Show that there is no finite ordered integral domain.

Question

Show that there is no finite ordered integral domain.

I know the definition of integral domain and an ordered set, but how do I prove that no such integral domain can be finite? Also, i think $\mathbb{Z}_2$ is a finite integral domain and as well as ordered,since $0$ is the smallest element. So how to prove this fact isn't clear to me. Can someone please help? Thanks in advance.

Answer 1

A finite ring cannot be ordered (domain or not). If $R$ is a ring with $1$, with $n$ elements,then by Lagrange $n=0$. Now, notice that $1>0$ or $-1>0$ (the standard definition asks for $1>0$, but we don't need it here).

But $0=1.1+\cdots+1.1=(-1)(-1)+\cdots+(-1)(-1)$ is then positive in any case, contradiction.

Answer 2

I will use the definition of ordered integral domain in the book Number Systems and Foundations of Analysis by Elliot Mendelson.

First let $(D,+,\times)$ be a integral domain, then this integral domain with an binary relation $<$ on $D$ satisfying:

• (O1): $x \not\lt x$ (Irreflexivity)
• (O2): $[x \lt y \land y \lt z] \Rightarrow y\lt z$ (Transitivity)
• (O3): $[x < y \lor x=y \lor y < x]$ (Trichotomy)
• (O4): $x<y \Rightarrow x+z < y+z$
• (O5): $[x<y \land 0<z] \Rightarrow x\times z < y \times z$

Then $(D,+,\times,<)$ is an ordered integral domain (with the stipulation that $1 \neq 0$).

Now we proceed by contradiction, assume that $D$ is a finite set, then it is impossible to form a strictly increasing sequence on $D$, from this follows that $D$ have some greatest element $m$ such that $(\forall x)([x \in D \land x \neq m] \Rightarrow x < m)$

We need to derive the simple fact that $x \neq 0 \Rightarrow 0<x^2$, thus first assume $x \neq0$, then by (O3) we have $0<x\lor x<0$.

Considering $0<x$ we can apply (O5) which gives $[0<x \land 0<x] \Rightarrow 0\times x < x \times x$, that is $0<x^2$.

Now considering $x<0$, apply (O4) using $-x$ as $z$, which gives $x+(-x) < 0+(-x)$, which is $0<-x$, now we apply (O5), $[0<-x \land 0<-x] \Rightarrow 0\times -x < -x \times -x$ from which follows again $0 < x^2$.

Since we stipulated that $1 \neq 0$, we have now that $0 < 1^2$, but as $1 \times 1 = 1$, it is just $0 < 1$, now apply (O4) with our greatest element $m$ in place of $z$ and we got $0+m < 1+m$ which is $m<1+m$, since $D$ is closed under addition we got $1+m \in D$ which contradicts the fact that $m$ is the greatest element.

Therefore, its possible to create a strictly increasing sequence with the element of $D$ since the set dont have a greatest element, then if an integral domain have an order relation satisfying those five properties it cant have a finite underlying set.