February 14th
Today I learned some basic facts about Artin $L$-functions. We define these by an Euler product, but it turns out that unramified primes are a bit annoying to deal with, so I will adopt the notation\[\prod_\mf pf(\mf p)\simeq\prod_\mf pg(\mf p)\]to mean that the two Euler products agree on all but finitely many terms.
Fix $L/K$ a finite Galois extension of number fields. (I think this works for arbitrary global fields, but I don't want to think about that.) For an unramified prime $\mf p$ of $K$ under a prime $\mf P$ of $L,$ we have a Frobenius element $\op{Frob}_\mf P\in\op{Gal}(L/K),$ but in fact $\op{Frob}_\mf P$ is determined up to conjugacy by $\mf p.$ So, abusing notation, we write $\op{Frob}_\mf p$ to mean the entire conjugacy class.
Now, what Artin $L$-functions bring in is representation theory: for a representation $\rho:\op{Gal}(L/K)\to\op{GL}(V),$ we note that $\rho(\op{Frob}_\mf p)$ is determined up to change of basis (conjugacy), so\[\op{charpoly}(\rho(\op{Frob}_\mf p))=\det(I-t\rho(\op{Frob}_\mf p))\]is completely determined by $\op{Frob}_\mf p.$ Note that this is a bit of a nonstandard definition of the characteristic polynomial, but it functions. Now, we define the Artin $L$-function by plugging in $\op N(\mf p)^{-s}$ into the characteristic polynomial, writing\[L(\rho,s)=\prod_{\mf p\text{ unramified}}\frac1{\det\left(I-\op N(\mf p)^{-s}\rho(\op{Frob}_\mf p)\right)}.\]Note that only finitely many primes are unramified, so this is a pretty good definition of the Artin $L$-function. I have been told there are things one can do to deal with ramification, but I don't know this yet.
We do the trivial example for concreteness. If we fix our extension to be $K/\QQ$ and use the trivial representation $\rho_{\text{trivial}}:g\mapsto I,$ then we get\[L(\rho,s)=\prod_{(p)\text{ unramified}}\frac1{\det(I-p^{-s}I)}=\prod_{(p)\text{ unramified}}\frac1{1-p^{-s}}.\]This is $\zeta(s)$ (up to finitely many Euler factors), so our Artin $L$-functions do give the typical $\zeta$ function.
The main theorem we're going to show is that Artin $L$-functions include Dedekind $\zeta$-functions. Namely, if we fix our extension $K/\QQ,$\[L(\rho_{\text{reg}},s)\simeq\zeta_K(s),\]where $\zeta_K$ is the Dedekind $\zeta$ function. Notably, the regular representation actually matters. Expanding out the Euler products, we need to show that\[\prod_{(p)\text{ unramified}}\frac1{\det\left(I-p^{-s}\rho_{\text{reg}}(\op{Frob}_p)\right)}\stackrel?\simeq\prod_{\mf p\subseteq K}\frac1{1-\op N(\mf p)^{-s}}.\]For ease, fix $n=[L:K]$ with $e_p$ and $f_p$ the inertial and ramification information of $(p).$ Collecting the Euler factors on the right-hand side above a particular rational $(p),$ we want to show\[\prod_{(p)\text{ unramified}}\frac1{\det\left(I-p^{-s}\rho_{\text{reg}}(\op{Frob}_p)\right)}\stackrel?\simeq\prod_{(p)}\left(\frac1{1-p^{-f_ps}}\right)^{n/(e_pf_p)}.\]In light of the $\simeq,$ it suffices to ignore ramification and show that\[\det\left(I-p^{-s}\rho_{\text{reg}}(\op{Frob}_p)\right)\stackrel?=\left(1-p^{-f_ps}\right)^{n/f_p}.\]The $p^{-s}$ is somewhat arbitrary, so we might as well show $\det\left(I-t\rho_{\text{reg}}(\op{Frob}_p)\right)=\left(1-t^{f_p}\right)^{n/f_p}.$
