basic facts about ordered rings

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

Lemma 1.

If $a,b,c\in R$ with $a<b$ , then $a+c<b+c$ .

Proof.

The contrapositive will be proven.

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

$\begin{array}[]{rl}b&=b+0\\ &=b+c+(-c)\\ &\leq a+c+(-c)\\ &=a+0\\ &=a.\qed\end{array}$

Lemma 2.

If $|R|\neq 1$ and $R$ has a characteristic, then it must be $0$ .

Proof.

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

Let $r\in R$ with $r>0$ . By the previous lemma, $\displaystyle 0<r\leq\ldots\leq\sum_{j=1}^{n-1}r\leq\sum_{j=1}^{n}r=0$ , a contradiction. ∎

Lemma 3.

If $a,b\in R$ with $a\leq b$ and $c\in R$ with $c<0$ , then $ac\geq bc$ .

Proof.

Note that $-c\in R$ and $0=c+(-c)<0+(-c)=-c$ . Since $a\leq b$ , $-(ac)=a(-c)\leq b(-c)=-(bc)$ . Thus,

$\begin{array}[]{rl}bc&=bc+0\\ &=bc+(ac+(-(ac)))\\ &=(bc+ac)+(-(ac))\\ &\leq(bc+ac)+(-(bc))\\ &=-(bc)+(bc+ac)\\ &=(-(bc)+bc)+ac\\ &=0+ac\\ &=ac.\qed\end{array}$

Lemma 4.

Suppose further that $R$ is a ring with multiplicative identity $1\neq 0$ . Then $0<1$ .

Proof.

Suppose that $0\not<1$ . Since $R$ is an ordered ring, it must be the case that $1<0$ . By the previous lemma, $1\cdot 1\geq 0\cdot 1$ . Thus, $1\geq 0$ , 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