Algebraic Number Theory

This is immediate for $n=1$. Suppose it for $n$. Then the difference of the right-hand side of the above equation for $n+1$ and $n$ is

as required. □

Let $t_0=\mathrm{Re}(s_0)$ and $t=\mathrm{Re}(s)$ for $s\in Z(s_0)$, so $t>t_0$. For $m\ge 1$, let $D_m>0$ be the maximum of all $|{\sum }_{k=1}^n{a}_k{k}^{-s_0}|\le D_m$ with $n\le m$. These are bounded by some $D>0$ since the partial sums converge. Applying Lemma 7.1.2 to the sequences ${(a_n{n}^{-s_0})}_{n\ge 1}$ and ${(n^{s_0-s})}_{n\ge 1}$, we obtain

Note that

from which it follows that

Taking the limit as $n\to \infty$, we obtain

Thus ${\sum }_{n=m+1}^{\infty }{a}_n{n}^{-s}$ approaches $0$ uniformly as $m$ increases for $s$ in any fixed compact subset of $Z(s_0)$. □

For $t=\mathrm{Re}(s)$, the computation of Lemma 7.1.4 with $s_0=0$ and $D_n\le C{n}^u$ gives

Taking the limit as $n\to \infty$, we obtain

which again converges uniformly in any bounded subset of $Z(s)$. □

The first statement is an immediate consequence of Lemma 7.1.5, the condition of which holds for $C=1$ and $t=1$. Consider ${\zeta }_2(s)={\sum }_{n=1}^{\infty }{(-1)}^{n-1}{n}^{-s}$. We can again apply Lemma 7.1.5 for $C=1$ and $t=0$ to see that the latter series converges uniformly and absolutely on $\mathrm{Re}(s)>0$. Note that

from which we see that ${\zeta }_2(s)=(1-2^{1-s})\zeta (s)$ for $s\in Z(0)$ aside from those with $2^{s-1}=1$, i.e., those $s$ of the form $1+\frac{2\mathit{𝜋𝑛𝑖}}{\log 2}$ for some $n\in \mathbb{Z}$. Similarly, the Dirichlet series ${\zeta }_3(s)$ attached to the sequence ${(a_n)}_{n\ge 1}$ with $a_{3k+1}=a_{3k+2}=1$ and $a_{3k+3}=-2$ for all $k\ge 0$ converges for $\mathrm{Re}(s)>0$ and satisfies ${\zeta }_3(s)=(1-3^{1-s})\zeta (s)$ on said region, aside from $s$ with $3^{s-1}=1$, those $s$ of the form $1+\frac{2\mathit{𝜋𝑛𝑖}}{\log 3}$. Thus $\zeta (s)$ is analytic outside $s=1$. Finally, we note that

Thus, $\underset{s\to 1}{\lim }(s-1)\zeta (s)=1$, so $\zeta (s)$ has a simple pole with residue $1$ at $s=1$. □

The logarithm of the Euler product is given by

with the latter sum taken over all prime numbers $p$. For $t=\mathrm{Re}(s)$,we have

so converges uniformly on an closed interval inside the interval $t>\frac{1}{2}$. In particular, the series ${\sum }_{n=1}^{\infty }{\sum }_p\frac{1}{n{p}^{\mathit{𝑛𝑠}}}$ converges absolutely and uniformly on any compact subset of $Z(\frac{1}{2})$ on which ${\sum }_p{p}^{-s}$ does. In particular, the Euler product defines an analytic function on $Z(1)$.

We can then compare finite products and sums. Let $S$ be a finite set of prime numbers and $I_S$ be the semigroup they generate. Then

We then have $\zeta (s)={\prod }_p{(1-p^{-s})}^{-1}$ for $\mathrm{Re}(s)>1$ by taking the limit over all $S$. □

We sketch part of the proof. Again, we consider the logarithm of the Euler product, noting that

and the latter sum is at most $[K:\mathbb{Q}]{\sum }_p{p}^{-s}\sim [K:\mathbb{Q}]\log {(s-1)}^{-1}$, from which we obtain convergence of the Euler product on $Z(1)$ to an analytic function. We can then compare the Euler product over a finite sum of primes with the partial sum in the Dirichlet series over ideals divisible only by those primes to obtain equality in said region. We omit the argument regarding its meromorphic continuation and simple pole. □

As before, we have that $\log {\zeta }_L(s)\sim {\sum }_{𝔓}N{𝔓}^{-s}$. At the same time, the sum over primes that are not completely split is bounded absolutely by the finite sum over primes of degree $\ge 2$ at $s=1$, since $|N{𝔓}^{-s}|$ is for such $𝔓$ is at least $p^2$ for the prime $p$ with $(p)=𝔓\cap \mathbb{Z}$. Since each prime over $K$ that is completely split in $K$ has $[L:K]$ primes over it of the same absolute norm, we then have

