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I will consider more simple situation. Let C be a nodal curve which is a union of two $\mathbb{P}^1$, $C_1,C_2$. Which meets at a node $p$. Consider $C$ is embedded in a smooth variety $Y$. Assume that tangent directions $T_pC_1,T_pC_2$ are transversal in $T_pY$.

Then, is $C$ embedded in $Y$ regularly? Which means $I_{C/Y}/I_{C/Y}^2$ is locally free.

I think this is true but i cannot make a proof directly.

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  • $\begingroup$ This is just a question for the formal neighborhood of the node, right? And then it's enough to say that $C$ is a local complete intersection? Is it true that $C_1 \cup C_2$ lies in a smooth surface inside this formal neighborhood in $Y$? $\endgroup$
    – Allen Knutson
    Commented Oct 1, 2015 at 4:47
  • $\begingroup$ If $C_1$ and $C_2$ meets transversally at $p$, is it possible that there is no smooth surface containing $C$ in a formal neighborhood?? $\endgroup$
    – free-object
    Commented Oct 1, 2015 at 5:55
  • $\begingroup$ No, it's not possible: look at the generators of the ideal of C, and choose among those the elements whose differentials at p are linearly independent. They will cut out a subvariety that is smooth at p, and its tangent space is the same as C, which makes it a surface. $\endgroup$
    – t3suji
    Commented Oct 1, 2015 at 7:06
  • $\begingroup$ t3suji//thanks, then you think that it is true that $C$ is embeded regularly in this situation?? $\endgroup$
    – free-object
    Commented Oct 1, 2015 at 7:17
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    $\begingroup$ In other words: locally around $p$ (for the classical or étale topology), the embedding $C\hookrightarrow Y$ is isomorphic to $C \hookrightarrow T_p(C)\hookrightarrow T_p(Y)$, which is regular because it is the composition of two regular embeddings. $\endgroup$
    – abx
    Commented Oct 1, 2015 at 9:55

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