Prove there are infinitely many primes in $\mathbb{Z}[i]$

Question

I saw a proof online there are infinitely many primes in $\mathbb{Z}$. The Euler product let's us factor the harmonic series:

$$ \prod \left( 1 - \frac{1}{p} \right) = \sum \frac{1}{n}$$

I wonder if this extends to $\mathbb{Z}[i]$. Joan Baez Week 216 of This Week's Finds defines the zeta function of number field as the sum over non-zero ideals basically:

$$ \zeta_{\mathbb{Z}[i]}(s) = \sum \frac{1}{(m^2 + n^2)^s} = \prod_{p \in \mathbb{Z}[i]} \left( 1 - \frac{1}{|p|^s} \right) $$

Here $2 = (1+i)(1-i)$ means that $2$ is ramified in $\mathbb{Z}[i]$. When does this converge?

$$ \sum_{m, n \geq 1} \frac{1}{(m^2 + n^2)^s} \approx \frac{\pi}{4}\int \frac{ dr }{r^{2s-1}} < \infty$$

This zeta function converges for $s > 1$ since we are adding more numbers. So for $s = 1$

$$ \zeta_{\mathbb{Z}[i]}(1) = \sum \frac{1}{m^2 + n^2} = \prod_{p \in \mathbb{Z}[i]} \left( 1 - \frac{1}{|p|} \right) = \infty $$

We should also check that $\mathbb{Z}[i]$ has unique factorization, because it has a Euclidean algorithm

---

Did we just prove $\mathbb{Z}[i]$ has infinitely many primes?

If that doesn't work, we show that $\zeta(2)$ is irrational. In fact $\zeta_{\mathbb{Z}}(2) = \frac{\pi^2}{6}$ but what about $\zeta_{\mathbb{Z}[i]}(2) $ ?

• Different ways to prove there are infinitely many primes?

• GCD of gaussian integers

Answer 1

Your proof does indeed work (with a few minor corrections to your definition of $\zeta_{\mathbb Z[i]}$), and whilst it is not the easiest way to prove that $\mathbb Z[i]$ has infinitely many primes (see the other answer), the proof works in a much more general setting:

Theorem: Let $K$ be a number field and $\mathcal O_K$ be its ring of integers. Then $\mathcal O_K$ has infinitely many prime ideals.

The proof is exactly the same as the proof in your question. Define the Dedekind zeta function $$\zeta_K(s) = \sum_{0\ne\mathfrak a \subset \mathcal O_K}N\mathfrak a^{-s},$$where $s\in \mathbb C$, the summation is over non-zero ideals of $\mathcal O_K$, and the norm of an ideal is defined by $$N\mathfrak a = \#\mathcal O_K/\mathfrak a.$$

Equivalently, $$\zeta_K(s) =\sum_{n=1}^\infty a_nn^{-s}$$ where $a_n=\#\{\mathfrak a \subset\mathcal O_K:N\mathfrak a =n\}$. When $K=\mathbb Q$ this is just the Riemann zeta function. When $K = \mathbb Q(i)$ as in your question, then we will have $a_n = 0$ whenever $n$ cannot be written as a sum of two squares. However, for example, when $(x+yi)$ and $(x-yi)$ are different ideals, we will have $a_{x^2+y^2}\ge2$, so this function is slightly different to your $\zeta_{\mathbb Z[i]}$.

This series converges absolutely in the half-plane $\mathrm{Re}(s)>1$ and has a simple pole at $s=1$.

Now $\mathcal O_K$ may not be a UFD. However, it will always be a Dedekind domain, so it will be the case that every ideal can be uniquely factorised as a product of prime ideals. It follows that we can write $$\zeta_K(s) = \prod_{\mathfrak p\subset\mathcal O_K}(1-N\mathfrak p^{-s})^{-1}$$where the product is over prime ideals, converges absolutely for $\mathrm{Re}(s)>1$ and has a simple pole at $s=1$. Your proof now applies.

---

Even further generalisation is possible. For example, by considering the Dirichlet characters$$\chi:(\mathbb Z/m\mathbb Z)^\times\to\mathbb C^\times$$and their corresponding L-functions $$L(s,\chi) = \sum_{n=1}^\infty\chi(n)n^{-s}=\prod_p(1-\chi(p)p^{-s})^{-1}$$(and some amount of hard work), one can prove Dirichlet's Theorem on Primes in Arithmetic Progressions: that for any $a$ with $(a,m) = 1$, there are infinitely many primes $p$ with $p\equiv a \pmod m$.

One can go even further, and consider what an analogue of this theorem would be for a general number field. One quickly finds oneself playing with some of the ideas of Class Field Theory. For a good introduction to this approach, see (especially chapter 2 of) Class Field Theory by N. Childress.

Answer 2

HINT:

Modify slightly the argument of Euclid by taking $q= 4 p_1\cdot \ldots \cdot p_n-1$ to show that $q$ has a prime factor $\equiv -1 \mod 4$ which is also different from all the $p_i$'s. Therefore, there exist infinitely many prime numbers of form $4 k-1$. These will also be prime elements in $\mathbb{Z}[i]$ ( see Gaussian integers)

Answer 3

Question: "Prove there are infinitely many primes in $Z[i]$"

You may also use Gauss reciprocity/quadratic reciprocity to classify all prime ideals $I$ in $A:=\mathbb{Z}[i]$: The prime ideals in $A$ are the following ideals:

$I:=(p)A$ with $p\in \mathbb{Z}$ an odd prime and $p \cong 2,3(mod(4))$.

$I:=(a+ib)$ with $(a,b)=1$ and $a^2+b^2=p$, with $p$ an odd prime and $p \cong 1(mod(4))$.

$I:=(1+i)$.

Any odd prime $p\in \mathbb{Z}$ is $\cong 1(mod(4))$ or $\cong 2,3(mod(4))$ and there are infinitely many primes $p$, hence $A$ has infinitley many prime ideals.

Rings of integers and extension of primes: When does a prime remain prime?