Knowing that any ordered ring $A \neq \{0\}$ is infinite, can I say that if an ordered ring is finite, it must be $\{0\}$ ?
By this result - If an ordered ring $A$ has an upper bound, does it have to have a lower bound? - a ring with an upper bound or a lower bound is finite.
Then, to show that if ordered ring $A$ has upper or lower bound, then $A = \{0\}$, I just have to say that:
Upper or lower bound implies upper AND lower bound which means ring is finite. Ring is finite, so, by my first paragraph, it must be {0}
Is this right?
Let's suppose that the ordered ring $A$ has an upper bound (an element greater or equal than any other element in $A$) and that $A$ is not trivial.
Then there exist $a\in A$ such that $a\neq 0$. If $a>0$, then $0$ is not the upper bound. If $a<0$, then $-a>0$ so by the same reasoning, $0$ is still not the upper bound.
Let $u$ be the upper bound of the ring $A$. We know that $u>0$. This implies that $u+u>u$ because the order relation $>$ has to be compatible with addition. But this contradicts the fact that $u$ is the upper bound because we have found a greater element in the ring $A$.
This proves that the ordered ring has to be trivial if it has an upper bound.