If K is field then why K(x) is not finetly generated K algebra?

Answer 1

First suppose that K is algebraically closed. Then, if K(x) is finitely generated, look at the roots of the denominators of the generators. There will be at most finitely many such roots. But now, if r is anything in K which is not one of these roots, 1/(x-r) is not in the finitely generated algebra.

For the general case, I think looking at the algebraic completion of K will do the trick, but I haven't checked it yet.