An operation preserving order on finite field

Question

Let $p\equiv 3(\mod 4)$ be a prime. How to show that there exists an order " $<$ " on the finite field $F:=\mathbb Z/p\mathbb Z$ such that the following three conditions hold :

(i) For every $a,b\in F$ , either $a=b$ , or $a<b$ , or $b<a$

(ii) For every $a,b,c\in F$ , $a<b \implies a+c <b+c$

(iii) For every $a,b,c\in F$ , $a<b$ and $0<c \implies ac<bc$

?

Answer 1

You can take advantage of basic facts about quadratic residues. Let $$Q=\{a^2\in F\mid a\neq0\}.$$ Then we know that

• $-1\notin Q$, here we need the assumption $p\equiv3\pmod4$,
• $Q$ is a subgroup of the multiplicative group, in other words $x,y\in Q\implies xy\in Q$,
• $|Q|=(p-1)/2$,
• Together the first and the third bullet imply that if $a\in F$ then exactly one of $a$ and $-a$ is an element of $Q$.

These properties of $Q$ suggest the following "order".

Definition. For $a,b\in F$, we define that $$a<b\Longleftrightarrow b-a\in Q.$$

Then

1. The fourth bullet implies that if $a\neq b$, then exactly one of the differences $a-b, b-a$ is an element of $Q$. This shows that our order relation has property (i).
2. Because $(b+c)-(a+c)=b-a$ the definition of $<$ immediately implies that our order relation has property (ii).
3. If we assume that $a<b$ and $0<c$, then both $c=c-0$ and $b-a$ are elements of $Q$. By the second bullet $c(b-a)\in Q$, but this implies $ac<bc$ verifying property (iii).

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I think it is not too difficult to see that this is the only possible order relation. Consider the set of positive elements $$ G=\{x\in F\mid 0<x\}. $$ Then

1. Property (i) implies that $0\notin G$.
2. Property (ii) implies that $a<b$ if and only if $0<b-a$ if and only if $b-a\in G$.
3. Property (iii) implies that $G$ is closed under products, and therefore a subgroup of $F^*$.
4. Items 1. and 2. together imply that for $a\neq0$ we have $a\in G$ iff $-a\notin G$.
5. Items 3. and 4. imply that $Q\subseteq G$ because either $0<a$ or $0<-a$ and $(-a)^2=a^2\in G$ in either case.
6. Item 4. says that $G$ has exactly $(p-1)/2$ elements, so item 5. tells us that $G=Q$.