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I am trying to prove the following:

Let $R=\mathbb{Z}[i]$, $a, b \in \mathbb{Z}, ab \neq 0$. Show that $a+bi$ is an irreducible element in $R$ if and only if $a^2 + b^2 ∈ \mathbb{Z}$ is prime.

Since $R$ is a Euclidean domain, irreducible elements are prime. I think this is relevant, but am stuck on where to go from here.

user26857
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  • One direction is easy. Suppose a+bi=(c+di)(e+fi). Take conjugates. Then a-bi=(c-di)(e-fi). Multiply these equations. Then a^2+b^2=(c^2+d^2)(e^2+f^2). – P. Lawrence Mar 09 '20 at 04:26
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    Note that $ab\ne0$ is necessary, since $3$ is prime in $\mathbb Z[i]$. – joriki Mar 09 '20 at 06:15

1 Answers1

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If $a+b\mathrm i$ is prime in $\mathbb Z[\mathrm i]$, then $a^2+b^2=(a+b\mathrm i)(a-b\mathrm i)$ has only two prime factors in $\mathbb Z[\mathrm i]$, and thus (by unique factorization in $\mathbb Z[\mathrm i]$) at most two prime factors in $\mathbb Z$. If it is prime in $\mathbb Z$, we are done. Otherwise (again by unique factorization in $\mathbb Z[\mathrm i]$) its two prime factors in $\mathbb Z$ must be associates of its two prime factors $a+b\mathrm i$ and $a-b\mathrm i$ in $\mathbb Z[i]$. Thus $a+b\mathrm i$ is an associate of an element of $\mathbb Z$, so $ab=0$.

Conversely, if $z$ is not prime in $\mathbb Z[\mathrm i]$, so that $z=rs$ with $r$ and $s$ not units, then $N(z)=N(r)N(s)$ (where $N(a+b\mathrm i)=a^2+b^2$ is the norm on $\mathbb Z[i]$), and $N(r)$ and $N(s)$ are not units, so $N(z)$ is not prime in $\mathbb Z$.

joriki
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