Number Fields

We keep our previous notation, that is, $K,L$ are number fields with $K\subset L$ and $R=𝔸\cap K,S=𝔸\cap L$. Suppose $P$ is prime in $R$.

The primes lying over $P$ are those that occur in the prime decomposition of $PS$. Their corresponding exponents are called the ramification indices. If $Q^e$ is the exact power of $Q$ dividing $PS$, then $e$ is the ramification index of $Q$ over $P$. We denote this by $e(Q\mid P)$.

Example: Let $R=\mathbb{Z},S=\mathbb{Z}[i]$. Then $(1-i)$ lies over $2$ (which we use as shorthand for $2\mathbb{Z}$). It can be checked that $(1-i)$ is prime, and hence $e((1-i)\mid 2)=2$ since $2S=(1-i{)}^2$. Also, we have $e(Q\mid p)=1$ for $p\ne 2$ and any $Q$ lying over $p$.

More generally, if $R=\mathbb{Z},S=\mathbb{Z}[\omega ]$ for $\omega =e^{2\pi i/m}$ where $m=p^r$ for some prime $p\in \mathbb{Z}$ then the principal ideal $(1-\omega )$ is $S$ is a prime lying over $p$, and $e((1-\omega )\mid p)=\varphi (m)=p^{r-1}(p-1)$ (TODO: proof). Also $e(Q\mid q)=1$ whenever $q\ne p$ and $Q$ lies over $q$, which will follow from a theorem we shall prove later.

Notice $R/P,S/Q$ are fields since $P,Q$ are maximal. Moreover, there is a natural way to view $R/P$ as a subfield of $S/Q$. Since $R\subset S$, there is a ring homomorphism from $R\to S/Q$ whose kernel is $R\cap Q$. Now $R\cap Q=P$ hence we have an embedding $R/P\to S/Q$. The fields $R/P,S/Q$ are called the residue fields associated with $P,Q$. The residue fields of a number ring are finite as was shown when proving every number field is a Dedekind domain, hence $S/Q$ is a finite extension of $R/P$. Let $f$ be the degree of this extension. We call $f$ the inertial degree of $Q$ over $P$, and denote it by $f(Q\mid P)$.

Example: Let $R=\mathbb{Z},S=\mathbb{Z}[i]$. Recall 2 in $\mathbb{Z}$ lies under the prime $(1-i)$ in $\mathbb{Z}[i]$. Now $\mid S/2S\mid =4$ and $(1-i)$ properly contains $2S$ so $\mid S/(1-i)\mid$ must be 2. Then $\mid R/P\mid =\mid S/Q\mid =2$, so $f=1$ in this case. On the other hand, $3S$ is prime in $S$, and $\mid S/3S\mid =0$. Thus $f(3S\mid 3)=2$.

It can be easily verified that $e,f$ are multiplicative in towers. That is, if $P\subset Q\subset U$ are primes in $R\subset S\subset T$ then

$$\begin{array}{ccc}e(U\mid P)& =& e(U\mid Q)e(Q\mid P)\\ f(U\mid P)& =& f(U\mid Q)f(Q\mid P)\end{array}$$

In general if $Q$ is a prime in $S$, then $Q$ lies over a unique prime $p\in \mathbb{Z}$. Thus $S/Q$ is a field of order $p^f$ where $f=f(Q\mid p)$. Since $pS\subset Q$ we have $p^f\le \mid S/pS\mid =p^n$ where $n=[L:\mathbb{Q}]$. Thus $f\le n$ when the ground field is $\mathbb{Q}$. In fact, we can show more.

For an ideal $I$ of $R$, define $\parallel I\parallel$ to be $\mid R/I\mid$. We shall prove the following two theorems simultaneously:

Theorem: Let $n=[L:K]$. Let $Q_1,...,Q_r$ be primes of $S$ lying over a prime $P$ of $R$. Let $e_1,...,e_r$ and $f_1,...f_r$ be the ramification indices and inertial degrees. Then ${\sum }_{i=1}^r{e}_i{f}_i=n$.

Theorem: Let $n=[L:K]$. Then

1. For ideals $I,J$ in $R$, $\parallel IJ\parallel =\parallel I\parallel \parallel J\parallel$.

2. Let $I$ be an ideal of $R$. Then $\parallel IS\parallel =\parallel I{\parallel }^n$.

3. Let $\alpha \in R$ be nonzero. Then $\parallel (\alpha )\parallel =\mid N_{\mathbb{Q}}^K(\alpha )\mid$.

Proof of (1): We first prove for the case when $I,J$ are coprime, and then show $\parallel P^m\parallel =\parallel P{\parallel }^m$ for all primes $P$. The result will then follow from uniqfue factorization of ideals.

Assume $I,J$ are coprime, so that $I+J=R,I\cap J=IJ$. Then by the Chinese Remainder Theorem we have an isomorphism

$$R/IJ\to R/I\times R/J$$

hence $\parallel IJ\parallel =\parallel I\parallel \parallel J\parallel$.

