true if $a=0$ and $b$ is an ordinary prime of the form $4k+3$.

Added: If $a^2+b^2$ is an ordinary prime, then $a+bi$ is a Gaussian prime. For suppose that $a+bi=(s+ti)(u+vi)$. By a norm calculation we have $a^2+b^2=(s^2+t^2)(u^2+v^2)$. So since $a^2+b^2$ is prime, one of $s+ti$ or $u+vi$ is a unit.

For the other direction, suppose $a+bi$ is a Gaussian prime, where neither $a$ nor $b$ is equal to $0$. We show that $a^2+b^2$ is an ordinary prime. Proof would be easy if we assume standard results that characterize the Gaussian primes. So we try not to use much machinery.

If $a$ and $b$ are both odd, then $a+bi$ is divisible by $1+i$. Then $a+bi$ is an associate of $1+i$, and $a^2+b^2=2$.

So we may assume that $a$ and $b$ have opposite parities. In that case, $a+bi$ and $a-bi$ are relatively prime. For any common divisor $\delta$ must divide $2a$ and $2b$. Since $a$ and $b$ have opposite parity, any common divisor divides $a$ and $b$, so must have norm $1$ if $a+bi$ are prime.

Now suppose that $p$ is a prime that divides $a^2+b^2$. Then $p$ divides $(a+bi)(a-bi)$. Note that $p$ cannot be a Gaussian prime, else it would divide one of $a+bi$ or $a-bi$,

Let $\pi$ be a Gaussian prime that divides $p$. Then $\pi$ divides one of $a+bi$ or $a-bi$. So $\pi$ must be an associate of one of these, and the conjugate of $\pi$ is an associate of the other. Since $a+bi$ and $a-bi$ arer relatively prime, we conclude that $(a+bi)(a-bi)$ divides $p$, which forces $p=a^2+b^2$.

Solution 2:

Here is something that is true: $a+bi$ is prime in $\def\Z{\Bbb Z}\Z[i]$ implies that $(a+bi)\Z[i]\cap\Z=p\Z$ for some (ordinary) prime $p$ of $\Z$. This is because, (1) since $a+bi$ is irreducible and $\Z[i]$ is a Euclidean and therefore unique factorisation domain, $(a+bi)\Z[i]$ is a prime ideal of $\Z[i]$, so (2) its intersection with $\Z$ is a prime ideal of $\Z$ (as is always true for the intersection of a prime ideal and a subring), and (3) the intersection is not reduced to$~\{0\}$ because $(a+bi)(a-bi)=a^2+b^2\in\Z\setminus\{0\}$. (Here, like in the question, one does not have "if and only if": here the converse fails for $a+bi$ equal or associated to a prime number $p\not\equiv3\pmod4$, such as $p=2$ or $p=5$; then $(a+bi)\Z[i]\cap\Z=p\Z$, but $p$ and therefore $a+bi$ are composite in $\Z[i]$.)

The case $a+bi$ is prime in $\Z[i]$ splits into two subcases. By the above there exists a prime number $p$ and $z\in\Z[i]$ with $p=(a+bi)z$; then $p^2=N(p)=N(a+bi)N(z)$, and either $z$ is non-invertible, in which case $N(a+bi)=p$ and $z=a-bi$, or $z$ is invertible, in which case $a+bi\in\{p,ip,-p,-ip\}$ and it can be shown that $p\equiv3\pmod4$. Indeed, the irreducibility of $p$ in the UFD $\Z[i]$ means that $\Z[i]/p\Z[i]$ is an integral domain (it is a field), so the kernel $(X^2+1)$ of the ring morphism $(\Z/p\Z)[X]\to\Z[i]/p\Z[i]$ sending $X\mapsto i$ is a prime ideal, so $X^2+1\in(\Z/p\Z)[X]$ is irreducible, which excludes both $p=2$ (for which $X^2+1=(X+1)^2$) and $p\equiv1\pmod4$ (in which case $X^2+1=(X+a)(X-a)$ for some element $a$ of order $4$ in the cyclic group $(\Z/p\Z)^\times$ of order $p-1$.

Solution 3:

Let me add another argument that I believe is simpler. First I need to prove a little proposition.

Let $p \in \mathbb{N}$ be a prime and $N$ be the norm function $N(\alpha) = \alpha \overline{\alpha}$.
If $\alpha \in \mathbb{Z}[\sqrt{n}]$ is prime then $N(\alpha) = p,-p,p^2,-p^2$.
Furthermore, if $N(\alpha) = p^2,-p^2$ then $\alpha \sim p$ where $\sim$ is the associates relation on $\mathbb{Z}[\sqrt{n}]$.

Proof:

If $\alpha$ is prime in $\mathbb{Z}[\sqrt{n}]$, then $N(\alpha) \neq 0,1,-1$ this is because $\alpha$ is not a unit and the fact that $N(\alpha) = \alpha \overline{\alpha} \neq 0$ comes because we are working in an integrity domain so that $\alpha = 0 \lor \overline{\alpha} = 0$ but since $n$ is square-free (by definition) it is necessary that if $\alpha = a+bi$ then $a = b = 0$, therefore $\alpha = 0$ and cannot be prime.

So let $N(\alpha) = \prod p_i$ be its prime factorization over the integers. Then $\alpha\mid p_i$ since $\alpha$ is prime. Therefore, $\alpha \beta = p \implies N(\alpha) N(\beta) = p^2 \implies N(\alpha) \in \{p,-p,p^2,-p^2\}$ as we wanted. Furthermore, if $N(\alpha) = p,-p^2$ then $N(\beta) = 1,-1$ which means that $\beta$ is a unit and therefore $\beta \sim p$. $\tag*{$\blacksquare$}$

So now come to our problem over $\mathbb{Z}[i]$:

$\alpha = a+bi$ is prime in $\mathbb{Z}[i] \approx p = N(a+bi) = a^2+b^2$ is prime in $\mathbb{Z}$.

$\impliedby)$ Since $\mathbb{Z}[i]$ is a UFD, $\alpha$ is prime if and only if it is irreducible. But if, $\alpha = \beta \gamma$ is a proper factorization of $\alpha$ then $N(\alpha) = N(\beta) N(\gamma)$ and you would have a proper factorization of $p$. Contradiction.

$\implies)$ $\alpha$ prime implies that $N(\alpha) = p,-p,p^2,-p^2$. Since the norm is positive $N(\alpha) = p,p^2$. If $N(\alpha) = p$ then we are done. If $N(\alpha) = p^2$ then $\alpha \sim p \in \mathbb{N}$ prime. Recalling the units of $\mathbb{Z}[i]$ are $U(\mathbb{Z}[i]) = \{i,-i,1,-1\}$ and multiplying you will see that necessarily $\alpha \in \{p,-p,pi,-pi\}$.

However, you cannot rule out this last case. For instance, take $3$ in $\mathbb{Z}[i]$ which following a norm-argument can be shown to be prime, yet its norm is $p^2 = 9$ which is not prime in $\mathbb{Z}$.