Equivalence relation on ordered field Let p be a prime number. Define an equivalence relation ∼ on Z as: n ∼ m if n−m is divisible by p. For n ∈ Z, let [n] be the equivalence class of n with respect to this equivalencerelation. LetZp ={[n]:n∈Z}. Define+:Zp×Zp →Zp and·:Zp×Zp →Zp by
[n] + [m] = [n + m] and [n] · [m] = [nm].
(ii) Show that (Zp, +, ·) is a field. You may use the fact that if k1, k2 are coprime nonzero integers, then there are integers r1, r2 such that r1k1 + r2k2 = 1.
(iii) Show that there is no way one can define a relation < on Zp so that (Zp,+,·,<) is an ordered field.
Any hints for ii) and iii)?
 A: Here are some hints:
For (ii), you are meant to show the field axioms are satisfied. This basically says that you can do arithmetic as freely as with, say the rational numbers.
Explicitly, for every $x,y,z \in \mathbb{Z}/p$:


*

*$x+y = y+x$

*$x \cdot y = y \cdot x$

*$(x+y)+z = x+(y+z)$

*$(x \cdot y) \cdot z = x \cdot (y \cdot z)$

*$x + [0] = x$

*$x \cdot [1] = x$

*for every $x$, there is a $-x$ such that $x+(-x) = [0]$

*for every $x \not = 0$, there is a $x^{-1}$ such that $x \cdot x^{-1} = [1]$

*$x \cdot (y+z) = (x \cdot y) + (x \cdot z)$
Can you show these facts?

For (iii), recall an Ordered Field is a field (so it satisfies the above) equipped with a relation $<$ satisfying properties which say $<$ isn't a bad choice of symbol:


*

*$x < y$ and $y < z$ implies $x < z$

*exactly one of $x = y$, $x < y$, $y < x$ holds for any $x,y$.


We also want to know that $<$ is compatible with the field structure, namely the following (again, obvious) equation should hold:


*

*$x < y$ implies $x + z < y + z$

*$0 < x$ and $0 < y$ implies $[0] < x \cdot y$
Can you see how to use $x < y$ implies $x + z < y + z$, coupled with $$\underbrace{x + x + x + \ldots + x}_{p \text{ times}} = 0$$ to show that no such order can exist? 
Feel free to comment on this answer if you have more specific issues.

I hope this helps ^_^
