Categorization of Gaussian primes

Question

I'm reading Neukirch's text on Algebraic Number Theory, and we are proving that any Gaussian primes are in the form of $1 + i$ or $a + bi$ where $a^2 + b^2 = $a prime $p$ or $p = 3 \mod 4$.

I understand the steps to show that these are indeed Gaussian primes, but I'm not very clear on his proof on any Gaussian prime take one of the forms. The argument was that for any prime $\pi \in \mathbb{Z}[i]$, $$ N(\pi) = \pi \cdot \overline{\pi} = p_1 \cdot p_r.$$ Then we claim that $\pi | p_i$ for some $i$. Why is this claim true? How do we know that $\pi$ must divide a rational prime?

Answer 1

The definition of $\pi$ being prime is that if $\pi$ divides $ab$ then $\pi$ divides $a$ or $\pi$ divides $b$. So if $\pi$ divides $p_1\cdot\cdot \cdot p_r \in \Bbb{Z}$, it must divide one of the $p_i$. I would say to use induction of $r>2$ but in the instant case you will only ever have $r=1$ or $r=2$.

Answer 2

Does Neukirch explain what $\overline \pi$ is? If $\pi = 1 - i$, then $\overline \pi = 1 + i$, and vice-versa. And since $(1 - i)(1 + i) = 2$, it follows that $$\frac{2}{1 - i} = 1 + i$$ and $$\frac{2}{1 + i} = 1 - i.$$

Along similar lines you can prove it for purely real primes that are composite in $\mathbb{Z}[i]$.

Answer 3

I'm reading Neukirch's text on Algebraic Number Theory, and we are proving that any Gaussian primes are in the form of $1 + i$ or $a + bi$ where $a^2 + b^2$ is a prime $p$ or $p \equiv 3 \bmod 4$.

I understand the steps to show that these are indeed Gaussian primes, but I'm not very clear on his proof on any Gaussian prime take one of the forms.

Assuming you have quoted Neukirch exactly, he's missing some primes of the above forms multiplied by units other than 1. Namely:

• $(1 + i)i = -1 + i$, which is a prime of norm 2,
• $(1 + i)(-1) = -1 - i$, which is another prime of norm 2, and...
• $(1 + i)(-i) = 1 - i$, which is yet another prime of norm 2.

If $a$ and $b$ are both purely real nonzero integers, then $a^2$ and $b^2$ are both purely real positive integers. So we could say that $-a + bi$, $-a - bi$ and $a - bi$ are all implied. If $a$ and $b$ are of opposite parity, then $a^2 + b^2 \equiv 1 \bmod 4$ and $a^2 + b^2$ is purely real and positive.

This means that if $\pm a \pm bi$ are primes, then they all have a norm that is a prime among the purely real integers, a positive prime that is 1 more than a multiple of 4. But of course that prime is not prime among the Gaussian integers, but, well, um... composite (snobbery is the only reason I can think of to not use that word in a unique factorization domain).

So if $p$ is a positive prime among the purely real integers, and 1 less than a multiple of 4, then there's no solution to $a^2 + b^2 = p$, and so $p$ is also a prime among the Gaussian integers. And we see that $pi$, $-p$ and $-pi$ are also primes. e.g., 7, $7i$, $-7$, $-7i$.

The argument was that for any prime $\pi \in \mathbb{Z}[i]$, $$ N(\pi) = \pi \cdot \overline{\pi} = p_1 \cdot p_r.$$ Then we claim that $\pi \mid p_i$ for some $i$. Why is this claim true? How do we know that $\pi$ must divide a rational prime?

Because, like the others have already said, $\pi \overline \pi$ is a purely real number, 1 more than a multiple of 4, that is prime among the purely real integers. For example, $(2 - i)(2 + i) = 5$. Then obviously both $2 - i$ and $2 + i$ are Gaussian divisors of 5.

I think you also need to look at Gaussian integers with even norms. If $a$ and $b$ are both even, then obviously $a^2 + b^2$ is a multiple of 4 and $a + bi$ is divisible by 2. In instead $a$ and $b$ are both odd, then the norm is divisible by 2 but not by 4, and so much less obviously $a + bi$ is divisible by $1 + i$ but not by 2.