Let $F$ be a field, let $k$ be a subfield of $F$, then $F$ is a finitely generated $k$-algebra iff it's a finitely generated $k$-module.
The $\impliedby$ part is obviuos, so I'm only interested in the $\implies$ part.
I don't want to use the NSS Theorem because I want to use this result to prove the (weak) NSS.
This is a sketch of a proof I came up with.
Let $F$ be a finitely generated algebra over $k$, then there exists $f_1,\dots,f_n \in F$ (all non-zero) such that $F = k[f_1,\dots,f_n]$
Because $F$ is a field I know that $1/f_1 \dots 1/f_n \in F = k[f_1,\dots,f_n]$ so there exists $P_1,\dots,P_n \in k[x_1,\dots,x_n]$ such that
$$\frac{1}{f_i} = P_i(f_1,\dots,f_n) \; \forall i=1,\dots,n$$
Now if $n = 1$ by doing euclidean division with $x_1 \cdot P_1(x_1) - 1$ I find that the set $\{1,f_1,\dots,f^{deg(P_1)}_1\}$ generates $F$ as a $k$-module.
Now the reasonable thing to do would be to apply an induction step, but I didn't managed to do It.
What I should do is, for any $P \in k[x_1,\dots,x_n]$, to find some $Q_1,\dots,Q_n,R \in k[x_1,\dots,x_n]$ such that
$$P = Q_1\cdot (x_1 P_1 - 1) + \dots + Q_n \cdot (x_n P_n - 1) + R$$
Where $deg(R) < \max{deg(P_i) \, i =1,\dots,n\} =: M$
In this case I would get that $F$ is the $k$-module generated by the set
$$\{ f^{\alpha_1}_1 \cdot \dots \cdot f^{\alpha_n}_n \; : \; \alpha_1 + \dots + \dots_n < M \}$$
The problem is that in more then one variable I can't to euclidean division unless the leading coefficient is invertible. Applying induction I get stuck since the leading coefficient of $x_i \cdot P_i - 1$ with $i < n$ could not be in $k[x_1,\dots,x_{n-1}]$