One of the recurring themes of modern number theory seems to be a nagging feeling that almost all the things we know (and many more things we don’t know) can be expressed as elementary properties of ${\zeta}$ and ${L}$-functions.

For example, the zeta function of a number field ${K}$ is defined by the product

$\displaystyle \zeta_K(s) = \prod_{\mathfrak{p} \text{ prime ideal of } \mathcal{O}_k} (1 - (N\mathfrak{p})^{-s})^{-1}.$
If we expand this as a Dirichlet series

$\displaystyle \zeta_K(s) = \sum_n a_n n^{-s}$
the coefficients ${a_n}$ give detailed information about the factorisation of ideals in ${K}$. For example, if ${a_p=1}$ then ${p}$ is totally ramified, whereas if it is equal to the degree of the field, ${p}$ is totally split.

It is also well-known that the Riemann zeta function (the case ${K={\mathbb Q}}$ above) controls the distribution of the prime numbers in ${{\mathbb Z}}$. By modifying this ${\zeta}$ function we can obtain the Dirichlet L-functions which give us even more information. To define these, we start with a homomorphism ${\chi: {\mathbb Z} \rightarrow {\mathbb Z}/q{\mathbb Z} \rightarrow {\mathbb C}^*}$ called a Dirichlet character, and twist’ the ${\zeta}$-function using it:

$\displaystyle L(\chi, s) = \sum_n \chi(n) n^{-s} = \prod_p (1 - \chi(p)p^{-s})^{-1}.$
For example, the trivial character ${\chi_1 = 1}$ recovers the original ${\zeta}$-function. Dirichlet famously used this, together with observations about the (lack of) zeroes of these (mostly) convergent power series at ${Re(s)>1}$ and ${s=1}$, to deduce that any arithmetic progression which might plausibly contain infinitely many primes does. Note that if ${q}$ is the least quotient ${\chi}$ factors through, it is called the conductor of ${\chi}$.

It seems to be true that certainly for ${E/{\mathbb Q}}$ an abelian extension, the above two objects are related, and in fact ${\zeta_E}$ can be written as a product of Dirichlet L-functions. I don’t know enough class field theory to go into the general cases, but want to look at a specific case that seems quite interesting. Apparently the following result is true. Let ${q}$ be a fixed prime, and ${\chi}$ any quadratic (taking values ${1}$ and ${-1}$) character with conductor ${q}$. Now consider the quadratic field ${K = {\mathbb Q}(\sqrt{\chi(-1) q})}$. We have the following identity:

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

Looks innocuous enough…. surely even if this is true, it can’t say much interesting.

Well, let’s unravel what it says:

$\displaystyle \prod_{p \text{ ramified}} (1-p^{-s})^{-1} \prod_{p \text{ split}} (1-p^{-s})^{-2} \prod_{p \text{ inert}} (1-p^{-2s})^{-1} = \zeta(s) L(\chi, s).$
So cancelling the familiar factors from ${\zeta}$, it says that

$\displaystyle L(\chi, s) = \prod_{p \text{ split}} (1-p^{-s})^{-1} \prod_{p \text{ inert}} (1+p^{-s})^{-1}.$
So comparing product expansions, it is equivalent to

$\displaystyle \chi(p) = \begin{cases} 1 \text{ iff }p\text{ is split,}\\ 0 \text{ iff }p\text{ is ramified,}\\ -1 \text{ iff }p\text{ is inert.} \end{cases}.$

Interesting. This splitting into cases reminds us of a more classical area of algebraic number theory. Write ${K={\mathbb Q}(\sqrt{d})}$, and assume ${d \equiv 1 \mod 4}$. Then the minimal polynomial is ${X^2 + X + (d-1)/4}$, which factorises modulo an odd prime ${p}$ iff it does after we’ve multiplied by 4 and completed the square to get ${(2X+1)^2 + d}$, whose factorisation modulo ${p}$, together with the Kummer-Dedekind theorem for the splitting of primes in number fields whose ring of integers is primitively generated, gives a similar criterion in terms of the Legendre symbol:

$\displaystyle \left(\frac{d}{p}\right) = \begin{cases} 1 \text{ iff }p\text{ is split,}\\ 0 \text{ iff }p\text{ is ramified,}\\ -1 \text{ iff }p\text{ is inert.} \end{cases}.$
Now, take ${\chi(n) = \left(\frac{n}{q}\right)}$ and, noting that ${\chi(-1) = (-1)^{(q-1)/2}}$, we see straight away, comparing the above two statements, that:

$\displaystyle \left(\frac{p}{q}\right) = \left(\frac{d}{p}\right) = \left(\frac{(-1)^{(q-1)/2}}{p}\right)\left(\frac{q}{p}\right) = (-1)^{(p-1)(q-1)/4}\left(\frac{q}{p}\right).$

So our identity in terms of ${\zeta}$ and L-functions was actually a (fairly heavily) disguised statement of Gauss’ famous law of quadratic reciprocity! Of course, I haven’t given you a proof: I’m not personally sure exactly how to prove the identity about the ${\zeta}$ and L-functions. Maybe I’ll let you know when I figure it out, but in the meantime there are plenty of proofs of quadratic reciprocity out there: http://www.rzuser.uni-heidelberg.de/~hb3/rchrono.html. I should also refer the interested reader to Kolowski’s opening chapter of an introduction to the Langlands program’ which was certainly where I read about this.