Radical ideal in $k[X_{1},...,X_{n}]$ Let $k$ be an field with $\mathrm{char}(k) = 0$ and $f_{1},...,f_{n} \in k[X_{1},...,X_{n}]$. Consider the jacobian matrix $A := (\frac{\partial f_{j}}{\partial X_{l}})$ and suppose that $\det(A) = 1$. Is it true that $\sqrt{(f_{1},...,f_{n})} = (f_{1},...,f_{n})$?
Note that if $n = 1$, then $\det(A) = 1 \Longrightarrow f_{1} = X+b$ with $b \in k$. So, $\sqrt{(f_{1})} = (f_{1})$. 
 A: Let $R=k[x_1, \dotsc, x_n]/(f_1, \dotsc, f_n)$ and $X=\operatorname{Spec}R$. We have to show that $R$ is reduced.
We can assume that $k$ is algebraically closed, because the assumption on the Jacobian is certainly invariant after passing to the algebraic clousure and if $R$ were not reduced, $R \otimes_k \overline k$ were also not reduced by the virtue of the injection $R \hookrightarrow R \otimes_k \overline k$.
If $X=\emptyset$, we have nothing to show. Otherwise, let $x \in X$. By this beautiful answer, we get $$\dim_{k(x)} \Omega_X \otimes k(x) = n-\operatorname{rank} J_x,$$
where $J_x$ is the Jacobian matrix evaluated at $x$. By your assumption on the Jacobian, we get $\dim_{k(x)} \Omega_X \otimes k(x)=0$, i.e. Nakayama yields $\Omega_{X,x}=0$. Since this holds for any $x \in X$, we get $\Omega_X=0$.
Next, let us show that $X$ is actually zero-dimensional. It suffices to show that any irreducible component of $X$ is zero-dimensional. By generic smoothness, any component admits a non-empty open $U$, where it is non-singular, i.e. $\mathcal \Omega_U$ is local free of rank $\dim U$. This shows $\dim U=0$ and since you check the dimension on any non-empty open subset in the finite type case, we have that any irreducible component is zero-dimensional, thus $X$ is zero-dimensional, i.e. a finite discrete set of points. This means $R = A_1 \times \dotsc \times A_s$ for local artinian rings $A_i$.
We are left to show that each $A_i$ is a field. Note that $A_i$ is a quotient of $R$, in particular it satisfies the assumption of Theorem II.8.8 in Hartshorne, thus $\Omega_{A_i}=0$ implies that $A_i$ is regular. A regular local artinian ring is a field.
