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I'd like to prove the following statement:

Given a well-ordered ring $A\neq \emptyset$ with identity $I$ and a homomorphism $\phi: \mathbb{Z}\to A$ such that $\phi(n) = nI$, then the set

$$\left\{ a\in A: 0<a<I \right\}$$

is empty.

If $\phi$ was surjective (or bijective), then this would obviously be true. (So, I suspect that that's the case.) But, how could I then prove that?

J. W. Tanner
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Rodrigo
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  • Exactly. Thanks for pointing that out. I've corrected it. – Rodrigo May 17 '20 at 12:10
  • Must the well-ordering obey any rules related to the ring operations? If not, you have the ring of Gaussian integers $\mathbb{Z}[i]$, with a spiral ordering like $0 < i < 1+i < 1 < 1-i < -i < \cdots$ – aschepler May 17 '20 at 12:29
  • @aschepler Would it be a ring if it didn't obey the rules defined for rings? Can we say that $i<1$? – Rodrigo May 17 '20 at 13:03
  • Given that well-ordering of the set ${a+bi ; a,b \in \mathbb{Z}}$ independently of the ring, sure we can have $i<1$. But it would be very reasonable to say that "well-ordered ring" means "ordered ring" plus "the total ordering is also a well-ordering", not just "ring" plus "well-ordering", just as "ordered ring" means more than "total order" plus "ring". – aschepler May 17 '20 at 13:08
  • Same as in this proof in the linked dupe. – Bill Dubuque May 17 '20 at 21:21

1 Answers1

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We'll argue by contradiction.

Suppose that the set $S=\{x \in A :0<a<I\}$ is non-empty. Then, since $A$ is well-ordered, $S$ has a minimum element, let's call it $m$.

We know that $0 < m < I$, which means that $0 < m^2 < m < I$.

But now we found an element $m^2 \in S$ that is less than the minimum $m$, contradiction.

Thus, the set $S$ must be empty.

J. C.
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  • Is $m^2<m, \forall 0<m<I$ true in for all rings? If so, why? – Rodrigo May 17 '20 at 12:38
  • What properties do you require for an ordering of a ring? This result should follow from them, but the exact details depend on the way things are defined. – J. C. May 17 '20 at 12:44
  • I believe I follow the most widespread definition of ordered ring: $a+b,ab\in A^+, \forall a,b\in A^+$ and that $\forall 0\neq a\in A$ we either have $a\in A^+$ or $-a\in A^+$. – Rodrigo May 17 '20 at 13:01
  • @J.C. I wonder how we can conclude from the axioms of partially ordering that $a<b$ implies $ac<bc$? For example, in the ring $\mathbb Z/n\mathbb Z$, it seems to be not valid. –  May 17 '20 at 13:05
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    @Rodrigo We know that $m \in A^+$ and $I-m \in A^+$, hence $m(I-m) = m - m^2 \in A^+$, which means $m^2<m$. – J. C. May 17 '20 at 13:09
  • @FranzNietzche As far as I know, when speaking of ordered rings we usually require some properties like those in Rodrigo's comment, not just a bare partial order defined on the ring. In that sense $\mathbb{Z}/n \mathbb{Z}$ is not an ordered ring. – J. C. May 17 '20 at 13:11
  • Could we have $0<m$ but $m^2=0$? – aschepler May 17 '20 at 13:14
  • @aschepler The set of positive elements is closed under multiplication, by definition, so that cannot happen. – J. C. May 17 '20 at 13:27
  • @J.C. Hmm, with Rodrigo's definition yes. But does that also follow from Wikipedia's definition $a \leq b \Rightarrow a+c \leq b+c$ and $0 \leq a \land 0 \leq b \Rightarrow 0 \leq a+b$? – aschepler May 17 '20 at 13:32