Can we arrange Noether normalization to be smooth at a given point? Given a finitely generated algebra $R$ over a field $k$ of characteristic $0$, Noether normalization says that $R$ contains a polynomial ring $S=k[x_1,\dots, x_d]$ in such a way that $R$ is a finitely generated $S$-module. Geometrically, this says that the affine scheme $X=\operatorname{Spec}(R)$ is a finite branched cover of the affine space $\mathbb{A}^d$. In fact, the proof of Noether normalization will tell you that if $X\subset \mathbb{A}^m$, then any "sufficiently general" projection $\mathbb{A}^m\rightarrow \mathbb{A}^d$ induces our finite map $\pi: X\rightarrow \mathbb{A}^d$. If $q\subset R$ is a maximal ideal with residue field $k$, i.e. $q$ corresponds to a $k$-point $Q$ of $X$, can we choose a projection in such a way that $\pi$ is smooth at $Q$? If it is necessary, we may assume that $X$ is smooth at $Q$ or that $X$ itself is smooth over $k$.
The normalization is constructed in stages (as seen here) : if $X\subset \mathbb{A}^m$, then we consider its closure $\bar{X}\subset \mathbb{P}^m$. We choose a point $P$ lying at infinity and a hyperplane $H$ that misses $P$, then we project away from $P$ to the hyperplane $H$. Since $P$ lies at infinity, this induces a map $X\rightarrow H\cap \mathbb{A}^m\cong \mathbb{A}^{m-1}$. If the map is surjective, then we win. Otherwise, repeat the argument with $\operatorname{im}(X)$ in place of $X$ and $H\cap \mathbb{A}^m$ in place of $\mathbb{A}^m$. Eventually we reach a stage where the projection is surjective.
Now, I believe we can ensure the map is smooth at our given point $Q$ by choosing at each stage a point at infinity $P$ for which the line spanned by $Q$ and $P$ is not tangent to $X$ at $Q$. Is this enough to guarantee smoothness?
Edit: Here's an example that leads me to believe this tangency condition is good enough: Let $S=k[x], R=k[x,y]/(x^2+y^2-1)$. $S$ is a Noether normalization for $R$ given by projecting from the point at infinity $P=[0:1:0]$ to the hyperplane $H=\{Y=0\}$. Writing $R=S[y]/(y^2+(x^2-1)$, the map $\pi: \operatorname{Spec}(R)\rightarrow \mathbb{A}^1$ is smooth if $2y$ is a unit in $R$ (see here). Now $2y$ isn't a unit in $R$, but it is a unit in $R[y^{-1}]$, i.e. $\pi$ is smooth on the open subscheme $D(y)\subset \operatorname{Spec}(R)$. This open subscheme is precisely the set of points $Q$ where the line from $P$ to $Q$ is not tangent to $X$.
 A: Let $X=V(I)\subset\Bbb A^n$ be our variety. Up to a change of coordinates, we may assume that the projection $\pi$ sends $(x_1,\cdots,x_{n-1},x_n)\mapsto(x_1,\cdots,x_{n-1})$. Letting $I_n=I\cap k[x_1,\cdots,x_{n-1}]$, the map $X\to\pi(X)$ corresponds to the map $$k[x_1,\cdots,x_{n-1}]/I_n \to k[x_1,\cdots,x_n]/I.$$
Letting $R=k[x_1,\cdots,x_{n-1}]/I_n$, we can rephrase this as $R\to R[x_n]/I$, and this map is smooth at $Q$ iff the Jacobian matrix of this rephrased map is of corank  zero at $Q$. What is the Jacobian matrix of this map? It's the column $$\begin{pmatrix} \frac{\partial f_1}{\partial x_n} \\ \vdots \\ \frac{\partial f_m}{\partial x_n}\end{pmatrix}$$ of the Jacobian matrix of $X$.
On the other hand, since the tangent space to $X$ at $Q$ is given by the kernel of the Jacobian matrix at $Q$, we see that the line spanned by $(0,\cdots,0,1)$ is in the kernel iff this column is zero. This exactly gives that the projection map $\pi$ is smooth at $Q$ iff the line $PQ$ is not tangent to $X$ at $Q$.
