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 $h$. 



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> 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?
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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 $\mathbb{C}\{x\_1, \ldots, x\_n\}\_0$.



Regular sequence from prime ideal
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). 


