# Primes in Gaussian Integers

Let $p$ be a rational prime. It is is well known that if $p\equiv 3\;\;mod\;4$, then $p$ is inert in the ring of gaussian integers $G$, that is, $p$ is a gaussian prime. If $p\equiv 1\;mod\;4$ then $p$ is decomposed in $G$, that is, $p=\pi_1\pi_2$ where $\pi_1$ and $pi_2$ are gaussian primes not associated. The rational prime $2$ ramifies in $G$, that is $2=u\pi^2$, where $u$ is a unit in $G$ and $\pi$ a prime in $G$.

where can I find a proof of this fact? I want a direct proof, not a proof for the quadratic integers and then deduce this as a particular case.

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There are several places where you can find a direct proof of this. For instance, you can find it in the first 4 pages of Jurgen Neukirch's Algebraic Number Theory about the Gaussian integers. Also, LeVeque's Elementary Theory of Numbers has a short chapter dedicated to the Gaussian integers, where he proves this fact (see section 6.5).

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Well, hopefully a proof right here will suffice instead of some resource. This is a rather nice proof of the case of Gaussian Integers, but it fails to generalize for the other quadratic integer domains without putting in more work at least.

Define the norm $N(a+bi) = a^2 + b^2$. It is straightfoward to check it is multiplicative. Now, consider a prime $p \equiv 3 \pmod{4}$. Suppose it is not inert, i.e. it factors as $p = \alpha \beta$ for some $\alpha, \beta \in \mathbb{Z}[i]$ and neither of them are units.. Then it is easy to see $N(\alpha) = N(\beta) = p$. However, the equation $a^2 + b^2 = p$ has no solutions modulo $4$, therefore we have reached a contradiction so primes $3 \pmod{4}$ remain inert.

Now to prove primes $1 \pmod{4}$ split. It suffices to show $p = a^2 + b^2$ has a solution where $a,b$ are integers. Let $z$ denote the least value of $\sqrt{-1} \pmod{p}$. Now define $\mathcal L = \{(a,b) \in \mathbb{Z}^2 | a \equiv zb \pmod{p}\}$. It is straightfoward to check $\mathcal L$ is a lattice whose fundamental parallelogram has area $p$. Now by Minkowski's theorem one has $\mathcal L$ contains a nontrivial lattice point inside the circle $x^2 + y^2 < 2p$. Call this point $(a,b)$. But then $a^2 + b^2 \equiv 0 \pmod{p}$ based on the definition of the lattice, thus it must be $a^2 + b^2 = p$. But then $p = (a+bi)(a-bi)$, proving it factors so we are done.

Proving $2$ ramifies is trivial, since it's just $2 = -i(1+i)^2$.

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