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I want to find the Dedekind-Zeta function of $K=\mathbb{Q}[i]$. Here is what I have done:

$$\zeta_K(s) = \sum_{\mathfrak{a}} \frac{1}{(N(\mathfrak{a}))^s} = \prod_\mathfrak{p} \left(1 - \frac{1}{{ (N(\mathfrak{p}))}^s}\right)^{-1}; s>1.$$ We know that if $\mathfrak{q}$ is a prime ideal of $\mathcal{O}_K$ then $\mathfrak{q} \cap \mathbb{Z}$ is a prime ideal of $\mathbb{Z}$ which implies that $\mathfrak{q} \cap \mathbb{Z} = (p)$ for some prime $p\in \mathbb{Z}$. Also, since $p \in \mathfrak{q}$, $(p) \subseteq \mathfrak{q}$ and $N(\mathfrak{q}) \le N(p)$.

After some calculations, we end up with $(1-\cfrac{1}{N(\mathfrak{p_i})^s})^{-1} \ge (1-\cfrac{1}{N(p_i)^s})^{-1} $ where we have some prime $p_i \in \mathfrak{p_i}$ for each prime ideal $\mathfrak{p_i}$. Finally, I get

$$\prod_\mathfrak{p} \left(1 - \frac{1}{{ (N(\mathfrak{p}))}^s}\right)^{-1} \ge\prod_{p:prime}\left(1-\cfrac{1}{p^{2s}}\right)^{-1}$$.

What should I do next, am I going in the correct way? Thanks.

Ninja
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  • What do you want to "find"? In the second line is the Dedekind Zeta function. Do you want to have it as a product over primes? –  Jul 25 '17 at 07:00

2 Answers2

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It would be more helpful to systematically figure out the splitting behavior of rational primes in $K=\mathbb Q[i]$. 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 $.

Now let us build our zeta function.

The ramified prime yields one factor of $$\left(1-\cfrac{1}{N(\mathfrak{p})^s}\right)^{-1}=\left(1-\cfrac{1}{2^s}\right)^{-1} $$ The split primes yield two identical factors $$\left(1-\cfrac{1}{N(\mathfrak{p_1})^s}\right)^{-1}\left(1-\cfrac{1}{N(\mathfrak{p_2})^s}\right)^{-1}=\left(1-\cfrac{1}{p^s}\right)^{-2} $$ The inert primes yield one factor each $$ \left(1-\cfrac{1}{N(\mathfrak{p})^s}\right)^{-1}=\left(1-\cfrac{1}{p^{2s}}\right)^{-1}=\left(1-\cfrac{1}{p^s}\right)^{-1}\left(1+\cfrac{1}{p^s}\right)^{-1}$$

So if we multiply all these three kinds of factors, notice that we can collect one full copy of the usual Riemann-Zeta function and then we are left with $$\prod_{p\equiv 1\pmod 4}\left(1-\cfrac{1}{p^s}\right)^{-1}\prod_{p\equiv 3\pmod 4}\left(1+\cfrac{1}{p^s}\right)^{-1} =\prod_p \left(1-\cfrac{\chi(p)}{p^s}\right)^{-1} $$

which is the $L$-function $L(s,\chi)$ of the character $\chi:(\mathbb Z/4\mathbb Z)^*\to \mathbb C$ where $\chi(3)=-1$.

So $$\zeta_K(s)=\zeta_{\mathbb Q}(s)L(s,\chi).$$

Sebastian Schoennenbeck
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Arkady
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2

To add on Ravi's answer :

  • If $\mathfrak{p}$ is a prime of $\mathbb{Z}[i]$ then $\mathfrak{p} \cap \mathbb{Z} = p \mathbb{Z}$ is a prime of $\mathbb{Z}$. The (field) norm is $N(a+ib)=a^2+b^2$, which is also the ideal norm $N(I) =\#\{c \in \mathbb{Z}[i]/I\}$.

    $N(\mathfrak{p}) \in \mathbb{Z}$, and since $N(p) = p^2$, there is one or two primes of $\mathbb{Z}[i]$ above $p$, with norm $p$ or $p^2$. Here $\mathbb{Z}[i] = \mathbb{Z}[x]/(x^2+1)$ (is monogenic) so it reduces to know that

  • $x^2+1 \equiv (x+1)^2 \bmod 2$

  • $x^2+1\bmod p$ is irreducible if $p \equiv 3 \bmod 4$

  • otherwise if $p \equiv 1 \bmod 4$ then by the Fermat two squares theorem (the simplest case of quadratic reciprocity) $-1 \equiv a^2 \bmod p$ and $x^2+1 \equiv (x-a)(x+a) \bmod p$

All together, since $\mathbb{Z}[i]$ is a PID, it shows $$\zeta_{\mathbb{Q}(i)}(s) = \sum_{I \subset \mathbb{Z}[i]}N(I)^{-s} = \!\!\sum_{\alpha \in \mathbb{Z}[i]^*/\mathbb{Z}[i]^\times} N(\alpha)^{-s} =\frac{1}{4} \sum_{\alpha \in \mathbb{Z}[i]^*} N(\alpha)^{-s} =\frac14 \sum_{n,m \in \mathbb{Z}^2 \setminus(0,0)}\!\! (n^2+m^2)^{-s}\\=\prod_{\mathfrak{p} \in \mathbb{Z}[i]} \frac{1}{1-N(\mathfrak{p})^{-s}} =\frac{1}{1-2^{-s}}\prod_{p \equiv 1 \bmod 4} \frac{1}{(1-p^{-s})^2}\prod_{p \equiv 3 \bmod 4} \frac{1}{1-p^{-2s}} = \zeta(s)L(s,\chi_4)$$ (this last property of $\zeta_K$ being a product of Dirichlet L-functions is common to abelian extensions of $\mathbb{Q}$)

The next step is to prove the functional equation and to look at the non-PID case $\mathbb{Z}[\sqrt{-5}]$) (for this, using Sage or Magma is a good idea).

reuns
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