Bourbaki

Remark

Tag 0DOA, permanent. Cite it as [Bourbaki, Tag 0DOA].

Algebra, chapter IX, § 10. Angles

When –1 belongs to the class mod. (A∗)2(A^*)^2 in question, one generally chooses ω\omega in such a way that ω2=−1\omega^2 = -1, which determines it up to sign. When this class contains 1 but not –1, one generally chooses ω\omega in such a way that ω2=1\omega^2 = 1, which again determines it up to sign.

No. 2

ANGLES

This being so, every element φ\varphi of S+S^+ can be written, in one and only one way, in the form

(1)

φ=cw(φ)+sw(φ)ω \varphi = c_w(\varphi) + s_w(\varphi)\omega
where cw(φ),sw(φ)c_w(\varphi), s_w(\varphi) belong to AA; the element sw(φ)/cw(φ)s_w(\varphi)/c_w(\varphi) of the projective field A~\tilde{A} (Chap. II, 2nd ed., App. III, No. 5) is denoted by tw(φ)t_w(\varphi); it depends only on the class of φ\varphi mod. HH; thus twt_w defines, by passing to the quotient, a mapping of S+/HS^+/H into the projective field A~\tilde{A}, which we shall again denote by twt_w by abuse of language. We shall often write c,sc, s and tt instead of cw,swc_w, s_w and twt_w. We have c−w=cw,s−w=−swc_{-w} = c_w, s_{-w} = -s_w and t−w=−twt_{-w} = -t_w.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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