Proof that the nonnegativy of squares does not follow from the rest of the prepositive cone definition

Question

A prepositive cone $P$ is a subset of a field $F$ defined as exhibiting the following properties:

1. Additive closure: $p+q\in P$;
2. Multiplicative closure: $p\cdot q\in P$;
3. No negative identity: $-1\notin P$;
4. Nonnegativity of squares: if $x\in F$ then $x\cdot x\in P$.

My intuition wants me to believe the fourth axiom is redundant. I wanted to try to prove this by contradiction, but all my ideas involved assuming that a field is ordered. If a field is ordered, then $x^2\ge0$, but this seems like circular reasoning.

How can I prove that 4 does not follow from 1 through 3?

Answer 1

There are a lot of examples of subsets of fields that fulfill 1 to 3 but not 4. $$ F=\mathbb{R}\;\;,\;\; P=[1,\infty ) \;\;,\;\;\text{Axiom 4 fails for }x\in(-1,1) \\ F=\mathbb{R}\;\;,\;\; P=(0,\infty ) \;\;,\;\;\text{Axiom 4 fails for }x = 0 \phantom{xxxx}\\ $$ There are also more exotic examples like $F=\mathbb{Q}$ and $P$ is the set of all non-negative fractions with denominators that are not divisible by a given prime $p.$