TITLE: The rank of Jacobian matrix at a point of affine variety is independent of choice of generators
QUESTION [8 upvotes]: This is a statement made in Hartshorne Chpt 1, Sec 5.
Let $Y\subset A^n$ be an affine variety. Let $\{f_i\}$ be a set of generators of ideal $I(Y)\subset k[x_1,\dots, x_n]$. Let $p\in Y$. Define $r=\mathrm{rk}(\frac{\partial f_i}{\partial x_j}|_p)$. Then $p$ is smooth iff $r=n-\dim(Y)$. 
Q: How does one see this definition of rank is independent of the choice of generators? I guess he did not mention tangent space is isomorphic to $m_p/m^2_p$ at this part and I cannot use this intrinsic definition.

REPLY [8 votes]: Suppose $g$ is a $k[x_1,\cdots,x_n]$-linear combination of the $f_i$. Then we may write $$\frac{d}{dx_j}g=\frac{d}{dx_j}\left(\sum c_if_i\right)=\sum\frac{d}{dx_j}(c_if_i)=\sum\left(c_i\frac{df_i}{dx_j}+f_i\frac{dc_i}{dx_j} \right)=\left(\sum c_i\frac{df_i}{dx_j} \right)+\left(\sum f_i\frac{dc_i}{dx_j}\right). $$
It is clear that the first term on the right-most side is a $k[x_1,\cdots,x_n]$-linear combination of the $\frac{df_i}{dx_j}$, and since $f_i$ evaluate to zero at $p$, the second term on the rightmost side vanishes at $p$. So adding any element which is already in the ideal to the generating set doesn't change the rank. 
So if you'd like to compare the ranks associated to the two finite sets of generators $\{f_i\}$ and $\{f_i'\}$, you see that you can obtain the generating set $\{f_i\}\cup\{f_i'\}$ through the above process, and the $r$ obtained from that set of generators will be the same as the $r$ obtained from both the sets $\{f_i\}$ and $\{f_i'\}$.