Hello
Here is the statement if Zariski lemma:
If K is a field that is finitely generated as a F algebra, then it is a finite field extension of F.
It is not exactly as the statement we proved today. Today we assumed that K=k(a_1...a_n) with a_i slgebraic.
Zariski assume that K=k[a_1...a_n] with a_i any element in K (but square braket!!!)
The proof if Zariski lemma is similar to what we have seen today. Is by induktion on n.
You have that K is a finite field extension of k(x_1) now you gave to show that x_1 is algebraic /k. This is the hard part involving the theory of Integral dependence.
Does this answer your doubts?
Mvh
Sofia