Splitting of Primes in Extensions

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 $2,3$ 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 $R = \mathbb{A} \cap K$ 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 $S = \mathbb{A} \cap L$, i.e. $P S = \{\alpha_1\beta_1 +...+\alpha_r\beta_r : \alpha_i\in P \beta_i\in S\}$.

For now we shall fix some notation. Let $K, L$ be number fields with $K \subset L$ and let $R = \mathbb{A}\cap K, S =\mathbb{A}\cap L$, 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. $Q | PS$

  2. $Q \supset PS$

  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 $S$, $(4) \implies (3)$ trivially, $(4) \iff (5)$ since $Q \subset \mathbb{A}$. We need only show $(3) \implies (4)$. 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 $N^Q_R$.) Lastly, if $r s \in Q$, 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 $P S$, so we need only show $P S \ne S$, that is, $P S$ has at least one prime divisor. We shall show that $1 \notin P S$ by using lemma proved earlier: there exists a $\gamma \in K - R$ with $\gamma P \subset R$. thus $\gamma P S \subset R S = S$. Then if $1 \in P S$, 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 $P S$. Their corresponding exponents are called the ramification indices. If $Q^e$ is the exact power of $Q$ dividing $P S$, then $e$ is the ramification index of $Q$ over $P$. We denote this by $e(Q|P)$.

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 $e((1-i)|2) = 2$ since $2S = (1-i)^2$. Also, we have $e(Q|p) = 1$ 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 $e((1-\omega)|p)=\phi(m)=p^{r-1}(p-1)$ (TODO: proof). Also $e(Q|q)=1$ 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 $R\rightarrow S/Q$ whose kernel is $R\cap Q$. Now $R\cap Q=P$ hence we have an embedding $R/P\rightarrow S/Q$. 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 $f(Q|P)$.

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