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?

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 $$1$$.

• Thank you. Very well explained. Oct 15, 2019 at 21:29

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.

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

• Slightly confused because we have not studied groups (in fact, this was totally skipped), but why is $$[X+X^2] = [X+X^2 - p(X)]$$? Oct 15, 2019 at 21:25
• $[q(X)]$ is a symbol of a residue class modulo $p(X)$. This means $[q(X)] = [q(X) + kp(X)]$ for all $k \in \mathbb Z$. This is the same as modular arithmetic in the integers: The elements in $\mathbb Z/n\mathbb Z$ ("modulo $n$") can be represented as $[x]$ where $[x] = [x+kn]$ for all $k\in \mathbb Z$: If we do computations modulo $4$ then $[1] = [5] = [-3] = [9] = ...$. Oct 16, 2019 at 6:53
• And I have to apologize, it is sometimes different to judge what people know already and what they do not, but please feel free to ask if you have further questions about this! Oct 16, 2019 at 6:54

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 $$l$$.