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So far, I know if $p$ is a rational prime, then
$(1)$ if $p\equiv 3\mod4$, then $p$ is prime in $\mathbb{Z}[i]$.
$(2)$ If $p\equiv1\mod4$ then $p=π_1 π_2$ where $π_1 $ and $π_2$ are conjugate, Then $π_1 $ and $π_2$ are primes in $\mathbb{Z}[i]$.
$(3)$ $2=(1+i)(1-i)$, then $(1+i)$and$(1-i)$ are primes in $\mathbb{Z}[i]$.

What's are all the prime elements in Gaussian integers $\mathbb{Z}[i]$? For example, $-3$ are prime in $\mathbb{Z}[i]$, but not in the above $3$ cases.

Yeyeye
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  • $-3$ is associated to $3$, $-3$ is also a prime element in $\mathbb{Z}$. Generally, for every prime $\pi$ and every unit $\varepsilon$, $\varepsilon\cdot \pi$ is also a prime, associated to $\pi$. And the characterisation only holds if you restrict to positive rational primes. – Daniel Fischer Aug 03 '13 at 14:20
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    In additoin to @DanielFischer's remark, $1+i$ and $1-i$ are also associated, so in this special case the ideal $(2)$ is the square of a prime ideal. – Hagen von Eitzen Aug 03 '13 at 14:22
  • Further to @HagenvonEitzen's comment - explicitly $(1+i)=i(1-i)$ – Mark Bennet Aug 03 '13 at 14:42

1 Answers1

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Instead of prime elements one should rather talk about prime ideals. Your list then becomes

  1. $(p)$ is a prime ideal if $p\equiv 3\pmod 4$ is a rational prime
  2. $(p)=(\pi_1)(\pi_2)$ with conjugate (and distinct) prime ideals $(\pi_1)$ and $(\pi_2)$ if $p\equiv 1\pmod 4$
  3. $(2)=(1+i)^2$ is the square of a prime ideal.

No other ideals in $\mathbb Z[i]$ are prime. Since $\mathbb Z[i]$ is a principal ideal domain, we may call any generator of a prime ideal a prime element, and such generators are detemined only up to a unit, the units in $\mathbb Z[i]$ being $\{1,-1,i,-i\}$. So $(3)=(3i)=(-3)=(-3i)$, i.e. the primes $3, 3i, -3, -3i$ are associated primes. On the other hand $$\begin{align}&(3+4i)=(-4+3i)=(-3-4i)=(4-3i)\\\ne&(3-4i)=(4+3i)=(-3+4i)=(-4-3i),\end{align}$$ so we obtain two times four associated primes dividing $5$. Finally $(1+i)=(1-i)=(-1+i)=(-1-i)$, so we get four associated primes above $2$.

Hagen von Eitzen
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    I was wondering, how does this prove that no other ideals in $Z[i]$ are prime? Or is this just something we are taken as given? How would one prove that these are all the primes in $Z[i]$? – gowrath Feb 01 '17 at 01:44
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    I know it has been a lot time but I wonder the same question. Any answers? – Ninja Oct 20 '17 at 10:46
  • @gowrath: For a prime $z\in\mathbb Z[\mathrm i]$, we have $z\overline z\in\mathbb Z$, and $\overline z$ is also prime in $\mathbb Z[\mathrm i]$. Thus (by unique factorization in $\mathbb Z[i]$) $z\overline z$ has at most two prime factors in $\mathbb Z$. If it is prime, that's one of the listed cases. If it has two prime factors, these must be associates of $z$ and $\overline z$ (again by unique factorization in $\mathbb Z[i]$), so this is also one of the listed cases. See also this question. – joriki Mar 09 '20 at 09:36
  • @Ninja: see the above comment. – joriki Mar 09 '20 at 09:37