This may be a rather silly question, but I'm puzzled that (at least so far as I can tell) all finite fields use modular arithmetic. Is there no other way to construct a finite field than by defining addition and multiplication in terms of the modulo? Why is this so? Is there no other arrangement that will map a set to itself?
I've probably butchered terminology, but stay with me, I'm still rather new to this!
On
Finite fields are uniquely determined by their order, and every finite field of prime order $p$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$. The field $\mathbb{F}_q$ of higher prime power order $q = p^n$ with $n > 1$ is not $\mathbb{Z}/q\mathbb{Z}$, though, (which contains zero divisors); for example, the field $\mathbb{F}_4$ with elements $0, 1, \alpha, \alpha +1$ with $2x = 0$ for all $x\in \mathbb{F}_4$ and multiplication given by $\alpha^2 = \alpha + 1$.
If you're asking whether it's necessarily the case that a finite field $k$ has some integer $n$ with $nx = 0$ for all $x\in k$, then that's true; the minimum positive such $n$ is the characteristic of $k$. The proof is simply that $n.1$ must vanish for some $n$, or else $k$ would contain infinitely many distinct elements $1, 2, \dots$; and thus $n.x = (n.1).x = 0$ for all $x\in k$. (In general, an infinite field need not have nonzero characteristic; take $\mathbb{Q}$, for example.)
On
Every finite field has size $p^n$, where $p$ is a positive prime and $n$ is a positive integer. Moreover, any two finite fields of the same size are isomorphic.
It's possible to show for any positive integer $n$, there is an irreducible polynomial $\pi(x)$ of degree $n$ in $\mathbb Z_p[x]$. It is then fairly straightforward to show that $\mathbb Z_p[x]/\pi(x)$ is a finite field of size $p^n$. So all finite fields can be constructed in this manner, using modular arithmetic.
All fields of prime order are isomorphic to the fields $\mathbb Z_p$. However, there are also other finite fields. Given $p$ prime and $n \in \mathbb N$, there is a field $\mathbb F_{p^n}$ of $p^n$ elements - this is quite advanced to prove, and some of these fields can be tricky to construct. We can also show that all finite fields have this form.
We can construct $\mathbb F_4$ as follows: take $\mathbb Z_2$ and add an element $\omega$ such that $\omega^2 + \omega + 1 = 0$ and $\omega^3 = 1$. We can show that the set of elements generated by $\mathbb Z_2 $ and $\omega $ form a field. However, it will not use modular arithmetic, as $\omega$ is not an integer.
$\mathbb F_4$ has elements $\{0,1,\omega, \omega^2\}$; note that $\omega + 1 = - \omega^2 = \omega^2$ (as $1 = -1$).
Recall that $\mathbb Z_p$ is a field if and only if $p$ is prime. There is no "modular arithmetic" field of order $4$.