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 $\mathbb{C}\{x_1, \ldots, x_n\}_0$.

  • 2
    $\begingroup$ Last sentence, you meant: contains a power of the maximal ideal? $\endgroup$ – Hailong Dao Nov 22 '18 at 19:30

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).

| cite | improve this answer | |
  • $\begingroup$ 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 '18 at 19:39
  • $\begingroup$ You wrote that $g_1,...,g_h$ is a regular sequence. $\endgroup$ – Hailong Dao Nov 22 '18 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 '18 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 '18 at 20:00
  • 1
    $\begingroup$ 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 '18 at 20:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.