How to show that $\mathbb{Q}_p$ cannot be ordered?

Question

I've seen many references on this site and others to the fact that the $p$-adic numbers cannot be ordered, but the closest I've seen to a proof of this is Wikipedia's vague reference to "$\mathbb{Q}_2$ contains a square root of −7 and $\mathbb{Q}_p$ ($p > 2$) contains a square root of $1 − p$" on the page for "Ordered Field". How does this statement imply that $\mathbb{Q}_p$ cannot be ordered? Or is there a more intuitive approach to the proof?

Answer 1

If $\alpha$ is a square root of $-7$, then $-7$ is a square of $\alpha$.

Answer 2

All nonzero squares in an ordered field must be positive. To see this, suppose $F$ is an ordered field and let $0\ne x\in F$. If $x$ is positive, then clearly $x^2$ must be positive. If $x$ is negative, then $0=x+(-x)<0+(-x) =-x$ so $-x$ is positive, thus $x^2=(-x)^2$ is positive.

Answer 3

It can be ordered, it just won't obey.

An ordered field has a set of positive numbers $P,$ such that $1 \in P,$ the sum or product of any two elements of $P$ is once again in $P.$ Finally, there is trichotomy, given a field element $x,$ exactly one of three is true: $x = 0,$ or $x \in P,$ or $-x \in P.$

These numbers $-7$ or $1-p$ would be "negative," as they are $-1$ times sums of copies of $1.$ But, if they are squares, they are squares of something positive, as $(-x)^2 = x^2,$ a contradiction.

The Artin-Schreier Theorem says that a field an be ordered if and only if $-1$ is not the sum of finitely many squares.