Let be R a ring and let $f:R[\! [ X_1, ..., X_n ]\! ]\rightarrow R[\! [ X_1, ..., X_n ]\! ]$ an homomorfism of $R$-algebras. Called $J\in M_{n\times n}(R)$ the jacobian matrix $J=\left( \frac{\partial f_i}{\partial X_j}(0)\right)$ where $f_i:=f(X_i)$. Show that $f$ is an isomorfism iff det$(J)$ is invertible in $R$.
I've tried in this way but i have some doubts.
Called $A:=R[\! [x_1, ..., x_n]\! ]$. I observe that $f$ induce a map called $J_f$ on $\Omega_{A/R}$:
$$ \begin{array} &A \longrightarrow & A \\ \downarrow &\downarrow\\ \Omega_{A/R} \longrightarrow &\Omega_{A/R} \end{array} $$
I've got that $J_f(d_A(x_i))=d_A(f(x_i))=d_A(f_i)=\Sigma_{j=1...n} \partial f_i/\partial x_j dx_j$. I observe that $J_f$ is linear so can be rapresented by a matrix $J$ that send $dx_i\to df_i$ so it's the jacobian matrix. Now i'm stucked.