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The Setup

I am investigating properties of the following definition of ordered rings:

Let $R$ be a ring. We say $R$ is an ordered ring under the total order relation $\leq$ if, for all $a,b,c \in R$, $$ (1). a \leq b \text{ implies } a+c \leq b+c, \text{ and} $$ $$(2). 0 \leq a \text{ and } 0\leq b \text{ implies } 0\leq ab. $$

We can also say $R$ is strictly ordered by $\leq$ if the above conditions hold with the strict order $<$ replacing $\leq$ throughout. It's been noted on this post and other places that the two notions of strict/non-strict are separate.

I found and proved the following theorem while looking at the order topology of $R$:

Let $(R,\leq)$ be a (nonzero) ordered ring with unity. If $0$ has an immediate successor $\epsilon \in R$, then $\epsilon\leq1$, and either $\epsilon^2 = 0$ or $\epsilon^2 = \epsilon$.

The proof is simple: we must have $\epsilon \leq 1$ because otherwise $1 \in (0,\epsilon)$, a contradiction. But then $0 < \epsilon \leq 1$ implies $0 \leq \epsilon^2 \leq \epsilon$ upon multiplication by $\epsilon$.

The Question

In the theorem, $\mathbb{Z}$ is a simple example of the latter case, because the "$\epsilon$" is just $1$. One can show that this is always so in a discrete strictly ordered ring. I also have an example of the former case: the dual integers $\mathbb{Z}[\epsilon] = \{a+b\epsilon : a,b\in\mathbb{Z}, \epsilon^2 = 0\}$, together with the lexicographic order. Here $\epsilon$ is the immediate successor of zero.

My question is this: Is there an ordered ring $R$ (necessarily non-strictly ordered) such that $0$ has an immediate successor $\epsilon \in R$ with $\epsilon^2 = \epsilon$ and $\epsilon \neq 1$?

My intuition tells me that the answer is no, but I don't have any evidence to back it. Any insight is appreciated!

Nick F
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1 Answers1

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In fact, it is impossible for an ordered ring to have any element $\epsilon\neq 0,1$ such that $\epsilon^2=\epsilon$. Note first that $\epsilon(1-2\epsilon)=-\epsilon$ so $1-2\epsilon$ must be negative and thus $\epsilon>1-\epsilon$. But note that $(1-\epsilon)^2=1-\epsilon$ and $1-\epsilon\neq 0,1$ as well so the same argument applies with $1-\epsilon$ in place of $\epsilon$ to conclude that $1-\epsilon>\epsilon$. This is a contradiction.

Eric Wofsey
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  • Thanks! Looking at $\epsilon(1-2\epsilon)$ seems somewhat "magic" to me--is there a more concrete way of arriving there? – Nick F Jun 05 '23 at 15:02
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    I came up with it by just messing around. But here's a more natural motivation: it's $\delta-\epsilon$, where $\delta=1-\epsilon$. The elements $\delta$ and $\epsilon$ are both idempotent with $\epsilon\delta=0$. The result is that $\epsilon(\delta-\epsilon)=-\epsilon$ whereas $\delta(\delta-\epsilon)=\delta$, so $\delta-\epsilon$ has to be both negative and positive. – Eric Wofsey Jun 05 '23 at 15:16
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    Or for another way to think about it, since $\delta\epsilon=0$, if $0\leq\delta\leq\epsilon$ then $0\leq\delta^2\leq\delta\epsilon=0$ so $\delta^2=0$, and similarly $\epsilon\leq\delta$ would give $\epsilon^2=0$. – Eric Wofsey Jun 05 '23 at 15:18
  • Cool, thanks again! I forgot how important it is to start with combining $\epsilon$ and $1-\epsilon$ when dealing with idempotents – Nick F Jun 05 '23 at 17:12