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I'm having problems finding an appropriate reference for this question.

Given two elements $f, g \in \mathbb{C}[x_1, \dots, x_n]$, consider their greatest common divisor, $\gcd_{\mathbb{C}[x_1, \dots, x_n]}(f, g) = h \in \mathbb{C}[x_1, \dots, x_n]$.

Now, consider this two elements $f$ and $g$ as elements of $\mathbb{C}[[x_1, \dots x_n]]$, i.e., formal power series with finitely many terms. Compute $\gcd_{\mathbb{C}[[x_1, \dots, x_n]]}(f, g) = h' \in \mathbb{C}[[x_1, \dots, x_n]]$.

Does exists a unit $u \in \mathbb{C}[[x, y]]$ such that $h = uh'$? This is equivalent to asking if $h$ is a $\gcd$ in $\mathbb{C}[[x_1, \dots, x_n]]$.

I'm sure this can be rephrased in terms of irreducible components of algebraic vs. analytic varieties.

Thanks.

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    $\begingroup$ The g.c.d. is defined up to a unit. In $\mathbb{C}[x]$ this means up to a scalar, but in $\mathbb{C}[[x]]$ up to any formal series with nonzero constant term, so you can always find a g.c.d. of hte form $x^n$. Thus the comparison does not make much sense. $\endgroup$
    – abx
    May 12, 2016 at 19:26
  • $\begingroup$ What is meant is: Is a gcd in $\mathbb C[x_i]$ also a gcd in $\mathbb C[[x_i]]$? $\endgroup$
    – Friedrich Knop
    May 12, 2016 at 19:39
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    $\begingroup$ As abx says, there are some issues of comparison. But, it can be made sensible as follows. Writing $f=ah, g=bh$, with $a, b$ polynomials and $\mathrm{gcd}(a,b)=1$, suffices to show that the same holds in the power series. If $a(0)\neq 0$ or $b(0)\neq 0$, they are units in the power series and so this is alright. If both are zero, then use the fact that they form a regular sequence in the polynomial ring and so remain a regular sequence in the power series and thus have no common factor in the power series. $\endgroup$
    – Mohan
    May 12, 2016 at 19:42
  • $\begingroup$ Right, modulo multiplication by a unit. Editing the question. $\endgroup$
    – user91576
    May 12, 2016 at 20:21

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Maybe one can argue as follows: By dividing by $h$ one can assume $h=1$. Then the ring $\mathbb C[x_i]/(f,g)$ is of dimension $n-2$. Its completion is $\mathbb C[[x_i]]/(f,g)$ and therefore should also be of (Krull) dimension $n-2$. This implies that $f$ and $g$ have no common divisor in $\mathbb C[[x_i]]$ either.

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  • $\begingroup$ This argument only deals with the case that $(0,0,\ldots,0) \in V(f,g)$, i.e. $f(0) = g(0) = 0$. You need to say a bit more (cf. Mohan's comment), but it looks like it's ultimately ok. $\endgroup$
    – R. van Dobben de Bruyn
    May 13, 2016 at 4:51

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