Now consider $\parallel P^m\parallel$ where $P$ is prime. We have the chain $R\supset P\supset P^2\supset ...\supset P^m$. We shall show that for all $k$, $\parallel P\parallel =\mid P^k/P^{k+1}\mid$ where the $P^k$ are viewed as additive groups. In fact we shall exhibit a group isomorphism $R/P\to P^k/P^k/P^{k+1}$.

Take any $\alpha \in P^k-P^{k+1}$. We have the isomorphism $R/P\to \alpha R/\alpha P$. The inclusion $\alpha R\subset P^k$ induces the homomorphism $\alpha R\to P^k/P^{k+1}$, with kernel $(\alpha R)\cap P^{k+1}$ and image $((\alpha R)+P^{k+1}/P^{k+1}$. It remains to show that $(\alpha R)\cap P^{k+1}=\alpha P$ and $(\alpha R)+P^{k+1}=P^k$, which can be done easily by viewing these as the lcm and gcd of $\alpha R,P^{k+1}$, and noting that $P^k$ is the exact power of $P$ dividing $\alpha R$.

Proof of first theorem, special case We prove the first theorem for $K=\mathbb{Q}$. Thus $P=p\mathbb{Z}$ for some prime $p\in \mathbb{Z}$. We have $pS={\prod }_{i=1}^r{Q}_i^{e_i}$, hence

$$\parallel pS\parallel =\prod _{i=1}^r\parallel Q_i{\parallel }^{e_i}=\prod _{i=1}^r(p^{f_i}{)}^{e_i}$$

Since we know $\parallel pS\parallel =p^n$, we have proved the theorem for $K=\mathbb{Q}$.

Proof of (2): Since (1) has been proved, we need only show this fact for prime ideals $P$. It can be checked that $S/PS$ is a vector space over $R/P$, and in fact $S/PS$ is a ring containing $R/P$. We shall show that the dimension is $n$.

First take any ${\alpha }_1,...,{\alpha }_{n+1}\in S$. We wish to show that the corresponding elements in $S/PS$ are linearly dependent. They are linearly dependent over $K$ and hence also over $R$. Hence ${\alpha }_1{\beta }_1+...+{\alpha }_{n+1}{\beta }_{n+1}=0$ for some ${\beta }_1,...,{\beta }_{n+1}\in R$, not all zero. We shall show that not all the ${\beta }_i$ lie in $P$, so after reducing modulo $P$ they do not all become zero. First we need the following generalization of a previous lemma:

Lemma: Let $A,B$ be nonzero ideals in a Dedekind domain $R$, with $B\subset A$ and $A\ne R$. Then there exists $\gamma \in K$ with $\gamma B\subset R,\gamma B\subset ⃒A$.

Proof: Recall there exists a nonzero ideal $C$ such that $BC$ is principal, say $BC=⟨\alpha ⟩$. Then $BC\subset ⃒\alpha A$. Fix any $\beta \in C$ such that $\beta B\subset ⃒\alpha A$ and set $\gamma =\beta /\alpha$. $\blacksquare$

Apply the lemma with $A=P$ and $B=⟨{\beta }_1,...,{\beta }_{n+1}⟩$, Thus $S/PS$ is at most $n$-dimensional over $R/P$.

To establish equality, let $P\cap \mathbb{Z}=p\mathbb{Z}$ and consider all primes $P_i$ lying over $p$. We know $S/P_{i}S$ is vector space over $R/P_i$ of dimension $n_i\le n$. We show equality holds for all $i$ and in particular $P_i=P$. Set $e_i=e(P_i\mid p),f_i=f(P_i\mid p)$. Then from above ${\sum }_i{e}_i{f}_i=m$ where $m=[K:\mathbb{Q}]$.

Thus $pR=\prod P_i^{e_i}$ hence $pS=\prod (P_{i}S{)}^{e_i}$, whence

$$\parallel pS\parallel =\prod \parallel P_{i}S{\parallel }^{e_i}=\prod \parallel P_i{\parallel }^{n_i{e}_i}=\prod (p^{f_i}{)}^{n_i{e}_i}$$

Also we have $\parallel pS\parallel =p^{mn}$ so $mn={\sum }_i{f}_i{n}_i{e}_i$. Since all $n_i\le n$ and ${\sum }_i{e}_i{f}_i=m$ and we have $n=n_i$ for all $i$. $\blacksquare$

Proof of first theorem, general case: We have $PS=\prod Q_i^{e_i}$, hence

$$\parallel PS\parallel =\prod \parallel Q_i{\parallel }^{e_i}=\prod \parallel P{\parallel }^{f_i{e}_i}$$

by (1) and the definition of $f_i$. By (2) we have $\parallel PS\parallel =\parallel P{\parallel }^n$. Thus $n={\sum }_i{e}_i{f}_i$. $\blacksquare$

