Is an elliptic curve over $\mathbb{F}_p$ order preserving for the points $(x,y) \in \mathbb{Z}_p$? Let $E\colon y^2=x^3+ax+b$ be an elliptic curve, and consider its realisation over the finite field of prime order $p$: $E(\mathbb{F}_{p})$.

Then if $0&lt;a,b$ is the following true?

$$\forall x,y\in[0,p) \text{ s.t. }E(x)=y \text{ : }(x_i&lt;x_j)\implies E(x_i)&lt;E(x_j)$$
The question as currently stated is true if we assume the equation takes place in $\mathbb{Z}$, since all the values are small integers.

Proof: If $x&lt;x'$, then $x^3&lt;(x')^3$ and $ax&lt;ax'$, so
$$
E(x)= x^3 +ax+b &lt; (x')^3 +ax'+b = E(x')
$$



The problem with trying to answer the more general issue you appear to be considering is working out what it means to say $x&lt;y$ for $x,y\in\mathbb{F}_p$. There isn't a reasonable ordering on elements of a finite field, or indeed on $\mathbb{Z}/p\mathbb{Z}$.  Intuitively, one might wish that $x&lt;y$ if and only if $x&lt;y$ when considered as integers. However, this doesn't work because there is no mapping that will respect the required properties. Consider the following [attempt] at defining this:


  For any $x\in\mathbb{Z}$, let $\bar{x}$ denote the image of $x$ under the projection into $\mathbb{F_p}$. Then,[try to] define an ordering $\prec$ on $\mathbb{F}_p$ by $\bar{x}\prec\bar{y}\iff x&lt;y$.


The biggest problem with this is that it isn't even well defined. For example, let $a,b\in\mathbb{F}_7$, with $a=\bar{3}$ and $b=\bar{6}$. Then since $3&lt;6$ we deduce $a\prec b$, but since $a=\bar{3}=\bar{10}$ we must also have $b\prec a$ (since $6&lt;10$). This would imply $a=b$, which is clearly false.

For more information, this wikipedia page might help.