Height of a prime ideal and number of generators. Let $k$ be a field and $A=k[x_1,\dots,x_n]\big/(I)$ where $I$ is a prime ideal and $\operatorname{dim}(A)=1$, i.e $\operatorname{Spec}(A)$ is a curve. I want to construct a dense open set $U \subset \operatorname{Spec}(A)$ such that $U$ is a normal scheme. To do this I want to take as open set,
$$U=\{ x \in \operatorname{Spec}(A)| x \text{ is a smooth point}\}$$
since the notion of smoothness and normality coincide in dimension $1$. Furthermore the generic point $\eta \in \operatorname{Spec}(A)$ corresponding to the zero ideal is clearly smooth so density of $U$ follows. I'm having problems finishing the proof that $U$ is actually open.
Since $k[x_1,\dots,x_n]$ is catenary it follows by the dimension formula,
$$ \operatorname{dim}(k[x_1,\dots,x_n])= \dim(A)+ \operatorname{ht}(I)$$
that $I$ has dimension $n-1$. Now this implies that the minimal number of generators of $I$ must be greater of equal to $n-1$. Let us denote the cardinality of this set $\mu(I)$ and assume that $\mu(I)=n-1$. Then we can use the Jacobian criterion for smoothness to show that $x \in \operatorname{Spec}(A)$ is smooth if and only if the rank of the Jacobian at that point is maximal. Therefore a point is not smooth if and only if the rank of the Jacobian is not maximal and this corresponds to the point $x$ satisfying some polynomial equations that clearly constitute a closed set in $\operatorname{Spec}(A)$.
However if $\mu(I)=n$ for example then this argument fails. Is there any way to fix this?
 A: Let me first mention that this is not always possible in positive characteristic as the curve $V(x^p+y^p+a)\subset \mathbb{A}^2_{k}$ for $k$ of characteristic $p$ and $a\in k$ not a $p^{th}$ power shows. I'll be working in characteristic zero for the rest of the answer.
The claim "the singular points of a variety in characteristic zero form a closed subset of that variety" is true, and it may be proven in the following manner: show that any variety is birational to a hypersurface in $\mathbb{P}^n$, and then compute there using the Jacobian criterion.
Your proof ought to be fixable by asking for the rank of the Jacobian to be at least $n-1$ on an open subset. This open subset would be determined by the nonvanishing of the determinants of all $n-1\times n-1$ minors of the Jacobian matrix.
You need not worry about imposing equations to make the Jacobian have exactly rank $n-1$: it already has rank at most $n-1$ no matter what (you can test this with the twisted cubic, $V(z-x^3,z^2-y^3,y-x^2)\subset\mathbb{A}^3$, for example- the Jacobian has determinant $6x(y^2-xz)$ which vanishes on the twisted cubic and therefore the Jacobian has rank at most 2 at all points of this curve). Here is a proof:
Let $I\subset k[x_1,\cdots,x_n]$ be the ideal defining our curve $X$ inside $Y=\operatorname{Spec} k[x_1,\cdots,x_n]$. Then the following exact sequence is standard:
$$I/I^2\to \Omega_Y|_X\to \Omega_X\to 0$$
We may tensor by $k(x)$ for $x\in X$ to get 
$$I/I^2\to \Omega_Y\otimes k(x)\to \Omega_X\otimes k(x)\to 0$$
The first map sends $f$ to $df\otimes 1$, so if $f_i$ generate $I$, then $df_i$ generate the image of $I/I^2$ in $\Omega_Y\otimes k(x)\cong k(x)^n$. Let this image be called $Im$. Then the Jacobian $J_x$ associated to $f_i$ at the point $x$ has columns corresponding to the images of $df_i$ in $Im\subset \Omega_Y\otimes k(x)$. This shows us that $\dim_{k(x)}(\Omega_X\otimes k(x))=n-\operatorname{rank} J_x$, and given that the tangent space to a curve is always at least one dimensional, $\operatorname{rank} J_x$ is always at most $n-1$.
