The rank of the Jacobian of an algebraic curve Suppose $C \subset \mathbb{A}^n$ is an algebraic curve, with $C = V(I)$ where $I = \langle f_1,...,f_r\rangle\subset \mathbb{C}[x_1,...,x_n]$ is an ideal. Let $J$ be the Jacobian matrix where $J_{i,j} = \frac{\partial f_i}{\partial x_j}$. Why is it necessarily true that $\operatorname{Rank}(J) \leq n-1$?
What I know:
Since $C$ is an algebraic curve, I know that $I$ is generated by at least $(n-1)$ polynomials, so the maximum possible rank of $J$ is $n$, which could potentially happen if $I$ is generated by $n$ polynomials and all of the rows of the Jacobian are linearly independent over $\mathbb{C}$. Therefore, I imagine the goal is to show that this is actually impossible, say by contradicting the fact that $C$ is a curve.
What I have tried:
I know that one can change the generators of an ideal and will find that the rank of the Jacobian arising from each of these sets of generators will be the same at each point. So I hoped that one could try to find a Gröbner basis for $I$ which would only have $(n-1)$ generators, making the question trivial. However, I am not very familiar with Gröbner bases, I only know that in some sense they're "the best you can do" when it comes to finding generators for an ideal, so I have not been able to make much progress along this route.
A second approach:
As mentioned above, taking $I$ and $J_{i,j}$ as previously defined, the greatest possible rank of $J$ is $n$, we want to show that this is in fact impossible.
Suppose for the sake of contradiction that $\operatorname{rank}(J|_p) = n$ at some point $p$, then there is some $n \times n$ submatrix of $J|_p$ which is of full rank, and is hence invertible. Therefore there is some sequence of row operations which transforms the matrix $J|_p$ into the form
$$\begin{pmatrix} 
I_{n\times n}\\
0_{n \times (r-n)}
\end{pmatrix}$$
From here I am unsure how to progress, but I'll present some logic here. The row operations performed on $J$ correspond to similar operations on the generators $f_1,..,f_r$ of $I$. Being able to bring $J$ into the above form therefore corresponds to being able to map $f_1$ to some function $g_1$ where $\frac{\partial g_1}{\partial x_1}(p) = 1$ while $\frac{\partial g_1}{\partial x_i}(p) = 0$ for $i\neq 1$ and similarly for the other functions. I would like to be able to conclude from this that my new generators give an ideal which define a zero dimensional variety, but I am unsure how to progress from this point. My first intuition is that the restriction above meant that $g_1 = x_1$, $g_2=x_2$ and so on, but there are numerous counter examples to this claim.
Is there a suitable way to proceed from one of the methods outlined here, or am I making this more complicated than it needs to be?
 A: I don't know how to use Gröbner bases to attack this problem, and I don't think your second method is appropriate. Let me present a solution from a different perspective.
The key concept here is that if your curve is embedded in to $\Bbb A^n$, the dimension of the tangent space to your curve at $p$ plus the rank of the Jacobian at $p$ equals $n$. So if the rank of the Jacobian is $n$, the tangent space to your curve at $p$ must be zero-dimensional, which is a contradiction: the tangent space to any variety of dimension $d$ must also be of dimension at least $d$.
This is an application of a fact from commutative algebra: if $(A,\mathfrak{m},k)$ is a noetherian local ring of dimension $d$, then $\dim A \leq \dim_k \mathfrak{m}/\mathfrak{m}^2$. This is a combination of Krull's Hauptidealsatz and Nakayama's lemma – any generating set for $\mathfrak{m}$ must project down to a spanning set of $\mathfrak{m}/\mathfrak{m}^2$, so it contains at least $\dim_k \mathfrak{m}/\mathfrak{m}^2$ elements. On the other hand, $\operatorname{height}\mathfrak{m}=\dim A$ is at most the number of generators of $\mathfrak{m}$.
