# The successor of 0 in an ordered ring

## 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!

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.
• Thanks! Looking at $\epsilon(1-2\epsilon)$ seems somewhat "magic" to me--is there a more concrete way of arriving there? Commented Jun 5, 2023 at 15:02
• 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. Commented Jun 5, 2023 at 15:16
• 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 $\epsilon^2=0$. Commented Jun 5, 2023 at 15:18
• Cool, thanks again! I forgot how important it is to start with combining $\epsilon$ and $1-\epsilon$ when dealing with idempotents Commented Jun 5, 2023 at 17:12