i=0 ai,kv i = 0. Recall that, since v is trancendental over K, the set {1, v, v, ...} is linearly independent over K. This implies that ai,k = 0 for all k > 0 (otherwise there would be a finite linear combination of powers of v which is zero, hence v would be algebraic over K). Thus each hi above is of the form hi(x) = ai,0 (a constant polynomial), namely hi ∈ K. But then (2) implies that v is ...