# basic facts about ordered rings

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

###### Lemma 1.

If $a,b,c\in R$ with $a, then $a+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 $n>1$.

###### 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 BasicFactsAboutOrderedRings 2013-03-22 16:17:21 2013-03-22 16:17:21 Wkbj79 (1863) Wkbj79 (1863) 11 Wkbj79 (1863) Result msc 06F25 msc 12J15 msc 13J25 MathbbCIsNotAnOrderedField