Showing that in a well ordered ring homomorphic to $\mathbb{Z}$, there are no elements between zero and the ring's identity.

Question

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?

Answer 1

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.