This is an idle question, but I give the example that motivated me below.


Say $X \subseteq {\mathbb A}^n\_k$ is irreducible and $k$ is infinite. Then by picking a regular point of $X$ and picking equations from $X$'s ideal that cut out $T\_x X$, we get a scheme containing $X$ as a component.



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> If we pick those equations generically, can we ensure that that scheme is a complete intersection with at most one extra component beyond $X$?
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The example that got me wondering this is where $X = ${$(A,B) : AB = BA$} is the space of pairs of commuting matrices. Then one case of the above construction is $Y = ${$(A,B) : AB-BA$ is diagonal}, which is a reduced complete intersection with two components. I thought this was interesting but now I'm guessing it's the expected behavior.



If $X$ is an affine variety, is $X$ one component of a complete intersection with two?
Perhaps I'm wrong but I thought this was ok by Bertini. 


Choose a general hypersurface $H\_1$ containing $X$ (ie, choose a general linear combination of the generators of the ideal of $X$, make sure this isn't a pencil). Choose another general hypersurface $H\_2$ containing $X$. Repeat this process. Eventually we end up with an intersection of $n - \dim X$ hypersurfaces containing $X$. Call this reducible variety $Y$. This has only one irreducible component besides $X$ by Bertini's theorem (the base locus was $X$ and its not a pencil). Is this what you had in mind?


I assume this must come up in linkage theory (discussed in Eisenbud's book). I believe I've seen this in work of Kawakita and also Ein-Mustata on singularities (see in particular, *lci defect ideals*).


