Showing the existence of a certain kind of extension of a finitely generated algebra I wanted to ask if someone had a proof for the following claim:

Given and integral domain $A$ and a finitely generated $A$-algebra $B$, show that there exists elements $x_1,...,x_n \in B$ algebraically independent over $A$ and a non-zero element $a \in A$ such that $B[\frac{1}{a}]$ is integral over $A[\frac{1}{a}][x_1,...x_n]$.

Thank you!
 A: This is a corollary to Noether normalization lemma.
Let $K = \operatorname{Frac}(A)$ be the fraction field of $A$, and let $B_K$ be the tensor product $B\otimes_A K$, which is a finitely generated $K$-algebra. There is a canonical map $\phi: B \rightarrow B_K$, sending any element $b\in B$ to the element $\frac b 1 \in B_K$.
Frau Noether tells us that there are elements $y_1, \dotsc, y_n \in B_K$, algebraically independent over $K$, such that $B_K$ is a finitely generated module over the ring $S = K[y_1, \dotsc, y_n]$. In particular, $B_K$ is integral over $S$.
Multiplying $y_i$ by a non-zero element of $A$ if necessary, we may assume that each $y_i$ is in $\phi(B)$. We then choose, for each $i$, an element $x_i\in B$ such that $\phi(x_i) = y_i$.
It follows that $x_1, \dotsc, x_n$ are algebraically independent over $A$: otherwise there is a polynomial $0\neq f \in A[X_1, \dotsc, X_n]$ such that $f(x_1, \dotsc, x_n) = 0$, which implies $f(y_1, \dotsc, y_n) = 0$, contradicting the algebraic independency of $y_1, \dotsc, y_n$ over $K$.
Consequently, the map $\phi$ induces an isomorphism from $A[x_1, \dotsc, x_n]$ to $A[y_1, \dotsc, y_n]$.

Suppose $B$ is generated as $A$-algebra by elements $b_1, \dotsc, b_m$. Then by integrality of $B_K$ over $S$, there exists a polynomial $0\neq g \in S[X]$, such that $g(\phi(b_j)) = 0$ for every $j$. Multiplying $g$ by an element of $A$ if necessary, we may assume that the polynomial $g$ is in $A[y_1, \dotsc, y_n][X]$.
Since $\phi$ induces an isomorphism from $A[x_1, \dotsc, x_n]$ to $A[y_1, \dotsc, y_n]$, we may define $f$ as the unique polynomial in $A[x_1, \dotsc, x_n][X]$ such that $\phi(f) = g$.
We then look at each $f(b_j)\in B$. Since $\phi(f(b_j)) = g(\phi(b_j)) = 0$ in $B_K$, we see that there exists $0\neq a_j \in A$ such that $a_j f(b_j) = 0$ in $B$. Taking the product of all $a_j$ gives us $0\neq a \in A$ such that $af(b_j) = 0$ for all $j$.
I claim that $B[\frac 1 a]$ is integral over $A[\frac 1 a][x_1, \dotsc, x_n]$. Since $B[\frac 1 a]$ is generated as $A$-algebra by the elements $b_j$ and $\frac 1 a$, it suffices to show that every $b_j$ is integral. But this is just by construction: we have $af(b_j) = 0$, which means $f(b_j) = 0 \in B[\frac 1 a]$.
