# Can there be a Finite Field That Does Use Not Modular Arithmetic?

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!

• en.m.wikipedia.org/wiki/Finite_field see the classification section. The reason you see finite fields constructed using modular arithmetic is that all fields of size $p$ are isomorphic to $\mathbf{Z}/p\mathbf{Z}$ – A.E Sep 10 '14 at 23:54
• The arithmetic operations of finite fields of order $2^{2^n}$, i.e. a very limited subcase, CAN be defined differently. This trick is due to Conway. The definition is a cute extension of the Nim-game, but not very useful for, say, computer implementations. I'm a bit curious. Do you have something against modular arithmetic? – Jyrki Lahtonen Sep 11 '14 at 6:47

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 $4$.

• But the additive group on your $\mathbb F_4$ is isomorphic to $\mathbb Z_2 \times \mathbb Z_2$, so I would argue that it does use modular arithmetic. Similarly, $\mathbb F_{p^n}$ uses addition mod $p$. – Santiago Canez Sep 11 '14 at 0:03
• It is isomorphic as an additive group to $\mathbb Z_2 \times \mathbb Z_2$, but the latter is not a field, and has no corresponding multiplicative group. You could argue it uses modular addition, but certainly not modular multiplication – Mathmo123 Sep 11 '14 at 0:05
• You can transport the multiplication on your $\mathbb F_4$ to $\mathbb Z_2 \times \mathbb Z_2$, thereby turning $\mathbb Z_2 \times \mathbb Z_2$ into a field. Yes, this multiplication is not "multiplication mod $p$'', but modular arithmetic is still present in the additive construction regardless. EDIT: Seems we're on the same page ;) – Santiago Canez Sep 11 '14 at 0:08
• I think the fundamental point is that whilst there will be some modular arithmetic present in $\mathbb F_4$, that would not be the correct way to think about this field. – Mathmo123 Sep 11 '14 at 0:11
• @littleO - of course. I was assuming that the OP was referring to regular modular arithmetic over the integers - primarily because when fields are first introduced, the examples given are usually $\mathbb R, \mathbb Q, \mathbb C$ and $\mathbb Z_p$. A beginner to the subject would not have seen the fields $\mathbb F_q$. – Mathmo123 Sep 11 '14 at 0:20

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.)

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.