Can a finite set with a non prime number of elements be a field?

Question

I understand that as typically defined (using modular arithmetic) finite fields require a prime number of elements.

But I recall hearing someone say that if you modify the way addition and multiplication is defined on a set with a non-prime number of elements, say 4 elements, then it could still be a field.

Is this true? How would this set look and how would you define the addition and multiplication?

Answer 1

For any prime $p$ and integer $k\geq 1$, there is, up to isomorphism, exactly one field of order $p^k$.

In the case of $2^2$ elements, one usually denotes the elements as $0,1,x,x+1$ (or something similar), with addition done modulo $2$. The multiplication table looks like this: $$ \begin{array}{|c|cccc|}\hline &0&1&x&x+1\\\hline 0&0&0&0\\1&0&1&x&x+1\\ x&0&x&x+1&1\\ x+1&0&x+1&1&x\\\hline\end{array} $$ In general, you can find a multiplication table the following way: Start with $\Bbb Z_p$, the integers modulo $p$ (also known as the field with $p$ elements), and an irreducible polynomial $f$ of degree $k$ with coefficients in $\Bbb Z_p$. Then take the polynomial ring $\Bbb Z_p[x]$, and divide out by the ideal generated by $f$. Any element of our $p^k$-element field will correspond to a polynomial of degree less than $k$, with addition as normal. Multiplication is defined by reducing modulo $f$.

In our example, we have $\Bbb Z_2$, $k=2$ and $f(x)=x^2+x+1$. The elements are as given above, and addition is done as for regular polynomials with coefficients in $\Bbb Z_2$. As for multiplication, let's look at $x(x+1)$ as an example. With regular polynomials we have $x(x+1)=x^2+x$. Then reducing modulo $f$ basically means either

• Subtract multiples of $f$ until the degree is lower than $k=2$.
• $f(x)=0$ means $x^2=x+1$. Substitute this, repeatedly if necessary, until the degree is lower than $k=2$.

In either case, $x^2+x$ is reduced to $1$.

Answer 2

In fact for any power $p^n$ of a prime $p$ you can find a finite field, usually denoted by $\mathbb F_{p^n}$ or $GF(p^n)$. You can construct these by finding an integer polynomial $p \in \mathbb Z[X]$ of degree $\deg p = n$ that is irreducible over $\mathbb Z_p := \mathbb Z/p\mathbb Z$.

Then $\mathbb F_{p^n} := \mathbb Z[X]/(p(X))$ (where $(p(X))$ denotes the principal ideal $p(X)\mathbb Z[X]$. One can even show that there is exactly one field of order $p^n$ (up to isomorphism) and that these are all the finite fields.

Example: For $p=n=2$ (so $p^n=4$) there is exactly one such polynomial and it is $p(X) = X^2 + X + 1$. This means all elements in $\mathbb Z[X]/(p(X))$ are represented by the residue classes $[0], [1], [X], [1+X]$. Now we can actually do computations: E.g.

$$[X] \cdot [1+X] = [X+X^2] = [X+X^2 - p(X)] = [-1] = [1]$$

Answer 3

Actually, one proves the number of elements in a finite field is always some power $p^k$ of a prime $p$.

Conversely, for any natural number $k$ and any prime, there exists a field, denoted $\mathbf F_{p^k}$, with $p^k$ elements, and this field is unique up to a field isomorphism. Furthermore, the field $\mathbf F_{p^k}$ is (isomorphic to) a subfield of the field $\mathbf F_{p^l}$ if and only if $k$ divides $l$.