Bourbaki

Examples

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

Algebra, chapter VI, § 2. Ordered fields

1) The field Q\mathbf{Q} of rational numbers is an ordered field. 2) A subfield of an ordered field, with the induced ordering, is an ordered field. 3) \ The field of real numbers is an ordered field. \*

Let KK be an ordered field. For all x∈Kx \in K we put

sgn⁡(x)=1if x>0,sgn⁡(x)=−1if x<0,sgn⁡(x)=0if x=0. \begin{align*} \operatorname{sgn}(x) &= 1 & \text{if } x > 0, \\ \operatorname{sgn}(x) &= -1 & \text{if } x < 0, \\ \operatorname{sgn}(x) &= 0 & \text{if } x = 0. \end{align*}

Then we have sgn⁡(xy)=sgn⁡(x)sgn⁡(y)\operatorname{sgn}(xy) = \operatorname{sgn}(x) \operatorname{sgn}(y); we call sgn⁡(x)\operatorname{sgn}(x) the sign of xx. The map x↦sgn⁡(x)x \mapsto \operatorname{sgn}(x) from K∗K^* to the multiplicative group {−1,+1}\{-1, +1\} is a surjective homomorphism whose kernel, the set of strictly positive elements of KK, is a subgroup of K∗K^* of index 2.

Conversely, if KK is a commutative field and s:K∗→{−1,+1}s : K^* \to \{-1, +1\} is a surjective homomorphism whose kernel is closed under addition, then ss is the sign map for a unique ordered field structure, where the set of strictly positive elements is the kernel of ss.

For all xx and yy in KK we have x=sgn⁡(x)∣x∣x = \operatorname{sgn}(x)|x| and ∣xy∣=∣x∣∣y∣|xy| = |x||y|.

On the other hand every ordered field is of characteristic zero (Prop. 1).

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

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Cites

Transcribed from the printed edition, 2003, Springer. Not translated.