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

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

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 $1$.