By this point, the number theory is actually gone. For $g\in G=\op{Gal}(K/\QQ)$ with $|G|=n,$ we claim that\[\det(I-t\rho_{\text{reg}}(g))\stackrel?=\left(1-t^f\right)^{n/f},\]where $f$ is the order of $g.$ (Note $f_p$ is the order of $\op{Frob}_p.$) In order to make the characteristic polynomial more familiar, we send $t\mapsto\frac1\lambda$ and multiply by $\lambda^n$ so that we want\[\det(\lambda I-\rho_{\text{reg}}(g))\stackrel?=\left(\lambda^f-1\right)^{n/f}.\]Showing this is the matter of pick the right basis for $\rho_{\text{reg}}(g),$ for there is general theory about how to compute characteristic polynomials of permutation matrices. We organize $G$ by cosets $G/\langle g\rangle$: naming our cosets $h_1\langle g\rangle,h_2\langle g\rangle,\ldots,h_{n/f}\langle g\rangle,$ we order our basis for $\rho_{\text{reg}}(g)$ as\[e_{h_1},e_{h_1g},\ldots,e_{h_1g^{f-1}},\quad e_{h_2},\ldots,e_{h_2g^{f-1}},\quad\ldots\quad e_{h_{n/f}},\ldots,e_{h_{n/f}g^{f-1}}.\]This has the advantage that, for any of these coset-blocks of $f$ vectors, the definition of the regular representation tells us that\[\rho_{\text{reg}}(g):e_{h_\bullet}\mapsto e_{h_\bullet g}\mapsto\cdots\mapsto e_{h_\bullet g^{f-1}}\mapsto e_{h_\bullet}.\]Thus, under this basis, $\rho_{\text{reg}}(g)$ looks like\[\begin{bmatrix} C_f \\ & \ddots \\ & & C_f\end{bmatrix},\]where $C_f$ is the $f\times f$ matrix\[C_f=\begin{bmatrix} 0 & 0 & \cdots & 0 & 1 \\ 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0\end{bmatrix}.\]We can now compute the characteristic polynomial with few tears. Because of the blocky form of $\rho_{\text{reg}}(g),$ we already have $\det(\lambda I-\rho_{\text{reg}}(g))=\det(\lambda I-C_f)^{n/f}.$ So we want to show\[\det(\lambda I-C_f)\stackrel?=\lambda^f-1.\]We could show this by computing eigenvalues of $C_f$ or just noting it is\[\det\begin{bmatrix} \lambda & 0 & \cdots & 0 & -1 \\ -1 & \lambda & \cdots & 0 & 0 \\ 0 & -1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & -1 & \lambda\end{bmatrix},\]which is\[\lambda\det\begin{bmatrix} \lambda & \cdots & 0 & 0 \\ -1 & \cdots & 0 & 0 \\ \vdots & \ddots & \vdots & \vdots \\ 0 & \cdots & -1 & \lambda\end{bmatrix}-(-1)^f(-1)\det\begin{bmatrix} -1 & \lambda & \cdots & 0 \\ 0 & -1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & -1\end{bmatrix}\]after doing expansion along the top row. The left-hand matrix is lower-triangle with $\lambda$ along the diagonal, so that term is $\lambda^f.$ The right-hand matrix is upper-triangle with $-1$ along the diagonal, so that term is $(-1)^f(-1)(-1)^{f-1}=1.$ In total, we see\[\det(\lambda I-C_f)=\lambda^f-1,\]which is what we wanted. We remark that this method can find the characteristic polynomial of a general permutation matrix.
We give a small application to close. Very quickly, we note that $L(\rho_1\oplus\rho_2,s)=L(\rho_1,s)L(\rho_2,s)$ because, after expanding Euler products, it suffices to show that\[\det(I-p^{-s}(\rho_1\oplus\rho_2)(\op{Frob}_\mf p))=\det(I-p^{-s}\rho_1(\op{Frob}_\mf p))\det(I-p^{-s}\rho_2(\op{Frob}_\mf p)).\]This is true by, for example, expanding out what $\oplus$ does in a basis to see that the characteristic polynomial does indeed multiply. Anyways, because\[\rho_{\text{reg}}=\bigoplus_{\rho\text{ irred}}\rho^{\dim\rho}\]by properties of the regular representation (which we showed last month), we conclude\[\zeta_K(s)\simeq L(\rho_{\text{reg}},s)=\prod_{\rho\text{ irred}}L(\rho,s).\]This is called Artin decomposition and is quite nice. For example, this tells us that $\zeta_\QQ\simeq L(\rho_{\text{trivial}},s)$ "divides'' into $\zeta_K(s)$ in some natural way.