# basic facts about ordered rings

Throughout this entry, $(R,\le )$ is an ordered ring.

###### Lemma 1.

If $a\mathrm{,}b\mathrm{,}c\mathrm{\in}R$ with $$, then $$.

###### Proof.

The contrapositive will be proven.

Let $a,b,c\in R$ with $a+c\ge b+c$. Note that $-c\in R$. Thus,

$\begin{array}{cc}\hfill b& =b+0\hfill \\ & =b+c+(-c)\hfill \\ & \le a+c+(-c)\hfill \\ & =a+0\hfill \\ & =a.\mathit{\u220e}\hfill \end{array}$

###### Lemma 2.

If $\mathrm{|}R\mathrm{|}\mathrm{\ne}\mathrm{1}$ and $R$ has a characteristic^{}, then it must be $\mathrm{0}$.

###### Proof.

Suppose not. Let $n$ be a positive integer such that $\mathrm{char}R=n$. Since $|R|\ne 1$, it must be the case that $n>1$.

Let $r\in R$ with $r>0$. By the previous lemma, $$, a contradiction^{}.
∎

###### Lemma 3.

If $a\mathrm{,}b\mathrm{\in}R$ with $a\mathrm{\le}b$ and $c\mathrm{\in}R$ with $$, then $a\mathit{}c\mathrm{\ge}b\mathit{}c$.

###### Proof.

Note that $-c\in R$ and $$. Since $a\le b$, $-(ac)=a(-c)\le b(-c)=-(bc)$. Thus,

$\begin{array}{cc}\hfill bc& =bc+0\hfill \\ & =bc+(ac+(-(ac)))\hfill \\ & =(bc+ac)+(-(ac))\hfill \\ & \le (bc+ac)+(-(bc))\hfill \\ & =-(bc)+(bc+ac)\hfill \\ & =(-(bc)+bc)+ac\hfill \\ & =0+ac\hfill \\ & =ac.\mathit{\u220e}\hfill \end{array}$

###### Lemma 4.

Suppose further that $R$ is a ring with multiplicative identity^{} $\mathrm{1}\mathrm{\ne}\mathrm{0}$. Then $$.

###### Proof.

Suppose that $$. Since $R$ is an ordered ring, it must be the case that $$. By the previous lemma, $1\cdot 1\ge 0\cdot 1$. Thus, $1\ge 0$, a contradiction. ∎

Title | basic facts about ordered rings |
---|---|

Canonical name | BasicFactsAboutOrderedRings |

Date of creation | 2013-03-22 16:17:21 |

Last modified on | 2013-03-22 16:17:21 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 11 |

Author | Wkbj79 (1863) |

Entry type | Result |

Classification | msc 06F25 |

Classification | msc 12J15 |

Classification | msc 13J25 |

Related topic | MathbbCIsNotAnOrderedField |