K contains the elements 0 (take f = 0 and g = 1) and 1 (take f = 1 and g = 1).

K is closed under addition:

f(a)/g(a) + h(a)/l(a) = (f(a)l(a) + g(a)h(a)) /g(a) l(a);

here the polynomials fl + gh and gl are used.

K is closed under multiplication:

f(a)/g(a) * h(a)/l(a) = f(a)h(a)/g(a) l(a);

here the polynomials fh and gl are used.

The additive inverse of f(a)/g(a) is -f(a)/g(a); here the polynomials -f and g are used.

Every nonzero element in K has its multiplicative inverse in K: the inverse of f(a)/g(a) is g(a)/f(a). Note that f(a) 0.

To show that K is the smallest field containing a, we note that any field containing a also contains f(a) for every polynomial f L[X], since f(a) arises by repeated addition and multiplication starting from a and elements of L.

But if the subfield contains f(a) and g(a), with nonzero g(a), then it also contains the product of f(a) and the inverse 1/g(a), that is, the quotient f(a)/g(a). In conclusion, the subfield must contain K.