The Arithmetic of Quaternion Algebra

6 CHAPTER 1. QUATERNION ALGEBRA OVER A FIELD
Corollary 1.2.5. (Theorem of Frobenius) A non-commutative field containing
R in its center, of finite dimension over R, is isomorphic to the field of Hamilton
quaternion H.
. The proof of the theorem relays on the fact that the field of complex
numbers C is the unique commutative extension of finite dimension over the
real field R. The argument of the proof is similar essentially with that in the
following corollary (over a finite field there is no any quaternion field). Assume
d ∈ D − R, the field R(d) is commutative, hence it is true for R(i) with i 2 = −1.
It is different from D since it is not commutative. Suppose d ′ ∈ D, such that
R(d ′ ) = R(u) is different from R(i) and u 2 = −1. The new element u does not
commute with i, and we can replace it by an element of zero trace: j = iui + u,
such that ij = −ji. The field R(i, j) is isomorphic to the quaternion field of
Hamilton, H, and it is contained in D. If it is different from D again, by the
same reason we are allowed to construct d ∈ D but not belonging toR(i, j), such
that di = −id and d 2 ∈ R. But then, dj commute with i, so belongs to R. It is
absurd.
Corollary 1.2.6. (Theorem of Wedderburn).There is no any finite quaternion
field.
There is a weak form of the theorem of Wedderburn : every finite field is
commutative. The proof for the special case gives well the idea of that in the
general case. It uses the fact that every finite field Fq (the index q indicates
the number of the elements of the field) has an extension of a given degree
uniquely determined up to an isomorphism. If H is a quaternion field, its
center is then a finite field Fq, and all of its commutative maximum subfield are
isomorphic to Fq ′, where q′ = q 2 . It allows us to write H as a finite union of the
conjugates of hFq ′h−1 . Let us compute the number of H : q 4 = n(q 2 − q) + q,
where n is the number of the maximum commutative subfield of H. From (2),
n = (q 4 − 1)/2(q 2 − 1). It leads to a contradiction.
Now we show the theorem of automorphism. We begin to prove a preliminary
result. If V is the vector space over K of the subjacent space H. We are
going to determine the structure of the K-algebra End(V ) formed by the Kendomorphisms
of V . We remind here that the tensor product is always taken
over K if without a contrary mention.
Lemma 1.2.7. The mapping of H ⊗ H to End(V ) given by h ⊗ h ′ ↦→ f(h ⊗ h ′ ),
where f(h ⊗ h ′ )(x) = h × h ′ , for h, h ′ , x ∈ H, is an algebraic K-isomorphism.
Proof. It is obvious that f is a K-homomorphism of K-vector spaces . The fact
that the conjugation is an anti-isomorphism (i.e. hk = hk, h, k ∈ H ) implies
that f is a K-homomorphism as a K-algebra. Since the dimensions of H ⊗ H
and Edn(V ) over K is equal, for showing f is a K-isomorphism it is sufficient
to verify that f is injective. We can take an extension HF of H such that HF
is isomorphic to M(2, F ). The extended mapping fF is injective since it is not
zero: its kernel ( a two-sided ideal of HF ⊗F HF ) is zero since HF ⊗F HF is
isomorphic to M(4, F ) which is simple (exercise 1.4).
The proof of the theorem of automorphism. Let L be a commutative Kalgebra
contained in H but different from K, and let g is a non-trivial Kisomorphism
of L to H. We want to prove that g is the restriction of an inner