The Arithmetic of Quaternion Algebra

26 CHAPTER 2. QUATERNION ALGEBRA OVER A LOCAL FIELD
Proof. We shall lead an absurdity. If Lnr was not contained in H, then for every
x ∈ O, x /∈ R, the extension K(x)/K would be ramified. There exists then
a ∈ R such that x − a ∈ P ∩ K(x). We could then write x = a + uy with y ∈ O.
Iterating this procedure, the element x could be written as �
n≥0 anun , an ∈ R.
The field K(u) being complete would be closed. We thus had O ⊂ K(u). It is
absurd.
Corollary 2.1.7. The quaternion field H is isomorphic to {Lnr, π}.Its prime
ideal P + Ou satisfies P 2 = Oπ. Its integer ring O is isomorphic to RL + RLu.
The reduced discriminant d(O) of O equals n(P ) = Rπ.
Proof. According to I, Corollary 2.2 and 2.4, we have H � {Lnr, x} where
x ∈ K × but x /∈ n(L × nr). From (1), (2) of the beginning of this chapter it
follows x = πy 2 where y ∈ K × . We can suppose x = π , then the first part of
the corollary follows. Now suppose H = {Lnr, π}. The element u ∈ H satisfying
I.(1) is the non zero minimal valuation, thus P = Ousatisfies P 2 = Oπ. The
prime ideal Rπ is then ramified in O. according to Lemma 1.4, we have O =
{h ∈ H, n(h) ∈ R}. similarly, RL = {m ∈ Lnr, n(m) ∈ R}. We can verify easily
that, if h = m1 + m2u with mi ∈ Lnr, the property n(h) ∈ R is equivalent
to n(mi) ∈ R, i = 1, 2. We can show too that O = RL + RLu. Using the
formula involving the determinant in I, Lemma 4.7, we compute the reduced
discriminant d(O). Since d(RL) = R, we see easily that d(O) = Rπ. Hence it
follows d(O) = n(P ) or the different of O is O ⋆−1 = P .
Definition 2.4. Let Y/X be a finite extension of field equipped with a valuation,
and ring of it is AY , AX = X ∩ AY . Let PY , PX = PY ∩ AX be the prime ideals
and kX, kY be the corresponding residue field. The residue degree f of Y/X is
the degree [kY : kX] of the residue extension kY /kX. The ramification index of
Y/X is the integer e such that AY PX = P e Y .
We then deduce that the unramified quadratic extension Lnr/K has the
ramification index 1, and the residue degree 2. The quaternion field H/K has
the ramification index 2, and the residue degree 2.
Let F/K be a finite extension of commutative field with ramification index e
and residue degree f. We have ef = [F : K] because the cardinal of k is finite
and RF /πRF � RF /πF e RF , if πF is a uniform parameter of F .
Lemma 2.1.8. The following properties are equivalent:
(1) f is even,
(2) F ⊃ Lnr,
(3) F ⊗ Lnr in not a field.
Proof. For the equivalence (1) ←→ (2) see Serre [1],Ch. 1. For the equivalence
(2) ←→ (3), it is convenient to write Lnr as the K[X]/(P (X)) where (P (X)) is
a polynomial of degree 2. Thus F ⊗ Lnr equals F [X]/(P (X))F where (P (X))F
is the ideal generated by (P (X)) in the polynomial ring F [X]. Since P (X) is a
polynomial of degree 2, it is reducible if and only if it admits a root in F , i.e. if
F ⊃ Lnr.
Consider now HF � {F ⊗ Lnr, π}. If πF is a uniform parameter of F , it may
as well suppose that π = πF e . According to I. Corollary 2.4, and Lemma 1, we
find that, if e or f are even then we have HF � M(2, F ), hence F neutralizes
H. Otherwise, that is to say if [F : K] is odd, HF � {F ⊗ Lnr, πF } ,where