Why $av$ is integral over $k[Y\times \mathbb{A}^r]$ I am reading Shafarevich's Basic Algebraic Geometry 1(Third Edition).
I met some trouble when I read the proof of Theorem1.14 in page 62.
The theorem is 

Theorem 1.14 If $f:X\to Y$ is a regular map and $f(X)$ is dense in $Y$, then $f(X)$ contains an open set of $Y$.

The proof assumes $X$ is irreducible and affine, $r$ is the transcendence degreee of the field extension $k(X)/k(Y)$. Choose $r$ elements $u_1, \dots, u_r\in k[X]$ that are algebraically independent over $K(Y)$. Then
$$
K[X]\supset k[Y][u_1,\dots,u_r]\supset k[Y] \quad\text{and}\quad k[Y][u_1.\dots,u_r]=k[Y\times\mathbb{A}^r].
$$
In the seventh line of page 63, it says:

Any element $v\in k[X]$ is algebraic over $k[Y\times \mathbb{A}^r]$, hence there exists an element $a\in k[Y\times \mathbb{A}^r]$ such that $av$ is integral over $k[Y\times \mathbb{A}^r]$.

I don't know how to prove this statement.
 A: That is a general fact, we are going to prove it:
Let be $R$ a domain, $K=q.f(R)$ and let be $L$ a finite and algebraic extension over $K$. Then:
i) $\forall v\in L$ $\exists$ $a\in R-\{0\}:$ $av$ is integral over $R$.
ii) Exists a basis of $L$ over $K$ such that their elements are integral over $R$.
Proof of i)
Let be $v\in L$, we have $L/K$ is algebraic then exists $a_n,\ldots,a_1,a_0\in K$ such that
$$a_nv^{n}+a_{n-1}v^{n-1}+\cdots+a_1v+a_0=0$$
We have $K=q.f(R)$ so
$$\forall\hspace{0.05cm}i\in\{0,1\ldots,n\}\hspace{0.05cm} \exists\hspace{0.05cm}b_i\in R,c_i\in R-\{0\}: a_i=\frac{b_i}{c_i}$$
Let be $\gamma:=\prod_{i=0}^{n}c_i$, $\gamma_j=\prod_{0\leq i\leq n,i\neq j}c_i$ and $r_j:=\gamma_{j}b_{j}\in R$, then
$$a_nv^{n}+a_{n-1}v^{n-1}+\cdots+a_1v+a_0=0 \Rightarrow \gamma_{n}b_nv^{n}+\gamma_{n-1}b_{n-1}v^{n-1}+\cdots+\gamma_{1}b_1v+\gamma_{0}b_0=0\Rightarrow$$
$$r_nv^{n}+r_{n-1}v^{n-1}+\cdots+r_1v+r_0=0$$
Now we multiply by $r_n^{n-1}$ the last expression :
$$0=r_n^{n-1}r_nv^{n}+r_n^{n-1}r_{n-1}v^{n-1}+\cdots+r_n^{n-1}r_1v+r_n^{n-1}r_0=$$
$$(r_nv)^n + r_{n-1}(r_nv)^{n-1}+\cdots+r_1r_n^{n-2}(r_nv)+r_{n}^{n-1}r_0$$
Thus the polynomial $P(x) = x^n + r_{n-1}x^{n-1}+\cdots+r_1r_n^{n-2}x+r_{n}^{n-1}r_0$ has a zero in $r_nv$, it is monic and belong to $R[X]$, so $r_nv$ is integral over $R$
Proof of ii):
We have $L/K$ is finite then
$$m:=[L:K]=dim_{K}(L)<\infty$$
Thus exists $\{v_1,\ldots,v_m\}$ basis of $L$. If we use i) we have
$\forall\hspace{0.05cm}i\in\{1\ldots,m\}\hspace{0.05cm} \exists\hspace{0.05cm}a_i\in R-\{0\}$ such that $a_iv_i$ is integral over $R$. We have to show $\{a_1v_1,\ldots,a_mv_m\}$ is lineary independent over $K$:
Let be $\lambda_1,\ldots,\lambda_n\in K$ such that
$$\sum_{i=1}^{m}\lambda_i(a_iv_i)=0$$
Now we use $\{v_1,\ldots,v_m\}$ is lineary independent:
$$0=\sum_{i=1}^{m}\lambda_i(a_iv_i) = \sum_{i=1}^{m}(\lambda_ia_i)v_i \Rightarrow 0=\lambda_ia_i\hspace{0.15cm}\forall\hspace{0.05cm}i\in\{1\ldots,m\}$$
We have $a_i\neq 0$ for all $i\in\{1\ldots,m\}$ then $\lambda_i=0$ for all $i\in\{1\ldots,m\}$. Thus $\{a_1v_1,\ldots,a_mv_m\}\subset L$ is lineary indepent over $K$ and the set $\{a_1v_1,\ldots,a_mv_m\}$ is a basis because $m=dim_K(L))$.
