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 \(\mathbb{Z}[i]\).

Ben Lynn blynn@cs.stanford.edu 💡