# $a+bi$ is prime in $\mathbb{Z}[i]$ if and only if $a^2+b^2$ is prime in $\mathbb{Z}$

$a+bi$ 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?

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$a^2+b^2=(a+ib)(a-ib)$ – Mathematician Apr 21 '13 at 4:35
this is obvious.but I could not understand how should I use this result – habuji Apr 21 '13 at 4:40
False. Three is prime in $\mathbb{Z}[i]$, yet $3^2+0^2=9$ is not prime in $\mathbb{Z}$. – Jyrki Lahtonen Apr 21 '13 at 4:43
If the norm is prime, it can't be the product of Norms of two other numbers unless one is a unit. However the converse isn't always true. – Macavity Apr 21 '13 at 4:57
you would need restrictions on your statements, such as both a, b are nonzero, as @Jyrki has shown the hole there. Also, $a^2+b^2$ can not be of the form $4k+3$ (again, as is 3). – Eleven-Eleven Apr 21 '13 at 5:01

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 $p=a^2+b^2$.

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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 $p-1$.

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The first paragraph does not make clear precisely what is a consequence of said UFD property (that prime ideals are preserved under contraction is true in any ring). – Math Gems Apr 21 '13 at 16:20
@MathGems: The "since ... UFD" in the second sentence was awkwardly placed very early to indicate that that it is needed to justify the immediately following "is a prime ideal". And you are right, pulling back a prime ideal through a ring morphism always gives a prime ideal; maybe that is why they are so useful. – Marc van Leeuwen Apr 21 '13 at 16:45
I surmised what you intended. But I fear that those beginning their studies might have more difficulty inferring the intended meaning. Whenever I encounter things like that I leave comments in the hope that the author might improve the exposition to eliminate ambiguities etc. Thankfully many folks do the same for me when I too do likewise. – Math Gems Apr 21 '13 at 17:13
@MathGems: Fine, thank you. Did I succeed in reducing the ambiguity, or if not what would be better? – Marc van Leeuwen Apr 21 '13 at 18:39
Yes, that reads much more clearly. Thanks and +1. – Math Gems Apr 21 '13 at 19:05