Number Fields

Primes in $\mathbb{Z}$ are not necessarily irreducible in larger number rings. For example, $2=(1+i)(1-i),5=(2+i)(2-i)$ in $\mathbb{Z}[i]$. Similarly, although $\mathrm{2,3}$ remain irreducible in $\mathbb{Z}[\sqrt{-5}]$, the principal ideals $(2),(3)$ do not. This phenomenon is known as splitting. We say $3$ splits into the product of two primes in $\mathbb{Z}[\sqrt{-5}]$. We shall study how a prime splits in a given number ring.

More generally, if $P$ is a prime ideal in some number ring $R=𝔸\cap K$ for some number field $K$, and if $L$ is a number field containing $K$, we consider the prime decomposition of the ideal generated by $P$ in $S=𝔸\cap L$, i.e. $PS=\{{\alpha }_1{\beta }_1+...+{\alpha }_r{\beta }_r:{\alpha }_i\in P{\beta }_i\in S\}$.

For now we shall fix some notation. Let $K,L$ be number fields with $K\subset L$ and let $R=𝔸\cap K,S=𝔸\cap L$, and "prime" means "nonzero prime ideal".

Theorem: Let $P$ be a prime of $R$, and $Q$ a prime of $S$. The following are equivalent:

1. $Q\mid \mathrm{PS}$

2. $Q\supset \mathrm{PS}$

3. $Q\supset P$

4. $Q\cap R=P$

5. $Q\cap K=P$

Proof: $(1)\iff (2)$ was proved earlier. $(2)\iff (3)$ since $Q$ is an ideal in $S$, $(4)\Rightarrow (3)$ trivially, $(4)\iff (5)$ since $Q\subset 𝔸$. We need only show $(3)\Rightarrow (4)$. Now $Q\cap R$ contains $P$ and is an ideal in $R$. Since $P$ is maximal, we have $Q\cap R=P$ or $R$. If $Q\cap R=R$ then $1\in Q$ which implies $Q=S$, a contradiction.

When the above conditions hold, we say that $Q$ lies over $P$, or that $P$ lies under $Q$.

Theorem: Every prime $Q$ of $S$ lies over a unique prime $P$ of $R$. Every prime $P$ of $R$ lies under at least one prime $Q$ of $S$.

Proof: For the first statement, we need to show that $Q\cap R$ is a prime in $R$. Observe $1\notin Q$. Also, we must have $Q\cap R$ nonempty because in particular it contains $N(a)$ for all $a\in Q$, and norms of nonzero elements are nonzero in any integral domain. (Here $N$ means $N_R^Q$.) Lastly, if $rs\in Q$, then one of $r,s$ also lies in $Q$. If $r,s\in R$, then we have that one of $r,s$ lies in $Q\cap R$ showing that it is indeed prime.

Now for the second statement. The primes lying over $P$ are the prime divisors of $PS$, so we need only show $PS\ne S$, that is, $PS$ has at least one prime divisor. We shall show that $1\notin PS$ by using lemma proved earlier: there exists a $\gamma \in K-R$ with $\gamma P\subset R$. thus $\gamma PS\subset RS=S$. Then if $1\in PS$, we have $\gamma \in S$. But this would mean $\gamma$ is an algebraic integer, which is a contradiction.

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 $R=\mathbb{Z},S=\mathbb{Z}[i]$. Then 2 in $\mathbb{Z}$ lies under the prime $(1-i)$ in $\mathbb{Z}[i]$.