## 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 💡