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Let $f(x,y) \in \mathbb C\{x,y\}$ be a holomorphic function-germ at zero. Let $f_x, f_y$ denote its partial derivatives. What is the proof of the following statement?

If the $\mathbb C$-algebra $\mathbb C\{x,y\}/(f_x,f_y)$ is finite dimensional, then $(f_x,f_y)$ is a regular sequence.

Saying that $(f_x,f_y)$ is a regular sequence amounts to the following two statements:

  1. $\forall a \in \mathbb C\{x,y\}$, if $a\cdot f_x = 0$ then $a = 0$
  2. $\forall a \in \mathbb C\{x,y\}$, if $a \cdot f_y \in (f_x)$ then $a \in (f_x)$.

If possible I'd like to know both an elementary proof and a more sophisticated one (but still detailed, please), allowing for generalization to $n$ variables. Thank you!

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Here's the "sophisticated" proof: if you have a sequence of elements $r_1,\dots,r_n$ in a commutative ring $R$ and the Krull dimension of $R/(r_1,\dots,r_n)$ is $\dim R - n$, then call the sequence a system of parameters. In a Cohen-Macaulay ring, every system of parameters is a regular sequence. Cohen-Macaulayness is implied, for instance, by being a regular ring, which holds in your case (the ring of power series). For a reference for these facts, see Chapter 18 of Eisenbud's Commutative Algebra book.

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  • $\begingroup$ Why in our case the Krull dimension of $\mathbb C\{x,y\}/(f_x,f_y) = \dim \mathbb C\{x,y\} - 2$? (sorry for the dumb question) $\endgroup$
    – user336494
    Commented Oct 14, 2017 at 10:57
  • $\begingroup$ Finite-dimensional rings have Krull dimension 0, and the ring of power series in n variables over a field has Krull dimension n. $\endgroup$
    – Steven Sam
    Commented Oct 15, 2017 at 15:11

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