Question about the prime elements of $\mathbb{Z}[i]$.

Question

I am learning about gaussian integers and I have a few questions about the following argumentation.

What are the prime elements in $\mathbb{Z}[i]$? We remember that only the units $+1,-1,+i,-i$ in $\mathbb{Z}[i]$ have Norm equal to one. We will find the prime elements by using the norm. Let $\pi \in \mathbb{Z}[i]$ be prime. Since $N(\pi)=\pi \overline{\pi} \in$ $\mathbb{Z}$, $\pi$ divides a primenumber $p \in \mathbb{Z}$. Let $\pi z = p$, then $N(\pi)N(z)=N(p)=p^2$. So we either got $N(\pi)=p$ or $\pi u=p$ for a unit $u$.

Part 1:

Let $\pi \in \mathbb{Z}[i]$ be prime. Since $N(\pi)=\pi \overline{\pi} \in$ $\mathbb{Z}$, $\pi$ divides a primenumber $p \in \mathbb{Z}$

the above means that we start with a prime element $\pi$ in $\mathbb{Z}[i]$. But why does $N(\pi)=\pi \overline{\pi}$ imply that $\pi$ divides a prime number $p \in \mathbb{Z}$? I did think about it but the only thing I could come up with is the following: If $\pi$ is prime in $\mathbb{Z}[i]$, then for $r,s$ in $\mathbb{Z}[i]$ such that $\pi \mid rs$, then $\pi \mid r$ or $\pi \mid s$.

I know that if $a \mid b \in \mathbb{Z}$, then $N(a) \mid N(b) \in \mathbb{Z}$.

Thus we get that $N(\pi) \mid N(r)N(s)$, then $N(\pi) \mid N(r)$ or $N(\pi) \mid N(s)$. By this argumentation it follows that $N(\pi)$ is prime in $\mathbb{Z}$.

But that's not the same as in the quote above.

Question 1: So, why does $N(\pi)=\pi \overline{\pi} \in \mathbb{Z}$ imply that $\pi$ divides a prime $p\in\mathbb{Z}$?

Question 2: Does my argumentation work too?

Part 2:

Let $\pi z= p$, then $N(\pi)N(z)=N(p)=p^2$. So we either got $N(\pi)=p$ or $\pi u=p$ for a unit $u$.

If $\pi z= p$ and $p \in \mathbb{Z}$, then $N(\pi)N(z)=N(p)=p \overline{p}=p^2$. I am not sure if I understand why $N(\pi)N(z)=p^2$ does imply that $N(\pi)=p$ or $\pi=up$.

As far as I understand, if $N(\pi)N(z)=p^2$, we got the following cases:

Case 1: $N(\pi)=p$ and $N(z)=p$

Case 2: $N(\pi)=p^2$ and $N(z)=1$

Case one implies $N(\pi)=p$. And case two implies $N(z)=1 \Leftrightarrow z \in \mathbb{Z}[i]^{\times}$.

Question 3: Is my argumentation regarding part 2 correct?

Answer 1

It is maybe a good idea (for this answer) to always write in which ring an element divides an other one. We have a beautiful situation, both rings $\Bbb Z$ and $\Bbb Z[i]$ have unique factorization (unit times prime factors to powers, and the uniqueness is up to reordering factors).

Note that $\Bbb Z[i]$ is the ring of integers in $\Bbb Q(i)$, in other words, an element of $\Bbb Q(i)$ is in this ring, iff it is a root of an irreducible monic polynomial with integer coefficients. The norm map is a multiplicative map (we are excluding zero from its domain) $$ N:\Bbb Z[i]^\times \to\Bbb Z^\times\ , $$ the image is always positive. A maximal factorization in $\Bbb Z[i]^\times$ is mapped into a factorization in $\Bbb Z$, which can be or not maximal.

It may be good to have some examples to support the arguments.

Primes in $\Bbb Z$ are (up to units):

• the two, which ramifies like $(1+i)^2=2i$, so $2$ (times a unit), seen in $\Bbb Z[i]$ has two factors / one factor of multiplicity two.
• primes $p$ from $\Bbb Z$ of the shape $4k+3$, these are inert, when taken from $\Bbb Z$ and suddenly seen in $\Bbb Z[i]$. For instance, $3,7,11,19,...$ cannot be decomposed in $\Bbb Z[i]$. Quadratic reciprocity. We have, seeing $p$ as $p+0i$, $N(p+0i)=(p+0i)(p-0i)=p^2$. So the norm does not (in general) map primes to primes. So Question 2...
• primes $p$ from $\Bbb Z$ of the shape $4k+1$, these always decompose, when taken from $\Bbb Z$ and suddenly seen in $\Bbb Z[i]$, and the decomposition is of the shape $p=(a+bi)(a-bi)$, and these $a,b$ always exist. For instance, $5=(2+i)(2-i)$, $13=(3+2i)(3-2i)$, $17=(4+i)(4-i)$, $29=(5+2i)(5-2i)$, they can be (uniquely up to units) decomposed in $\Bbb Z[i]$, and the the pieces $(a\pm bi)$ are primes in $\Bbb Z[i]$. Quadratic reciprocity. We have $N(a+ib)=p$. So the norm maps such primes in primes.

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Start with $\pi=a+bi$, a prime in $\Bbb Z[i]$. Then $N(\pi)=(a+bi)(a-ib)$ is an integer, and it has a decomposition in prime factors $p_1p_2\dots p_r$. We can thus write: $$ \pi=(a+bi)\text{ divides in $\Bbb Z[i]$ }(a+bi)(a-bi)=(p_1+0i)(p_2+0i)\dots(p_r+0i)\ . $$ Since $\pi = (a+ib)$ is a prime, it divides at least one of the factors on the right side. That's all.

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It remains to use that $\pi$ divides some $p+0i$ in $\Bbb Z[i]$ for some prime $p$ from $\Bbb Z$. Write $p=\pi z$ in $\Bbb Z[i]$. So $p^2 =N(\pi)N(z)$, and for the two norms, of $\pi$ and of $z$ there are two possibilities, parallel to factorizations of $p^2$ in positive integers.

We have $N(\pi)=N(z)=p$ or $N(\pi)=p^2$ and $N(z)=1$.

In the last case, $N(z)$ means $z$ is a unit, so an element among $\pm 1$, $\pm i$ only.