Lectures on the Algebraic Theory of Fields - Tata Institute of ...

5. Transcendental extensions 13Let K/k be a transcendental extension and let S , S ′ be two subset ofK with the propertiesi) S algebraically independent over kii) S ′ algebraically independent over k(s)Then S and S ′ are disjoint subsets of K and S US ′ are algebraicallyindependent over k. That S and S ′ are disjoint is trivially seen. Let nowZ 1 ,...Z m ∈ S and Z ′ 1 ,...Z′ n ∈ S′ be algebraically dependent. This willmean that there is a polynomial f ,f=f (x 1 ,..., x m+n )in m+n variables with coefficients in k, such thatf (Z 1 ,...,Z m , Z ′ 1 ,...Z′ n )=0.Now f can be regarded as a polynomial in x m+1 ,..., x m+n with coefficientsin k(x 1 ,..., x m ). If all these coefficients are zero then Z 1 ,...Z m ,Z1 ′,...Z′ n are algebraically independent over k. If some coefficient is 0, then f (Z 1 ,...Z m , x m+1 ,..., x m+n ) is a non zero polynomial overk(S ) which vanishes for x m+1 = Z1 ′,... x m+n= Z n ′ which contradicts thefact that S ′ is algebraically independent over k(S ). Thus f = o identicallyand our contention is proved. 15The converse of the above statement is easily proved.An extension field K/k is said to be generated by a subset M of K ifK/k(M) is algebraic. Obviously K itself is a set of generators. A subsetB of K is said to be a transcendence base of K if1) B is a set of generators of K/k2) B algebraically independent over k.If K/k is transcendental, then, it contains algebraically independentelements. We shall prove that K has a transcendence base. Actuallymuch more can be proved as in