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 $|R|\eq 1$ 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 $|R|\eq 1$ , 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 $1\eq 0$ . Then $0<1$ .

Proof.

Suppose that $0\ot<1$ . 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.∎

单词	BasicFactsAboutOrderedRings
释义	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 $\|R\|\eq 1$ 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 $\|R\|\eq 1$ , 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 $1\eq 0$ . Then $0<1$ . Proof. Suppose that $0\ot<1$ . 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.∎
随便看	calculator and CAS support for various positional bases CalculatorTrigonometricFunctions calculator trigonometric functions CalculusOfSubgroupOrders calculus of subgroup orders CalculusOfVariations calculus of variations Calgebra C^-algebra CalgebraHomomorphismsAreContinuous C^-algebra homomorphisms are continuous CalgebraHomomorphismsHaveClosedImages C^-algebra homomorphisms have closed images CalgebraHomomorphismsPreserveContinuousFunctionalCalculus C^-algebra homomorphisms preserve continuous functional calculus CalgebrasHaveApproximateIdentities C^*-algebras have approximate identities CameronMartinSpace Cameron-Martin space CancellationIdeal cancellation ideal CancellationRing cancellation ring CancellativeSemigroup cancellative semigroup