Is a f.g. $B_f$ algebra also a f.g. $B$ algebra? All rings are unital commutative. Let $\pi:B \rightarrow A$ be a ring homomoprhism. Let the image of $f \in B$ be $f'$. Then $A_{f'}$ is a $B_f$-algebra. It seems to me that the following is true. 

$A_{f'}$ is f.g. as a $B_f$-algebra iff it is f.g. as a $B$-algebra.


My argument:  Suppose it is f.g. as $B_f$-algebra. Let the generators be $a_i/(f')^{k_i}$. The way $B_f$ acts on an element $A_{f'}$ is via 
$$ \frac{b}{f^k} \cdot \frac{a}{{f'}^l} = \frac{\pi(b) \cdot a}{{f'}^{k+l}}  .$$
In other words $A_{f'}$ is in fact generated by $a_i, \frac{1}{f'}$ as both $B$-algebra.
Equally, if it is f.g. as a $B$-algebra, then the same argument shows it is f.g. as a $B_f$-algebra. 

I would be happy to know if the claim is true and if my argument is correct, if not, what went wrong. 
 A: Your argument looks good to me. Here is another take: $B_{f} \cong B[X]/\langle fX - 1 \rangle$, and the data of a $B_{f}$-algebra $A$ is equivalent to the data of $B$-algebra $A$ such that the image of $f$ under the structural map $B \to A$ is invertible - this follows from the universal property of localization. 
Now, a $B$-algebra $A$ is finitely generated iff it admits a surjective $B$-algebra homomorphism $\varphi \colon B[X_{1}, \ldots, X_{n}] \to A$. If the image of $f \in B$ in $A$ is invertible, then by the universal property of localization, $\varphi$ extends to a (surjective) $B_{f}$-algebra morphism $\rho \colon B_{f}[X_{1}, \ldots, X_{n}] \to A$. 
On the other hand, suppose that $A$ is a finitely generated $B_{f}$-algebra, i.e. $A$ admits a surjective $B_{f}$-algebra morphism $\rho \colon B_{f}[X_{1}, \ldots, X_{n}] \to A$. Since $B_{f}[X_{1}, \ldots, X_{n}] \cong B[X_{0}, X_{1}, \ldots, X_{n}]/\langle fX_{0} - 1 \rangle$ as $B$-algebras, the composition of $\rho$ with the natural surjection of $B$-algebras $B[X_{0}, X_{1}, \ldots, X_{n}] \to B_{f}[X_{1}, \ldots, X_{n}]$ gives a surjective $B$-algebra map $\varphi \colon B[X_{0}, X_{1}, \ldots, X_{n}] \to A$. 
