A Brief Introduction to Classical and Adelic Algebraic ... - William Stein

94 CHAPTER 13. DECOMPOSITION AND INERTIA GROUPS
Taking ordpj on both sides we get ej = ek. Thus e1 = e2 = · · · = eg.
As was mentioned right before the statement of the theorem, for any σ ∈ G we
have OK/pi ∼ = OK/σ(pi), so by transitivity f1 = f2 = · · · = fg. Since OK is a
lattice in K, we have
which completes the proof.
[K : Q] = dimZ OK = dimFp OK/pOK
g
= dimFp OK/p ei
g
i = eifi = efg,
i=1
The rest of this section illustrates the theorem for quadratic fields and a cubic
field and its Galois closure.
13.2.1 Quadratic Extensions
Suppose K/Q is a quadratic field. Then K is Galois, so for each prime p ∈ Z we
have 2 = efg. There are exactly three possibilties:
• Ramified: e = 2, f = g = 1: The prime p ramifies in OK, so pOK =
p 2 . There are only finitely many such primes, since if f(x) is the minimal
polynomial of a generator for OK, then p ramifies if and only if f(x) has a
multiple root modulo p. However, f(x) has a multiple root modulo p if and
only if p divides the discriminant of f(x), which is nonzero because f(x) is
irreducible over Z. (This argument shows there are only finitely many ramified
primes in any number field. In fact, we will later show that the ramified primes
are exactly the ones that divide the discriminant.)
• Inert: e = 1, f = 2, g = 1: The prime p is inert in OK, so pOK = p is prime.
This happens 50% of the time, which is suggested by quadratic reciprocity
(but not proved this way), as we will see illustrated below for a particular
example.
• Split: e = f = 1, g = 2: The prime p splits in OK, in the sense that
pOK = p1p2 with p1 = p2. This happens the other 50% of the time.
Suppose, in particular, that K = Q( √ 5), so OK = Z[γ], where γ = (1 + √ 5)/2.
Then p = 5 is ramified, since pOK = ( √ 5) 2 . More generally, the order Z[ √ 5] has
index 2 in OK, so for any prime p = 2 we can determine the factorization of p in
OK by finding the factorization of the polynomial x2 − 5 ∈ Fp[x]. The polynomial
x2 − 5 splits as a product of two distinct factors in Fp[x] if and only if e = f = 1
and
g = 2. For
p =
2, 5 this is the case if and only if 5 is a square in Fp, i.e., if
5
5
p = 1, where p is +1 if 5 is a square mod p and −1 if 5 is not. By quadratic
reciprocity,
5
= (−1)
p
5−1
p−1
· p
p
+1 if p ≡ ±1 (mod 5)
2 2 · = =
5 5 −1 if p ≡ ±2 (mod 5).
i=1