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The Arithmetic of Quaternion Algebra

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Chapter 1<br />

<strong>Quaternion</strong> <strong>Algebra</strong> over a<br />

field<br />

In this chapter K always denotes a commutative field <strong>of</strong> arbitrary characteristics<br />

if no particular mention, and Ks is the separable closure <strong>of</strong> K.<br />

1.1 <strong>Quaternion</strong> <strong>Algebra</strong><br />

Definition 1.1. A quaternion algebra H with center K is a central algebra over<br />

K <strong>of</strong> dimension 4, such that there exists a separable algebra <strong>of</strong> dimension 2 over<br />

K, and an invertible element θ in K with H = L + Lu, where u ∈ H satisfies<br />

u 2 = θ, um = mu (1.1)<br />

for all m ∈ L, and where m ↦→ m is the nontrivial K-automorphism <strong>of</strong> L.<br />

We denote H sometimes by {L, θ}, but H does not determine this pair<br />

{L, θ} uniquely. For example, it is clear that one could replace θ by θmm, if<br />

m is an element <strong>of</strong> L such that mm �= 0. the element u is not determined<br />

by (1). If m ∈ L is an element with mm = 1, we could replace u by mu.<br />

the definition can be used in the case <strong>of</strong> arbitrary character. .we can verify<br />

easily that H�K is a central simple algebra, i.e. an algebra with center K<br />

and without any nontrivial two-sided ideal.conversely, we can prove that every<br />

cental simple algebra <strong>of</strong> dimension 4 over K is a quaternion algebra.<strong>The</strong> rule <strong>of</strong><br />

multiplication is deduced from (1). If m ∈ L, we have<br />

(m1 + m2u)(m3 + m4u) = (m1m3 + m2m4θ) + (m1m4 + m2m3)u. (1.2)<br />

Definition 1.2. <strong>The</strong> conjugation is the K-endomorphism <strong>of</strong> H: h → h, which<br />

is the extension <strong>of</strong> the nontrivial K-automorphism <strong>of</strong> L defined by u = −u.<br />

It is easy to verify it is an involutive anti-automorphism <strong>of</strong> H. It can be<br />

expressed by the following relations: if h, k ∈ H and a, b ∈ K, we have<br />

ah + bk = ah + bk, h = h, hk = kh.<br />

Definition 1.3. Assume h ∈ H. <strong>The</strong> reduced trace <strong>of</strong> h is t(h) = h + h. <strong>The</strong><br />

reduced norm <strong>of</strong> h is n(h) = hh.<br />

1

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