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I was trying to learn a little about the Dedekind zeta function. The first place I looked at was obviously the Wikipedia article above. So my question comes from a sentence by the end of the article in the section on relations to other L-functions. That section states that if you have an abelian extension $K/\mathbb{Q} \,$, then the Dedekind zeta function of $K$ is a product of Dirichlet L-functions.

In particular it states that if $K$ is a quadratic field, then the ratio

$$\frac{\zeta_K (s)}{\zeta(s)} = L(s, \chi)$$

or equivalently, that $$\zeta_K (s) = \zeta(s) L(s, \chi)$$

where $\zeta(s)$ is the Riemann zeta function and $L(s, \chi) \,$ is the Dirichlet L-function associated to the Dirichlet character defined by the Jacobi symbol as follows. If $K = \mathbb{Q}(\sqrt{d}) \,$, and $D$ is the discriminant of this number field, then

$$\chi(n) := \left ( \frac{D}{n} \right )$$

Then the Wikipedia article says the following:

That the zeta function of a quadratic field is a product of the Riemann zeta function and a certain Dirichlet L-function is an analytic formulation of the quadratic reciprocity law of Gauss.

This is were I was absolutely amazed, because even though I've seen in the past that the existence of the Euler product for the Riemann zeta function is equivalent to the unique factorization property of the integers, which apparently is also reflected more generally in the context of Dedekind zeta functions and number fields, this time realized as the unique factorization of ideals into products of prime ideals, I just find marvelous that these two things can be equivalent (in some sense which I don't know yet).

So my question is why is this analytic fact about the Dedekind zeta function of a quadratic number field,

$$\zeta_K (s) = \zeta(s) L(s, \chi)$$

an analytic reformulation of the quadratic reciprocity law?

And maybe if it is possible, to push it a little bit further, are there analogues of this, say for higher reciprocity laws, like for cubic or biquadratic reciprocity? Or is this a "peculiarity" that occurs just for quadratic fields?

Thank you very much for any help.

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1 Answer 1

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This is not an answer to all your questions, but anyway.

$\zeta_K(s)=\sum N(\mathfrak{a})^{-s}=\prod_\mathfrak{p} \frac{1}{1-N(\mathfrak{p})^{-s}}$. In the product we have $\frac{1}{1-p^{-s}}$ for every prime dividing $D$, $\frac{1}{(1-p^{-s})^2}$ for every prime that splits in $K$, i.e. such that $D$ is a square mod $p$, and $\frac{1}{1-p^{-2s}}= \frac{1}{(1-p^{-s})(1+p^{-s})}$ for every prime which doesn't split.

For every prime, the factor is thus $\frac{1}{1-p^{-s}}\frac{1}{1-(\frac{D}{p})p^{-s}}$. The product over all primes is thus $$\zeta(s)\,\prod_p \frac{1}{1-(\frac{D}{p})p^{-s}}$$ and your equation becomes $$L(s,\chi)=\prod_p \frac{1}{1-(\frac{D}{p})p^{-s}}.$$

The function $\chi$ in $L(s,\chi)=\sum_n \chi(n) n^{-s}$ is a (quadratic) Dirichlet character modulo $D$, i.e. it is a group morphism $(\mathbb{Z}/D\mathbb{Z})^\times\to\{+1,-1\}$, which is then extended to a function $\mathbb{Z}\to \{0,+1,-1\}$ by $\chi(n)=0$ if $(n,D)\neq1$. Since $\chi$ is multiplicative, we have $$L(s,\chi)=\prod_p \frac{1}{1-\chi(p)p^{-s}} .$$ We therefore indeed have $$\chi(p)=\left(\frac{D}{p}\right).$$ This implies that $(\frac{D}{p})$ depends on $p$ only modulo $D$ - something not evident at all from its definition, but an easy consequence of quadratic reciprocity. If in particular $D=(-1)^{(q-1)/2} q$ (where $q$ is a prime) then there is only one quadratic Dirichlet character, namely $\chi(n)=(\frac{n}{q})$. We therefore have $$\left(\frac{p}{q}\right)=\left(\frac{(-1)^{(q-1)/2}\, q}{p}\right)$$ i.e. quadratic reciprocity.

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Thank you very much for your answer. I have to admit that this is the direction in which I'm most interested about, namely that the identity for the Dedekind zeta function implies quadratic reciprocity. –  Adrián Barquero Apr 19 '11 at 20:58
    
Nevertheless I don't understand all the details in your argument, so I'm starting a bounty for this question. If you can provide more details to your answer and nobody else responds with a "better" answer I'd be very happy to award the bounty to you. For example, I do understand the first half of your answer but the second half is giving me some trouble to understand. I'm not quite clear on how you conclude that $\chi (p) = (D/p)$, and the part about quadratic reciprocity towards the end is not very clear to me. Thank you for your help. –  Adrián Barquero Apr 19 '11 at 21:11
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@Adrián: is one of the problems why we can conclude from the two expressions for $L(s,\chi)$ (one using $(\frac{D}{p})$ and the other $\chi(p)$) that $\chi(p)=(\frac{D}{p})$? It follows e.g. from Perron's formula. Expand both products (first the factors to geometric series and then multiply over $p$'s) to series of the form $\sum a_n/n^s$. If such a series converges for $Re\,s$ big enough, then $a_n$'s are determined uniquely by the sum. (the equality you want then uses only $a_p$'s). Here is Wikipedia page about Perron's formula: en.wikipedia.org/wiki/Perron%27s_formula –  user8268 Apr 20 '11 at 20:21
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+ you will probably have to tell me what is not clear with quadratic reciprocity at the end. Here it is again: $q$ is a prime, $K=\mathbb{Q}(\sqrt{q^*})$ with $q^*=(-1)^{(q-1)/2} q$, so that $D=q^*$. There is only one non-trivial (i.e. surjective) quadratic Dirichlet character modulo $q$, namely $\chi(x)=(\frac{x}{q})$ (as $\chi(x^2)=\chi(x)^2=1$, there is no other choice). $(\frac{p}{q})=\chi(p)=(\frac{q^*}{p})$ is the quadratic reciprocity law. –  user8268 Apr 21 '11 at 17:01
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some extra info: A (quadratic) Dirichlet character $(\mathbb{Z}/N\mathbb{Z})^\times\to\{+1,-1\}$ is called called primitive if it can't be reduced to $(\mathbb{Z}/M\mathbb{Z})^\times\to\{+1,-1\}$ for any proper divisor $M$ of $N$. For a given $N$ there is at most one primitive quadratic character (namely the Kronecker-Jacobi-Legendre symbol), and it exists iff $M=\pm D$ where $D$ is the discriminant of a quadratic extension of $\mathbb{Q}$. The $\chi$ in $\zeta_K(s)=\zeta(s)L(s,\chi)$ is primitive (for $N$ the discriminant of $K$). –  user8268 Apr 21 '11 at 17:12

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