In the second half of the proof of Proposition 7.9. in Atiyah-Macdonald (Introduction to Commutative Algebra) we have a field $F = k[y_1,...,y_s]$ as a $k$-algebra with each $y_j = f_j/g_j$ with $f_j$ and $g_j$ polynomials in $x_1,...,x_r$ algebraically independent over $k$.
Now (by extension of Euclid's proof) we have infinitely many irreducible polynomials in $k[x_1,...,x_r]$ so there exists $h$ coprime to $g_1,...,g_s$ for example $h = g_1\cdots g_s +1$.
Now its inverse $h^{-1}$ is in $F$, because $F$ is a field, but it is not supposed to be a polynomial in $y_1,...,y_s$ which leads to a contradiction because then $F$ isn't a finitely generated $k$-algebra, so we have $F$ to be a algebraic extension of $k$.
But why can't $h^{-1}$ be expressed as a polynomial in $y_1,...,y_s$? As a concrete example I understand why it is not possible for $h = x + 1$, but I don't understand the general case.
Thanks to the comment of Aaron, I see it now:
Let f be a polynomial in the $y_j$, $f = \sum{a_j*(f_j/g_j)^{d_j}}$, then you can get everything on a common denominator by expanding with powers of the $g_j$ , respectively $g_j^{d_j}$, to get all the fractions to have a common denominator, $\prod{g_j^{d_j}}$, and so the denominator of every element in F has to be divisible by the $g_j$, which the denominator of $h^{-1}$ is not by construction.