# How can you construct a field of a given order?

Assume that we want to have a field of $n$ elements (say 8 for now). And we want to explicetly show the multiplication table for the field.

What is the most general procedure to construct the field and it's multiplication table?

(Since all fields of the same order are isomorphic we can assume the elements of the field are reals then do some map into our desired set later)

• The addition operation in a field of order $8$ is not the same as the addition of the integers mod $8$. – Josh B. Apr 1 '18 at 23:18
• In general, take the polynomials with coefficients mod p, then mod out by some irreducible polynomial. – Josh B. Apr 1 '18 at 23:21
• @JoshB. yes I meant reals, not integers, I messed up – Makogan Apr 1 '18 at 23:33
• I don't think the reals will work either. – Josh B. Apr 1 '18 at 23:35
• I am in big trouble then it seems :P – Makogan Apr 1 '18 at 23:38

## 1 Answer

To get a field with $q$ elements, characteristic $p$. Example: $p=2$, $q=8$.

Start with the field $k=\Bbb F_2$ with two elements. Find an irreducible polynomial of degree $3=\log_2(8)$. There are only two of these, so take $g(X)=X^3+X+1$. Then form the vector space over $k$ with the three basis elements $\{1,a,a^2\}$, where $a^3+a+1=g(a)=0$. Therefore $a^3=-a-1=a+1$, because in characteristic two, plus is minus. Note that you’ve divided the polynomial $X^3$ by $X^3+X+1$ to get the remainder $X+1$ (you don’t care about the quotient).

In principle this tells you everything possible about your field with eight elements, but in practice, I recommend strongly that you find a primitive element $R$ of the multiplicative group, i.e. an element whose powers exhaust the nonzero elements of your field. Here it’s easy: anything different from $1$ will do, for instance $R=a$. Then write down the powers of $R$: \begin{align} R^0&=1\\R^1&=a\\R^2&=a^2\\R^3&=a+1\\R^4&=a^2+a\\R^5&=a^2+a+1\\R^6&=a^2+1\\R^7&=1\,. \end{align} This gives you a log table of the elements of your big field $\Bbb F_8$: for instance if you want to find $(a^2+a)(a^2+a+1)$, you don’t need to do anything hard. That’s $R^4R^5=R^9=R^2=a^2$.