Automorphism of power series

Question

Let us consider an endomorphism $\mathbb{C}[[x_1, ..., x_n]] \to \mathbb{C}[[x_1, ..., x_n]]$, $g(x_1,..,x_2) \mapsto g(f_1, ..., f_n)$ where $f_1, ..., f_n \in \mathbb{C}[[x_1, ..., x_n]]$. My homework is to show that it is an automorphism iff $ J = det(\frac{\partial f_i}{ \partial x_j})$ has no free term (which is to say iff it is a unit). I do not know where to begin so any hint would be appreciated.

Answer 1

I’m real handy with one-variable series, and any suggestion I give here may be off the mark for many-variable series.

But I would prepare the situation by composing with the linear inverse of the Jacobian matrix. That is, form $J(\mathbf 0)$, take the inverse of this, and then take the linear substitution whose matrix is this $J^{-1}$. By composing with the original, you get $\{f_i=x_i+\text{higher}\}$. Finding the inverse of this should be much easier, conceptually at least, than the task for the original.

Alternatively, are you comfortable with Newton-Raphson for maps $\Bbb C^n\to\Bbb C^n$? Your given $n$-tuple of series $f$ represents a self-mapping of $\Bbb C[[z_1,\cdots,z_n]]$, by $g\mapsto f\circ g$. You want a $g$ that solves the equation $f\circ g=\mathbf{id}$, where $\mathbf{id}$ is the identity $n$-tuple mentioned by @ancientmathematician, $\mathbf{id}_i=x_i$. For single-variable series, Newton-Raphson is an extremely quick method of getting $\,f^{-1}$.