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The following is a well-known result:

Theorem: If $a$ and $m$ are integers, with $(a,m)=1$, then there are infinitely many primes congruent to $a\pmod m$.

It turns out that Dirichlet’s Theorem is actually a special case of Artin’s Reciprocity Law. So, we’ll discuss how this works.

Let $L/K$ be an extension of number fields. (That is, $L$ and $K$ are finite extensions of $\mathbb{Q}$.) Let $A$ and $B$ be the rings of integers of $K$ and $L$, respectively. (This means that $A$ and $B$ are the integral closures of $\mathbb{Q}$ in $K$ and $L$, respectively.) Now, let $\mathfrak{p}$ be a nonzero prime ideal in $A$. Then $\mathfrak{p}B=\prod_{i=1}^g \mathfrak{P}_i^{e_i}$ for some primes $\mathfrak{P}_i$ of $B$ and some positive integers $e_i$. If $e_i>1$, we say that $\mathfrak{P}_i$ is ramified over $\mathfrak{p}$. We call $e_i$ the ramification index. The primes $\mathfrak{P}_i$ are said to lie above $\mathfrak{p}$.

Since $A$ and $B$ are Dedekind domains, $\mathfrak{p}$ and $\mathfrak{P}_i$ are maximal ideals. Hence $k=A/\mathfrak{p}$ and $\ell=B/\mathfrak{P}_i$ are finite fields, and $\ell/k$ is a field extension. Let $f_i=[\ell:k]$. We call $f_i$ the residue degree.

It is not too difficult to show that if $n=[L:K]$, then $n=\sum_{i=1}^g e_if_i$. If $L/K$ is a Galois extension, then $e_i$ and $f_i$ are independent of $i$ (since the Galois group of $L$ over $K$ acts transitively on the $\mathfrak{P}_i$), so we can write $n=efg$.

Now, let’s define a few subgroups of $Gal(L/K)$. We’ll assume from now on that $L/K$ is a Galois extension. Furthermore, we fix a prime $\mathfrak{p}$ of $A$, and some prime $\mathfrak{P}$ of $B$ lying above $\mathfrak{p}$. Now, define $D_{\mathfrak{P}}=\{\sigma\in Gal(L/K):\sigma(\mathfrak{P})=\mathfrak{P}\}$. We call $D_{\mathfrak{P}}$ the decomposition group. We now have a homomorphism $\phi : D_{\mathfrak{P}}\to Gal(\ell/k)$. To define $\phi$, we note that an element of $D_{\mathfrak{P}}$ permutes cosets of $B/\mathfrak{P}$ and thus gives the desired homomorphism. Furthermore, this homomorphism is surjective. The kernel $T_{\mathfrak{P}}$ of $\phi$ is called the inertia group. Hence $D_{\mathfrak{P}}/T_{\mathfrak{P}}\cong Gal(\ell/k)$.

It is not hard to determine the sizes of $D_{\mathfrak{P}}$ and $T_{\mathfrak{P}}$ in terms of quantities we already understand: $|D_{\mathfrak{P}}|=ef$ and $|T_{\mathfrak{P}}|=e$. In particular, if $\mathfrak{p}$ is an unramified prime, then $D_{\mathfrak{P}}\cong Gal(\ell/k)$.

That’s particularly nice, because Galois groups of extensions of finite fields are always cyclic, generated by the Frobenius automorphism. Thus in the unramified case, $D_{\mathfrak{P}}$ is cyclic and generated by an automorphism $\sigma$ satisfying the congruence $\sigma(x)\cong x^{|k|} \pmod{\mathfrak{P}}$ for all $x\in B$. (We can extend $\sigma$ to all of $L$ by multiplicativity.) Furthermore, this element $\sigma$ is unique. The common notation for $\sigma$ is $(\mathfrak{P},L/K)$.

If $\mathfrak{P}$ and $\mathfrak{Q}$ are two primes lying above $\mathfrak{p}$, then there is some element $\sigma\in Gal(L/K)$ so that $\sigma(\mathfrak{P})=\mathfrak{Q}$. It is easy to verify that $(\mathfrak{Q},L/K)=\sigma(\mathfrak{P},L/K)\sigma^{-1}$. Therefore, if $Gal(L/K)$ is abelian, then $(\mathfrak{P}/L/K)$ depends only on $\mathfrak{p}$. In this case, we may write $(\mathfrak{p},L/K)$ for this element.

Finally, we’re ready to state (part of) the Artin reciprocity theorem.

Let $L/K$ be an abelian Galois extension of number fields, and let $\sigma\in Gal(L/K)$ be fixed. Then there are infinitely many primes $\mathfrak{p}$ of $A$ that are unramified and so that $\sigma=(\mathfrak{p},L/K)$.

(In fact, only finitely many primes ramify, since primes ramify if and only if they divide the discriminant, which can be easily verified. The other part of the statement is more interesting.)

Let’s look at one example. Take $K=\mathbb{Q}$ and $L=\mathbb{Q}(\zeta_n)$, where $\zeta_n=e^{2\pi i/n}$. Then $Gal(L/K)\cong(\mathbb{Z}/n\mathbb{Z})^\times$, where the isomorphism is as follows: if $(m,n)=1$, then there is an automorphism $\sigma_m\in Gal(L/K)$ defined by $\sigma_m(\zeta_n)=\zeta_n^m$. Now, if $(p,n)=1$, then $((p),L/K)=\sigma_p$. In particular, this case of the theorem is equivalent to Dirichlet’s Theorem on primes in arithmetic progressions.

It turns out that the theorem isn’t too much more general than this, since any abelian extension is contained in a cyclotomic extension (this is the Kronecker-Weber Theorem), and it’s not hard to see what happens to Frobenius elements when we pass to sub-extensions.

All this material can be found (with many more details included) in Serre’s Local Fields.