Question

Suppose I have a smooth complex projective variety $X$ and a singular subvariety $Z$. Can I find a general complete intersection subvariety $W$ of the same dimension as $Z$, and another smooth subvariety $Y$ so that we have an algebraic equivalence of cycles $$a[W]+ b[Z] \sim [Y]$$ for some positive rational numbers $a, b$?

You can ask it for homological equivalence if that is somehow easier. What about if I ask to actually smooth the union $W\cup Z$?

Answer 1

I think the answer to your last question as posed is "no", but the first question may be actually quite difficult.

#1

If you hadn't required $W$ to be a general complete intersection, then the answer to both questions would be "yes": Let $I$ be the homogenous ideal of $Z$ in $S$ the homogenous coordinate ring of $X$ and pick $f_1,\dots,f_q\in I$ general homogenous elements where $q=\mathrm{codim}_XZ$. Let $V:=W\cup Z= V(f_1,\dots,f_q)$. Then $V$ is a complete intersection and hence smoothable to a $Y$ as required.

#2

The fact is, there are non-smoothable varieties. For instance let $X$ be a big enough projective space and $Z$ a cone over an abelian variety of dimension at least $2$ (any non-Cohen-Macaulay isolated log canonical singularity would work). Then $Z$ has a single singular point which is non-smoothable. Adding $W$ does not help, since $W$ being general, it will miss the singular point, so $W\cup Z$ is still non-smoothable.

#3

In your first question, since you are asking about the cycles, you can drop the word "general". The point is, that you can "add" a $W$ to make a non-smoothable singularity smoothable, but it would have to go through the singular set and not even just randomly. As in #1, you can find some $W$, but I am not sure how to guarantee that this $W$ be a complete intersection. My guess is still that there should be singularities with which you can't do this, but this is certainly a more subtle question.