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Let $I$ be a prime ideal in $\mathbb{C}\{x_1, \ldots, x_n\}_0$ (the localization at the maximal ideal that defines $0$) and suppose that the height of $I$ is $h$. Then, there is a standard trick to extract a regular sequence of length $h$ from $I$. And so one can always see $V:=V(I)$ (which has codimension $h$) as an irreducible component of a complete intersection of codimension $h$.

Can you choose the regular sequence $g_1, \ldots, g_h$ so that the intersection of $V$ with each of the other irreducible components of $V(g_1,\ldots,g_h)$ is just $\{0\}$? What if we assume that the local ring of $(V(I),0)$ is Cohen-Macaulay?

It seems to me that this is a strong condition on the ideal $g:=(g_1, \ldots, g_h)$, however, you have some freedom to choose the elements within $I$. Observe, that this would imply that there exists a primary decomposition of $g$ such that the sum of the primary ideal corresponding to $V$ and any other primary ideal contains (a power of) the maximal ideal of $\mathbb{C}\{x_1, \ldots, x_n\}_0$.

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    $\begingroup$ Last sentence, you meant: contains a power of the maximal ideal? $\endgroup$
    – Hailong Dao
    Nov 22, 2018 at 19:30

1 Answer 1

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It is impossible to do this if $\dim V(I)\geq 2$. Because then $g$ defines a complete intersection of dimension at least $2$. But for any Cohen-Macaulay local ring of dimension at least $2$, the punctured spectrum is connected. (Unless of course if $V(g)$ has only one component, whence $I$ is a set-theoretic complete intersection).

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  • $\begingroup$ I don't see why $\dim(V(I)) \geq 2$ implies that $g$ defines a complete intersection. This happens if the singular locus of $V(g)$ has codimension >1 by Hartshorne irreducibilty criterion, right? But the other irreducible components of $V(g)$ might intersect in codim 1 analytic sets. $\endgroup$
    – Paul
    Nov 22, 2018 at 19:39
  • $\begingroup$ You wrote that $g_1,...,g_h$ is a regular sequence. $\endgroup$
    – Hailong Dao
    Nov 22, 2018 at 19:40
  • $\begingroup$ I don't understand why this does not allow that the other irreducible components don't intersect in codim 1 sets. But on the other hand, your answer suggests that they have to intersect to $V(I)$ in something of codim > 2. Does this imply that $V(I)$ is neccesarily an embedded component in some other irreducible component? That $g_1, \ldots, g_h$ is a regular sequence just tells that $g_{i+1}$ does not vanish along a whole component of $V(g_1, \ldots, g_{i})$ but it may vanish along a part of $V(g_1, \ldots, g_{i})$. $\endgroup$
    – Paul
    Nov 22, 2018 at 19:56
  • $\begingroup$ I don't understand what you are saying. The height of $I$ is $h$, so it is not an embedded component. $\endgroup$
    – Hailong Dao
    Nov 22, 2018 at 20:00
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    $\begingroup$ Right. I didn't see that assuming that $g_1, \ldots, g_h$ was a regular sequence was playing against me, rather than with me. Do you know if it is possible if we don't require $g_1, \ldots, g_h$ to be a regular sequence? or if we don't assume that we are in a cohen-macaulay ring? $\endgroup$
    – Paul
    Nov 22, 2018 at 20:08

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