# Classifying prime ideals of $\mathbb{Z}[i]$

I am studying number fields and their rings of integer and I want to understand prime ideals of Gaussian integers $\mathbb{Z}[i]$, which is the ring of integers of $\mathbb{Q}[i]$. We have the following arguments for ramification primes:

In $\mathbb{Z}[i]$, there are three kinds, ramified, split and inert primes:

1. There is exactly one rational prime that ramifies, namely $2$, and the prime above it, say $\mathfrak p$ has norm $2$.
2. Each rational prime $p\equiv 1\pmod 4$ splits into a product of two primes $pO_K=\mathfrak p_1\mathfrak p_2$ both of which have norm $p$.
3. Finally the rational primes $p\equiv 3\pmod 4$ stay prime with norm $p^2$.

Using the arguments above, how can we classify prime ideals of $\mathbb{Z}[i]$, ?

• $(2) = (1+i)^2$. If $p \equiv 3 \bmod 4$ then $(p)$ is prime, otherwise (hardest case, quadratic reciprocity) then $(p) = (p,i-c)(p,i+c)$ where $c$ is a root of $x^2+1 \bmod p$ and since $\mathbb{Z}[i]$ is PID $(p) = (a+ib)(a-ib)$ Oct 19 '17 at 1:00
• They would be $(0)$; $(1+i)$; $(a+bi)$ where $a^2 + b^2$ is a prime integer $\equiv 1 \pmod{4}$ (can assume $a > |b| > 0$ to make representation unique); and $(p)$ where $p \equiv 3 \pmod{4}$ is a prime integer. Oct 19 '17 at 1:03
• I still can not relate prime ideals with rational primes. I showed that each prime ideal $\mathfrak{p}$ of the ring is generated by an irreducible element $a+bi$, $N(\mathfrak{p}) =a^2 +b^2 = p^2$ for some rational prime $p$... but that's all and I'm stuck. Oct 20 '17 at 11:15

You also need the fact that $\textbf{Z}[i]$ is a principal ideal domain. This means that any number in this domain is in one or more principal ideals. Then you only need to check if the three kinds of numbers you've listed belong in only in ideals generated by the kinds of numbers you've listed.

Do you consider $\langle 0 \rangle$ to be a prime ideal? If you don't, then you already have the three classifications:

1. $\langle 1 + i \rangle$ has norm 2 and contains $1 - i$, $-1 - i$ and $-1 - i$. So $\langle 2 \rangle$ is a ramifying ideal just like in $\mathbb Z[\sqrt{-5}]$, $\mathbb Z[\sqrt{-13}]$, $\mathbb Z[\sqrt{-17}]$, etc.
2. $\langle a + bi \rangle$ where $a$ and $b$ are purely real integers. Then $(a - bi)(a + bi) = a^2 + b^2$ is a positive purely real prime congruent to 1 modulo 4.
3. $\langle p \rangle$ where $|p|$ is a purely real prime congruent to 3 modulo 4. These primes are "inert."

Verify that each Gaussian integer is contained in an ideal from one of these three categories (the integer 0 is contained in all ideals, so it doesn't hurt to not consider $\langle 0 \rangle$ prime).

Also note that most diagrams of the Gaussian primes show $-3$, $3i$ and $-3i$, to name just three contained in a Category 3 prime ideal, as inert primes. And rightly so, because those three numbers are also primes, and not just because people want the diagram to be pretty and symmetrical.

• My first question is what is purely real prime? Do you mean the "usual" primes? Second, how can I start proving that any $a+bi \in \mathbb{Z}[i]$ is contained one of the three classes? I need a little hint, unfortunately. Oct 20 '17 at 10:10
• By purely real I mean having nonzero real part and zero imaginary part, primes like 3, $-7$, $-31$, etc. I wouldn't use the term "classes" here because of the ideal class group, which is a different but related concept. I'm using "categories" until someone tells me there's a problem with that term as well. Oct 20 '17 at 18:23
• As for a hint, factor these purely real positive numbers into Gaussian integers: 2, 5, 13, 17, 29. Oct 20 '17 at 18:24
• I checked them and I saw that if the prime is equivalent to $1$ modulo $4$ then it splits -I can find non-zero integers so that the sum of their squares is equal to the prime-. Oct 22 '17 at 15:45
• Exactly. Since $(a - bi)(a + bi) = a^2 + b^2 = p \equiv 1 \bmod 4$, we see that those primes split and they belong in the ideals $\langle a - bi \rangle$ and $\langle a + bi \rangle$ (Category 2). The case of 2 is slightly different since $$\frac{1 + i}{1 - i} = i,$$ so $\langle 1 - i \rangle = \langle 1 + i \rangle$. Oct 22 '17 at 16:08