At the same time, since both ${\zeta }_L(s)$ and ${\zeta }_K(s)$ have a simple pole at $s=1$, we have

and therefore $\delta (S_{L∕K})={[L:K]}^{-1}$. □

Let $n\ge 2$ be the order of $\chi$, fix an $n$th root of unity $\zeta$, and let $d=[K:\mathbb{Q}]$. The geometry of numbers can be used to show that the number of ideals $𝔞$ of ${𝒪}_K$ with $\mathit{𝑁𝔞}\le N$ for $N\ge 1$ and $\chi (𝔞)=\zeta$ is $\mathit{𝐶𝑁}+O(N^{1-d^{-1}})$, where $C$ is a constant independent of $\zeta$. Given this, we note that

By Lemma 7.1.5, we therefore have that ${\sum }_{𝔞\subset {𝒪}_K}\chi (𝔞)N{𝔞}^{-s}$ converges absolutely and uniformly on every compact subset of $Z(1-d^{-1})$.

For the nonvanishing, write $\log {\zeta }_L(t)={\sum }_{\chi }\log L(\chi ,t)$ and observe that, up to a bounded function as $t\to 1^{+}$, the latter sum has absolute value at least $(1-{\sum }_{\chi }{m}_{\chi })\log {(t-1)}^{-1}$, where $m_{\chi }$ is the order of vanishing of $L(\chi ,s)$ at $s=1$. But if some $m_{\chi }\ge 1$, then $1-{\sum }_{\chi }{m}_{\chi }\le 0$, which is impossible since $\log {\zeta }_L(s)\sim \log {(s-1)}^{-1}$. □

From Proposition 7.1.16 and Proposition 7.1.19, it follows that

for $\mathrm{Re}(s)>1-{[L:\mathbb{Q}]}^{-1}$. For a prime $𝔭$ of $K$ unramified in $L$, we have that ${\phi }_{𝔭}({\zeta }_m)={\zeta }_m^{\mathit{𝑁𝔭}}$ for a primitive $m$th root of unity ${\zeta }_m$, and therefore $\chi ({\phi }_{𝔭})$ depends only on $\mathit{𝑁𝔭}$ modulo $𝔪$.

Much as before, we have

for $\mathrm{Re}(s)>1$. Given $\sigma \in \mathrm{Gal}(F({\mu }_m)∕F)$ with $\sigma ({\zeta }_m)={\zeta }_m^a$ for some $a$ prime to $m$, we have

Now, on the one hand we have

whereas on the other we have

since we know that $L(\chi ,1)\ne 0$ for all nontrivial $\chi$. Comparing the two equations, we obtain that the Dirichlet density of $𝔭$ with ${\phi }_{𝔭}=\sigma$ is $1∕|G|$ □

With Proposition 7.2.1 already in hand, we divide the remainder of the proof into two steps.

Step 1. First, we shall show that the theorem for $L∕K$ and $C$ follows from the theorem for a cyclic subextension $L∕E$, where $E$ is the fixed field of an element of $C$. Let $S$ be the set of prime ideals of $K$ unramified in $L$ with class $C$. Let $\sigma \in C$ and $E=K^{⟨\sigma ⟩}$ so that $L∕E$ is cyclic of degree $f=|⟨\sigma ⟩|$. Note that the latter order is independent of $\sigma$.

Let $T_{\sigma }$ be the set of primes $P$ of $E$ unramified in $L$ and over $K$ with Frobenius ${\phi }_P$ at a prime of $L$ over $P$ equal to $\sigma$. If $P\in T_{\sigma }$, then ${\phi }_P=\sigma$ fixes $E$, so $P$ has degree one over $F$. As $P$ is by definition inert in $L$, there are exactly $|G|∕f$ primes of $L$ over $P\cap K$. As the Frobenius elements of such primes are distributed evenly among the elements of the conjugacy class $C$ of $\sigma$, exactly $|G|∕f|C|$ of these have Frobenius $\sigma$.

We may then compute the Dirichlet density of $S$:

recalling once again that ${\sum }_{𝔭}N{𝔭}^{-s}\sim {\sum }_{P}N{P}^{-s}$. Supposing the theorem for $K∕E$, we have $\delta (T_{\sigma })=\frac{1}{f}$, and we therefore obtain $\delta (S)=\frac{|C|}{|G|}$, as desired.

Step 2. It remains to prove the theorem for cyclic extensions, and we shall actually make the weaker hypothesis that $L∕K$ is abelian. Choose $m\ge 1$ not dividing the discriminant of $L$ so that $H=\mathrm{Gal}(L({\mu }_m)∕L)$ is isomorphic to ${(\mathbb{Z}∕\mathit{𝑚\mathbb{Z}})}^{\times }$ via the mod $m$ cyclotomic character, and $\mathrm{Gal}(L({\mu }_m)∕K)\cong G\times H$. For $\sigma \in G$ and $\tau \in H$, let $S_{\sigma }$ be the set of primes of $K$ unramified in $L$ with Frobenius $\sigma$ in $G$, and let $S_{\sigma ,\tau }$ be the set of primes of $K$ unramified in $L({\mu }_m)$ with Frobenius $(\sigma ,\tau )\in G\times H$. Then

Now suppose that $|G|$ divides the order of $\tau$. Then $⟨(\sigma ,\tau )⟩\cap (G\times \{1\})=1$, which implies that $L({\mu }_m)$ is given by adjoining ${\mu }_m$ to $F=K{({\mu }_m)}^{⟨(\sigma ,\tau )⟩}$. Since we have previously shown the theorem for cyclotomic extensions such as $F({\mu }_m)∕F$, we have that the Dirichlet density of the set of unramified primes in this extension with Frobenius $(\sigma ,\tau )$ is $\frac{1}{|⟨\tau ⟩|}$. From the argument of Step 1, we see that $\delta (S_{\sigma ,\tau })$ exists and equals $\frac{1}{|G||H|}$.

