# Regular sequence from prime ideal

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

• Last sentence, you meant: contains a power of the maximal ideal? – 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).
• 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. – Paul Nov 22 '18 at 19:39
• You wrote that $g_1,...,g_h$ is a regular sequence. – Hailong Dao Nov 22 '18 at 19:40
• 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 $V(g_1, \ldots, g_{i})$. – Paul Nov 22 '18 at 19:56
• I don't understand what you are saying. The height of $I$ is $h$, so it is not an embedded component. – Hailong Dao Nov 22 '18 at 20:00
• 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? – Paul Nov 22 '18 at 20:08