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

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?

• $a^2+b^2=(a+ib)(a-ib)$ – 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 $p=a^2+b^2$.

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 $p-1$.

• 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

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 $$\alpha|p_i$$ 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 $$\mathbb{Z}[i] \simeq p = N(a+bi) = a^2+b^2$$ 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 $$\mathbb{Z}$$.

• Could you explain the last line where you claim that $ab\neq 0$? Take $\alpha=0+pi$, then $a=0, b=p$, but $ab=0$ and we still have $\alpha\sim p$. – user549397 Jan 27 at 14:35
• @user549397 you're right it seems i made a mistake, indeed above they are pointing the conjecture is wrong, so let me correct it – Javier Jan 27 at 17:17