Bourbaki

Remark

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

Algebra, chapter IX, § 10. Angles

Formulas (6) remain true for t=∞t = \infty provided one replaces the rational functions occurring on the right-hand side by their canonical extensions to the projective field A~\tilde{A} (Chapter II, 2nd ed., App. III, No. 5); in fact, if t=∞t = \infty, φ\varphi is the class of ω\omega, and one has d(φ)=−1,s(d(φ))=0d(\varphi) = -1, s(d(\varphi)) = 0 and c(d(φ))=−1c(d(\varphi)) = -1; these are indeed the values taken by the canonical extensions of the right-hand sides for t=∞t = \infty. The same is true for (7) when t(θ)=∞t(\theta) = \infty, and for (8) when t(φ)=∞t(\varphi) = \infty or when 1+δt(φ)2=01 + \delta t(\varphi)^2 = 0. Analogously, formula (2) remains true when only one of the elements t(φ),t(φ′),t(φ)t(\varphi), t(\varphi'), t(\varphi), for example, is infinite, provided one considers its right-hand side as a rational function of t(φ)t(\varphi) alone: in fact the product of the similarities 1+t(φ′)ω1 + t(\varphi')\omega and ω\omega is δt(φ′)+ω\delta t(\varphi') + \omega, whereas the value taken by the canonical extension of the right-hand side of (2) is 1/δt(φ′)1/\delta t(\varphi'). Finally, when t(φ)t(\varphi) and t(φ′)t(\varphi') are finite and one has 1+δt(φ)t(φ′)=01 + \delta t(\varphi)t(\varphi') = 0, one has t(φ)+t(φ′)=0t(\varphi) + t(\varphi') = 0 (otherwise t(φ)2t(\varphi)^2 would be equal to 1/δ1/\delta, which is impossible (prop. 4)); one may therefore agree that the value of the right-hand side of (2) is ∞\infty, and this value is indeed that of the left-hand side. When t(φ)t(\varphi) and t(φ′)t(\varphi') are both infinite, the right-hand side of (2) is not defined.

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

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Cites

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Translated with machine help, not yet reviewed by a person.