Let \(\langle A,+,\cdot\rangle\) be a commutative ring with identity \(1\) and let \(<\) be a linear order on \(A\) such that for all \(a,b,x\) in \(A\) \[\begin{align*} \text{(I)}\quad &a<b\Rightarrow a+x<b+x,\\ \text{(II)}\quad &a<b,\ 0<x\Rightarrow a\cdot x<b\cdot x. \end{align*}\]
Prove that \(\langle A,+,\cdot\rangle\) is an integral domain.
Let \(A^+=\{a\in A:0<a\}\). Prove the following:
\(A^+\) is closed under multiplication and addition.
If \(a\in A\), then exactly one of the following holds: \(a\in A^+\), \(-a\in A^+\), \(a=0\).
\(1\in A^+\).
Solution
We use the standard convention implicit in the problem that the identity is nonzero, \(1\ne0\).
1. For every \(a\in A\), exactly one of\[ a>0,\qquad a=0,\qquad -a>0 \]holds.
By linearity of the order, exactly one of \(a>0\), \(a=0\), or \(a<0\) holds. If \(a<0\), adding \(-a\) to both sides by (I) gives \[ 0<-a. \] Conversely, if \(-a>0\), adding \(a\) gives \(a<0\). Hence the three alternatives above are exactly the trichotomy alternatives, so precisely one holds. This proves part (b)(ii).
2. The set\[ A^+=\{a\in A:0<a\} \]is closed under addition.
Let \(a,b\in A^+\). From \(0<a\), condition (I) with \(x=b\) gives \[ b<a+b. \] Since \(0<b\), transitivity yields \[ 0<a+b. \] Thus \(a+b\in A^+\).
3. The set \(A^+\) is closed under multiplication.
If \(a,b\in A^+\), then \(0<a\) and \(0<b\). Apply (II) to \(0<a\) with the positive multiplier \(b\): \[ 0\cdot b<a\cdot b. \] Hence \(0<ab\), so \(ab\in A^+\). Together with 2 this proves part (b)(i).
4. The ring \(A\) has no zero divisors.
Suppose \(ab=0\) and \(b\ne0\). By 1, either \(b>0\) or \(-b>0\); replacing \(b\) by \(-b\) if necessary, we may assume \(b>0\). If \(a\ne0\), then again by 1 either \(a>0\) or \(-a>0\). If \(a>0\), 3 gives \[ 0<ab=0, \] a contradiction. If \(-a>0\), then 3 gives \[ 0<(-a)b=-(ab)=0, \] again a contradiction. Therefore \(a=0\). Thus \(ab=0\) implies \(a=0\) or \(b=0\), so \(A\) has no zero divisors.
5. Therefore \(A\) is an integral domain.
By hypothesis \(A\) is a commutative ring with nonzero identity, and 4 shows that it has no zero divisors. This is exactly the definition of an integral domain. This proves part (a).
6. One has\[ 1\in A^+. \]
Since \(1\ne0\), 1 says that either \(1>0\) or \(-1>0\). Suppose \(-1>0\). By multiplicative closure from 3, \[ (-1)(-1)=1>0. \] Then both \(1\) and \(-1\) lie in \(A^+\), contradicting the exclusivity in 1 applied to \(a=1\). Hence \(-1\) cannot be positive, so \(1>0\). This proves part (b)(iii).