An affine scheme $X = Spec(A)$ is said to be smooth if for any closed embedding
$X\subset\mathbf A^n$, of ideal $I$, it is true that, locally on $x\in X$, the ideal $I$
can be generated by a sequence $f\_{r+1},\dots,f\_n$ such that their Jacobian has maximal rank.


My question is:


* Will the Jacobian of ANY set of $n-r$ generators of $I$ be of maximal rank?



Jacobian criterion for smoothness of schemes
Yes, the rank of the Jacobian matrix doesn't depend on the set of generators of $I$. The Jacobian matrix at $x$ represents the subspace generated by the differentials at $x$ of all $f\in I$. 


Note that the rank of the Jacobian matrix at $x$ is computed in the fiber where $x$ lives, it has nothing to do with the base scheme.


**Some more explanations** We work over a base field $k$. Let $I$ be the ideal defining $X$ in $Y:=\mathbb A^n$. Then we have a canonical exact sequence 
$$ I/I^2 \to \Omega\_{Y}|\_X \to \Omega\_X \to 0 $$ 
where the first map is $\bar{f}\mapsto df\otimes 1$. Tensoring by $k(x)$ we get
$$ I/I^2 \to \Omega\_{Y}\otimes k(x) \to \Omega\_X\otimes k(x) \to 0.$$ 
If $g\_1,...,g\_m$ are a system of generators of $I$, then $dg\_1,...., dg\_m$ generate the image of $I/I^2$ in $\Omega\_{Y}\otimes k(x)\simeq k(x)^n$. Call this image $C$. Let $J\_x$ be the Jacobian matrix associated to $g\_1,...,g\_m$ at $x$ in a system of coordinates of $Y$. Then the columns of $J\_x$ correspond to the images of $dg\_1,...,dg\_m$ in $C\subseteq \Omega\_{Y}\otimes k(x)$. Therefore the rank of $J\_x$ is just the dimension of $C$ over $k(x)$, and this is independent on the choice of the system of generators $g\_1,...g\_m$. 


By the way, these discussions show that 
$$\dim\_{k(x)} (\Omega\_{X}\otimes k(x))=n - \mathrm{rank} J\_x.$$
As the smoothness at $x$ is equivalent to $ \dim\_{k(x)} (\Omega\_{X}\otimes k(x))= \dim\_x X$, we see that it is also equivalent to $\mathrm{rank} J\_x=n-\dim\_x X$ which is the Jacobian criterion of smoothness.


