How can a finite field have a cardinality that is not prime if every nonzero element is a unit?

Question

If a field requires every element that is not the additive inverse to have a multiplicative inverse, then how can a field have a cardinality that is not prime?
For instance 27?

So whether it is integers mod 27, or some direct product, there will be a nonzero zero divisor, hence not a unit. Does that not disqualify it from being a field?

Answer 1

Your argument shows that any field that is a quotient of the ring of integers must have a prime number of elements. Another way of saying this is that any finite field that has a single additive generator must have a prime number of elements, or more generally that the additive order of any individual nonzero element is prime.

However, not all finite fields are necessarily additively generated by a single element. The smallest example is $\Bbb Z_2[x]/(x^2+x+1)$, which has four elements. Here every element has order 2, and the additive group is the Klein Four group $\Bbb Z_2 \oplus \Bbb Z_2$. The multiplicative group of units is cyclic of order 3 in this case.

Answer 2

Take the matrix $$ A = \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 0 \end{array} \right) $$ and take the ring of all $$ xI + y A + z A^2 $$ with $x,y,z \in \mathbb Z / 3 \mathbb Z$

This is a field as $x^3 - x - 1$ is irreducible in the base field.