It's actually not that hard. An extension field of field F comes from an irreducible polynomial f(x), irreducible meaning that in any factorization f(x)=g(x)h(x

It's actually not that hard. An extension field of field F comes from an irreducible polynomial f(x), irreducible meaning that in any factorization f(x)=g(x)h(x), g or h is a constant. Say f(x)=x^n+cx^{n-1}+...+d. The extension field is then denoted by F[x]/(f(x)), which means that every polynomial with coefficients in F represents an element of the extension field, and you set f(x)=0, so anywhere you see x^n, you can replace it with -(cx^{n-1}+...+d). For example, the complex numbers are represented in this notation as an extension over the reals R by R[x]/(x^2+1). In this notation, "x" is essentially acting the same as "i" in the conventional complex-number notation, because anywhere you see x^2, you can replace it with -1. Multiplication works by multiplying polynomials as usual, then reducing the x^n terms to lower-degree polynomials. Inversion of a polynomial g(x) with degree less than n works by solving the equation g(x)h(x)=j(x)f(x)+1. Since g(x) has lower degree than f(x), and f(x) is irreducible, they are coprime, so appropriate h(x) and j(x) can be found using Euclid's algorithm (https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm#Simple_algebraic_field_extensions https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm#S...). OK, maybe that last bit is a little advanced, but Euclid's algorithm is also very simple arithmetic.