I wish to prove that if $(f_1, \ldots, f_n) = (1)$, and for each $i$, $B \to A_{f_i}$ is of finite type, then so is $B \to A$.
I am following Vakil's hints.
Let $r_{ij}$ be the finitely many elements that show that $A_{f_i}$ is finitely generated over $B$. Fix some element $r \in A$. Then we have that in $A_{f_i}, r = p_i(r_{ij}, f_i)f^{-N_i}$, where $p_i$ is some polynomial over $B$ in the given variables.
I am supposed to use a "partition of unity" argument, so I write $1 = \sum_i a_i f_i$. I can in fact WLOG set $1 = \sum c_i f_i^{N_i}$.
Then $r = \sum rc_if_i^{N_i}$, and in each of the $A_{f_i}$, I can cancel the $f_i^{-N_i}$; i.e. I can kill the denominator in one of the summands, in each of the $A_{f_i}$: which results in the summand $a_ip_i(r_{ij}, f_i)$. This would be fine since this is a polynomial over $B$.
Vakil suggests, that if I write $r' = \sum_k a_k p_k(r_{kj}, f_k)$, then $r$ and $r'$ have the same restrictions in each of the $D(f_i)$, hence they are equal as elements of $A$. This would finish the proof.
The problem is, I cannot see how the restrictions of $r$ and $r'$ are the same, as having $a_kp_k(r_{kj}, f_k) = a_krf_k^{N_k}$ requires the cancellation of $f_k$, which I can only perform in $A_{f_k}$.
I believe I am doing the partition of unity part "wrong", and that's why I cannot finish the proof. Can you please help me out?
Let $r\in A$. Let $g_{ij}\in A$ (finite set) be st $g_{ij}, f_i^{-1}$ generates $A_{f_i}$ as a $B$-algebra. This exists by assumption and by the definition of localisation. Let $p_{i}(g_{ij},f_i)$ be a polynomial with coefficients in $B$ such that $r=p_{i}(g_{ij},f_i)/f_i^{N_i}$ in $A_{f_i}$. By choosing $p_i$ appropriately, we may assume that $N=N_i$ is independent of $i$. Write $c_i=c_i(r):=p_{i}(g_{ij},f_i)\in A$, so that $r=c_i/f_i^N$ in $A_{f_i}$. We then see that for any $i,j$, we have $f_j^N c_i=f_i^N c_j$ in $A_{f_if_j},$ so that $$(f_i f_j)^{M} f_j^N c_i=(f_i f_j)^{M} f_i^N c_j$$ for some $M$ which can be chosen independent of $i,j$. Let $a_i\in A$ be such that $1=\sum_i a_i f_i^{2N+M}$. Let $$h:=\sum_i a_i f_i^{N+M} c_i.$$ We compute $$ hf_j^{N+M}=\sum_i a_i f_i^{N+M} c_i f_j^{N+M}=\sum_i a_i(f_i f_j)^{M}f_j^{N}c_i f_i^{N}=\sum_i a_i(f_i f_j)^{M}f_i^{2N}c_j $$ $$= f_j^M c_j\sum_if_i^{2N+M}a_i=f_j^M c_j $$ and thus $h=r$ in $A_{f_j}$. Hence $r=h$ and we have thus shown that $A$ is generated by the $a_i, f_i$ and $g_{ij}$.