Consider $X$ as a regular scheme over $S$ together with an open embedding into a regular, proper scheme $\bar{X}$. Then my question is why $\bar{X}\backslash X$ is a NCD on $\bar{X}$?
Thanks.
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3$\begingroup$ As abx points out, the answer to this is no. But I guess from your comments below, you are asking for existence. This amounts to resolution of singularities! In the generality you seem to want, it is open. $\endgroup$– Donu ArapuraCommented Nov 17, 2013 at 14:06
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2 Answers
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It is certainly not unless you put it as a hypothesis! Just take for $X$ the complement of a closed point in $\bar{X}$.
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$\begingroup$ SO, is there always such a $\bar{X}$? $\endgroup$– SamCommented Nov 17, 2013 at 13:30
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$\begingroup$ I just need to work with integral, pure dimensional excellent base scheme. But, I like to know about the general case as well. I guess for curves it's true, right? Thanks. $\endgroup$– SamCommented Nov 17, 2013 at 13:47
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Let us look at varieties over an algebraically closed field of characteristic zero. You have $X$ open in $\overline{X}$, and $\Delta = \overline{X}\setminus X$. Then $\Delta$ could easily be not normal crossing. For instance take a cusp $C\subset\mathbb{P}^2$.
However, by Hironaka's theorem there exists a log resolution $\pi:Y\rightarrow\overline{X}$. This means that $Y$ is smooth and $\pi^{-1}(D)\cup Exc(\pi)$ is a normal crossing divisor. Furthermore since $X = \overline{X}\setminus\Delta$ itself is smooth $\pi_{|\pi^{-1}(X)}:\pi^{-1}(X)\rightarrow X$ is an isomorphism. Therefore we found an open embedding of $X$ in a smooth proper scheme $Y$ such that $Y\setminus X$ is normal crossing.
