Example

Tag 06YZ, permanent. Cite it as [Bourbaki, Tag 06YZ].

Algebra, chapter VI, § 2. Ordered fields

There is only one totally ordered ring structure on $\mathbf{Z}$ : indeed $1 > 0$ , whence $n > 0$ for every natural number $n \neq 0$ , by induction. In contrast there exist ordered ring structures on $\mathbf{Z}$ which are not totally ordered (see below).

Let $P$ be the set of positive elements of an ordered ring $\mathbf{A}$ . It is known (VI, p. 3, Prop. 3) that $P$ determines the ordering on $\mathbf{A}$ . To say that $\mathbf{A}$ is an ordered ring is equivalent to saying that $P$ satisfies the following properties :

\begin{align*} (\mathrm{AP}_I) &\quad P + P \subset P \\ (\mathrm{AP}_{II}) &\quad PP \subset P \\ (\mathrm{AP}_{III}) &\quad P \cap (-P) = \{0\} \end{align*}

Indeed $(\mathrm{AP}_I)$ and $(\mathrm{AP}_{III})$ state that the additive group of $\mathbf{A}$ is an ordered group (VI, p. 3, Prop. 3), while $(\mathrm{AP}_{,,})$ is a translation of (OR).

Recall that the following condition is necessary and sufficient for the order relation on $\mathbf{A}$ to be total :

(\mathrm{AP}_{IV}) \quad P \cup (-P) = \mathbf{A}.

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

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