Ideals of an ordered ring

Question

Suppose $R$ is a strictly ordered (non-commutative) ring, in particular $ab > 0$ for any $a,\, b > 0$, that is also discrete in that there are no elements between $0$ and $1$.

Now consider a two-sided ideal $I$ that does not contain $1$. For any two non-zero elements $a,\, b \in I$, we have that $\lvert ab \rvert > \lvert a \rvert,\, \lvert b \rvert$ as their absolute values are greater than $1$. Does it then follow that $\bigcap_{n \in \mathbb{N}} I^n = \{ 0 \}$ as each successive power of $I$ contains elements with larger and larger absolute values?

And therefore we can embed $R$ in its $I$-adic completion?

Answer 1

Here's an example showing that the answer is negative.

Consider the monoid algebra $\mathbf{Z}[\mathbf{R}_{\ge 0}]$. It thus consists of finitely supported sums $q=\sum_{t\ge 0}q_tX^t$. Say that such an element is positive if it nonzero and $q_u>0$ where $u=v(q)$ is the maximum of its support.

This defines a total order, compatible with the ring structure, and $1$ is the min of all positive elements.

Let $I$ be the set of elements $q$ with $q_0=0$. Then $I$ is an ideal and satisfies $I=I^2$. So $\bigcap I^n=0$ fails.