I'm trying to prove the following:
Given a finite field $\mathbb{F}_p = \{1,2,...,p-1\}$ with $p$ elements and $p$ is prime, defined as follows: both addition and multiplication are defined as the remainder of the outcome when divided by $p$. Prove that $\mathbb{F}_p$ is not ordered.
I'm specifically curious if induction would work by first considering $\Bbb{F}_p$ is ordered => there exists $P$ s.t. for all $x \in \Bbb{F}_p$ one of the following is true: $x = 0$, $x \in P$, or $-x \in P$. And $x,y \in P \Rightarrow x+y, xy \in P$
consider $1 \in \Bbb{F}_p$. Either $1$ or $-1$ is in $P$.
if $1$ is in $P$ then $1+1$ is also in $P$, and $1+1+1$, etc... until every element of $\Bbb{F}_p$ is in $P$ which would then force a contradiction $b/c$ if every value of $\Bbb{F}_p$ is in $P$ then for any given element, $z$, there are no value in $\Bbb{F}_p$ remaining to be $-z$.
If $-1$ is in $\Bbb{F}_p$, then $-1 = p - 1$, and I can do pretty much the exact same process as above $p-1$ is in $P$ => $(p-1) + (p-1) = p-2$ is in $P$, etc...
This obviously makes me want to induct, however I was pretty certain I could only do induction on a finite set if it's also well founded, and I can't find an R-minimal element in any subset of $\Bbb{F}_p$.
What does it mean to be ordered? Any nonzero characteristic is incompatible with being ordered, because $x+x+\ldots x = 0$.