# Does existence of an isolated solution imply the Jacobian determinant is non-zero?

Let $f_1,\dots,f_n$ be formal power series in $\mathbb{C}[[x_1,\dots,x_n]]$ whose constant terms are all zero (i.e. $f_1,\dots f_n$ are not units in the ring). Suppose further that the radical of the ideal $(f_1,\dots f_n)$ is the maximal ideal. Does this imply that the Jacobian determinant $$\det \left(\frac{\partial f_i}{\partial x_j}\right)$$ is not identically zero?

Note, that in the special case when $f_1,\dots f_n$ are polynomials, we get the following statement: If the Jacobian determinant associated to a system of $n$ polynomial equations in $n$ variables with complex coefficients is identically zero, then the system has no isolated solutions.

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The answer is affirmative over any field $k$ of characteristic 0. Indeed, let $f:k[\![y]\!] \rightarrow k[\![x]\!]$ be the local $k$-algebra map defined by $y_i \mapsto f_i$, so by hypothesis the special fiber is $k$-finite. Thus, by successive approximation, $f$ is module-finite, so $\Omega^n_f = \widehat{\Omega}^n_f = k[\![x]\!]/J_f$. But char. 0 implies generic etaleness of $f$, so $\Omega^1_f$ has vanishing generic fiber and passing to $n$th exterior powers gives that $k(\!(x)\!)/J_f = 0$. Thus, $J_f \ne 0$. –  user29720 Dec 21 '12 at 3:34
Obviously this fails in characteristic $p$ since we can just take the $p$th power of the generators. –  Will Sawin Dec 21 '12 at 6:56
Why does it work in characteristic 0? What about the field of real numbers and $f_1=f_2=x_1^2+x_2^2$? The Jacobian is trivially zero, but the only solution of $f_1=f_2=0$ is $x_1=x_2=0$. Am I missing something trivial that has not been fully stated? –  Michael Renardy Dec 21 '12 at 19:07
@Michael: It works in char. 0 because I proved it. :) More specifically, the hypothesis that the ideal (f_1,\dots,f_n) in the formal power series ring has radical equal to the maximal ideal is not satisfied if $n > 1$ and $f_1 = \dots = f_n$. Your proposed example is confusing algebraic conditions with radical ideals and conditions on solution loci in the ground field for polynomials (the latter having no logical relevance in the formal power series case, nor in the polynomial case when the ground field isn't algebraically closed). –  user29720 Dec 21 '12 at 19:17
So, what you are saying is the statement added in the note only applies if the field is algebraically closed? –  Michael Renardy Dec 21 '12 at 19:23