Let $X$ be a smooth variety over a field $\Bbbk$, and let $Y, Z \subset X$ be closed reduced subschemes of the same dimension, both of which are local complete intersections. Is $Y \cup Z$ necessarily a local complete intersection?
2 Answers
The union of two planes in $\mathbb A^4$ which meet at a point is not CohenMacaulay, and so in particular not a local complete intersection.
More generally, any smooth subvariety of a smooth variety is a local complete intersection, so any nonCohenMacaulay subvariety whose components are smooth gives an example of a union of local complete intersections which is not itself a local complete intersection.
[Added: Your question has the caveat "with an appropriate scheme structure"; is there any other structure besides the reduced scheme structure on the union which you had in mind?]

$\begingroup$ Dear Emerton, you probably meant "any smooth subvariety of a smooth variety is a local complete intersection"? $\endgroup$ Commented Nov 5, 2011 at 21:23

$\begingroup$ Dear Francesco, Yes, thanks! Regards, Matthew $\endgroup$– EmertonCommented Nov 5, 2011 at 21:45

$\begingroup$ Two comments: First, the only other scheme structure I had in mind was the product ideal. Since they agree in your example, it's not an issue. Second, I was aware of your "more generally" paragraph when I wrote the question, but I was not familiar with examples of nonCohenMacaulay varieties with smooth components. Your word choice, together with the simplicity of your example, makes it sound like these should be quite easy to obtain. How does one go about it? $\endgroup$ Commented Nov 5, 2011 at 23:17

$\begingroup$ Please insert the word "equidimensional" at the appropriate place to make the previous comment reasonable. $\endgroup$ Commented Nov 6, 2011 at 2:39

$\begingroup$ Dear Charles, Being CohenMacaulay (or not) is a local, even infinitesimal, property of a singularity. So if you take any smooth surfaces in some ambient variety which meet at a single point in the same way that the two planes of my example meet at the origin, you will get a nonCM variety with smooth components. Obviously you will be able to make lots of these. Regards, $\endgroup$– EmertonCommented Nov 6, 2011 at 6:14
No. Let $X=\mathbb A^3_k$ with system of coordinates $t_1, t_2, t_3$, let $Y$ be the hyperplan $t_3=0$, $Z$ be the union of the $t_1$axe and the $t_3$axe. Then $Y$ and $Z$ are complete intersection, but $Y\cup Z$, whenever the structure you endowe with, can not be a local complete intersection at the origin because its irreducible components don't have the same dimension.

$\begingroup$ I meant to include the words "of the same dimension" in the question. $\endgroup$ Commented Nov 5, 2011 at 22:05