Proof of (3): Extend $K$ to a normal extension $M$ of $\mathbb{Q}$ and let $T=𝔸\cap M$. For each embedding $\sigma$ of $K$ in $ℂ$ we have $\parallel \sigma (\alpha )T\parallel =\parallel \alpha T\parallel$. This is because $\sigma (T)=T$ when extended to an automorphism of $M$. Set $N=N^K(\alpha )$. Then by (1) we have

$$\parallel NT\parallel =\prod _{\sigma }\parallel \sigma (\alpha )T\parallel =\parallel \alpha T{\parallel }^n$$

Clearly $\parallel NT\parallel =\mid N{\mid }^{mn}$ where $m=[M:K]$ and by (2) we have $\parallel \alpha T\parallel =\parallel \alpha R{\parallel }^m$. Thus $\parallel \alpha R\parallel =\mid N\mid$. $\blacksquare$

Example: We can use these theorems to show that $⟨1-\omega ⟩$ is prime in $\mathbb{Z}[\omega ]$ where $\omega =e^{2\pi i/m},m=p^r$: since $⟨1-\omega {⟩}^n=p\mathbb{Z}[\omega ]$ where $n=\varphi (m)$, any further splitting of $⟨1-\omega ⟩$ into primes would contradict the first theorem since $\mathbb{Q}[\omega ]:\mathbb{Q}=n$.

Alternatively we could have used the first part of the second theorem:

$$\parallel ⟨1-\omega ⟩{\parallel }^n=\parallel ⟨1-\omega {⟩}^n\parallel =\parallel p\mathbb{Z}[\omega ]\parallel =p^n$$

hence $\parallel ⟨1-\omega ⟩\parallel =p$. Since is $p$ is prime we have that $⟨1-\omega ⟩$ prime.

Example:

TODO

If $L$ is a normal extension of $K$, and $P$ is a prime of $R={\mathbb{Z}}_K$, then the Galois group $G=\mathrm{Gal}(L/K)$ permutes the primes lying over $P$. In fact, we have:

Theorem: Let $L$ be a normal extension of $K$ with corresponding number rings $S$ and $R$. Let $Q,Q\prime$ be primes of $S$ lying over the same prime $P$ of $R$. Then $\sigma (Q)=Q\prime$ for some $\sigma$ in the Galois group $G=\mathrm{Gal}(L/K)$.

Proof: Suppose $\sigma (Q)\ne Q\prime$ for all $\sigma \in G$. Then by the Chinese Remainder Theorem there is a solution the the system of congruences

$$\begin{array}{c}x=0(\mathrm{mod}Q\prime )\\ x=1(\mathrm{mod}\sigma (Q))\text{for all}\sigma \in G\end{array}$$

Let $\alpha \in S$ be such a solution. Then

$$N_K^L(\alpha )\in R\cap Q\prime =P$$

since one of the factors of $N_K^L(\alpha )$ is $\alpha \in Q\prime$. Since $\alpha \notin \sigma (Q)$ for all $\sigma$ we have ${\sigma }^{-1}(\alpha )\notin Q$. As $N_K^L(\alpha )$ is the product of ${\sigma }^{-1}(\alpha )$ for all $\sigma$ we see that $N_K^L(\alpha )\notin Q$, a contradiction since we have just shown $N_K^L(\alpha )\in P\subset Q$. $\blacksquare$

Corollary: If $L$ is normal over $K$, and $Q,Q\prime$ are two primes lying over $P$ then $e(Q\mid P)=e(Q\prime \mid P)$ and $f(Q\mid P)=f(Q\prime \mid P)$.

Proof: We have the unique factorization $PS=Q_1^{e_1}...Q_r^{e_r}$. For any $Q_i$, there exists $\sigma \in G$ that maps $Q_1$ to $Q_i$. We have

$$PS=\sigma (PS)=Q_i^{e_1}\sigma (Q_2{)}^{e_2}...\sigma (Q_r{)}^{e_r}$$

hence by unique factorization we must have $e_1=e_i$ for all $i$.

The theorem implies there is an isomorphism between $S/Q$ and $S/Q\prime$, which establishes the result for the inertial degrees. $\blacksquare$

Hence for normal extensions, a prime $P$ of $R$ splits into $(Q_1...Q_r{)}^e$ in $S$ where the $Q_i$ are distinct primes in $S$ of inertial degree $f$ over $P$. By a previous theorem we must have $ref=[L:K]$.

If $e(Q\mid P)>1$ then we say $P$ is ramified in $S$. (In other words, $PS$ is not squarefree.)

Theorem: Let $p$ be a prime in $\mathbb{Z}$. Suppose $p$ is ramified in a number ring $R$. Then $p\mid \mathrm{disc}(R)$.

Proof: Let $P$ be a prime of $R$ lying over $p$ with $e(P\mid p)>1$. For some ideal $I$ we have $pR=PI$. Note $I$ is divisible by all primes of $R$ lying over $p$. Let ${\sigma }_1,...,{\sigma }_n$ be the embeddings of the corresponding number field $K$ in $ℂ$.