Now let $H_n$ be the set of elements $\tau \in H$ of order divisible by $n$. By summing over all $\tau \in H_n$, we see that ${\delta }_{\inf }(S_{\sigma })\ge \frac{|H_n|}{|G||H|}$. Write $n=p_1^{k_1}\cdots p_r^{k_r}$ for distinct primes $p_1,\dots ,p_r$ and $k_1,\dots ,k_r\ge 1$. There exists a prime $m\equiv 1mod{n}^j$, since the Dirichlet density of completely split primes in $\mathbb{Q}({\mu }_{n^j})∕\mathbb{Q}$ is positive. For such an $m$, let $j_i=v_{p_i}(m-1)\ge j$. We then have

so $\frac{|H_n|}{|H|}$ tends to $1$ as $j$ increases. It follows that ${\delta }_{\inf }(S_{\sigma })\ge \frac{1}{|G|}$. Since the sum of these over all $\sigma \in G$ is then at least $1$, it must equal $1$. Thus, we have that $\delta (S_{\sigma })$ exists and equals $\frac{1}{|G|}$. □

The prime numbers with Frobenius $\phi$ in $\mathrm{Gal}(\mathbb{Q}({\mu }_n)∕\mathbb{Q})$ satisfying $\phi ({\zeta }_n)={\zeta }_n^a$ for a primitive $n$th root of unity ${\zeta }_n$ are exactly those in the arithmetic progression in question. Chebotarev’s theorem then tells us that the Dirichlet density is the reciprocal of the degree $\phi (n)$, as the extension is abelian. □

Since ${\mathrm{Cl}}_K^{𝔪}=I_K^{𝔪}∕P_K^{𝔪}$. We saw in Remark 7.3.8 that the natural map $I_K^{𝔪}\to {\mathrm{Cl}}_K$ is surjective with kernel $P_K\cap I_K^{𝔪}$. Moreover, the natural map

is by definition surjective. Since the elements of $K_{𝔪}$ that generate classes in $P_K^{𝔪}$ are those in ${𝒪}_K^{\times }{K}_{𝔪,1}$, we have the result. □

The kernel of the reduction modulo ${𝔪}_f$ map on $K_{𝔪}$ is $K_{{𝔪}_f,1}$ and those elements of $K_{{𝔪}_f,1}$ with trivial sign at all real places dividing ${𝔪}_{\infty }$ constitute $K_{𝔪,1}$. Therefore, we have an injective map as in the statement. That this map is surjective is an immediate corollary of weak approximation. □

Let $\alpha \in L_{𝔪,1}$. We have that $N_{L∕K}(\alpha )$ is the product of the $\tau (\alpha )$ over all embeddings $\tau$ of $L$ fixing $K$ in our fixed algebraic closure of $K$. Since the ideal ${𝔪}_f$ of ${𝒪}_K$ is fixed under these embeddings, we have $N_{L∕K}\alpha \equiv 1mod{𝔪}_f$.

As for the infinite part, if $\sigma :K\to ℝ$ corresponds to a place dividing ${𝔪}_{\infty }$ and $S$ is the set of embeddings $\tau :L\to ℂ$ extending $\sigma$, then

If $\tau \in S$ is real, then $\alpha \in L_{𝔪,1}$ tells us that $\tau (\alpha )>0$. If $\tau \in S$ is complex, then the complex conjugate embedding $\overline{\tau }$ is also in $S$, and

As a product of positive numbers, $\sigma (N_{L∕K}\alpha )$ is positive. □

Let $H=I_K^{𝔪}∕P_K^{𝔪}{N}_{L∕K}{I}_L^{𝔪}$, and let Let $\chi :H\to {ℂ}^{\times }$ be a nontrivial character. Much as with nontrivial characters of $\mathrm{Gal}(L∕K)$, we can define an $L$-function

which converges to an analytic function on $Z(1-{[K:\mathbb{Q}]}^{-1})$. Let $m_{\chi }$ be its order of vanishing at $s=1$. On the one hand, we have

while on the other hand, we have

As the primes $𝔭$ which are norms from $L∕K$ and unramified are the completely split primes $S_{L∕K}$, the latter sum is, up to a constant (the sum of the reciprocals of the ramified primes), at least

employing Theorem 7.1.14. We therefore have that every $m_{\chi }$ is zero and $1\ge \frac{|H|}{[L:K]}$, which is what we aimed to show. □