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Suppose that $X$ is a singular variety of dimension $n$, with singular locus of dimension $a$. Does there always exist a smooth subvariety $V$ of given dimension $m$ (where $a < m < n$) which contains $Sing(X)$? Of course it is sometimes possible: for example the ruling of a singular quadric cone passes through the singular point. But I am not sure what the general situation is.

What is the obstruction to finding a $V$? If there always is a $V$, can you do better, for example by requiring the tangent space to $V$ at a given singular point to contain a particular tangent direction?

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  • $\begingroup$ What if $a = n-1$ and $\operatorname{Sing}(X)$ is itself singular? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Apr 1, 2020 at 18:43
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    $\begingroup$ If $a = n-1$ then there is no $m$ with $a < m < n$. But anyway, assume $a = 0$ if you want; I'm curious about any results with this general flavor. Given a specific $X$ and $m$, how can I tell which way things go? $\endgroup$
    – Michael B
    Commented Apr 1, 2020 at 18:45
  • $\begingroup$ Oh, $m$ is given? That is not clear from the current formulation of the question. Please edit the question to clarify your exact assumptions. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Apr 1, 2020 at 18:46
  • $\begingroup$ I am pretty sure there are counterexamples with $n=2$, $a=0$ and $m=1$. $\endgroup$
    – Laurent Moret-Bailly
    Commented Apr 1, 2020 at 19:02
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    $\begingroup$ @SándorKovács: nice to receive your imprimatur. :) $\endgroup$
    – Pop
    Commented Apr 4, 2020 at 21:23

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