Let $X\subset\mathbb P^4$ a projective hypersurface with an ordinary double point at $o\in X$.


Blow-up $\mathbb P^4$ at $o$ and let $E\simeq\mathbb P^3$ the exceptional divisor of this blow-up. Consider the strict transform $\widetilde X$ of $X$. Then, it is easy to verify that $Y=\widetilde X\cap E$ is a quadric hypersurface in $E\simeq\mathbb P^3$, thus isomorphic to $\mathbb P^1\times\mathbb P^1$.


Now, take an hyperplane (if you want, generic) section $C$ of $Y$ in $E$, that is $C=Y\cap H(\simeq\mathbb P^1)$, where $H$ is a hyperplane in $E\simeq\mathbb P^3$. Obviously, the curve $C$ meets each line of the two rulings of $Y$ in exactly one point. So, choose one ruling and define the obvious morphism $\varphi\colon Y\to C$ obtained by collapsing any line of the choosen ruling to its intersection point with $C$.


**Here is my question.**


Does there exist a *smooth* projective variety $\overline X$, containing a curve isomorphic to $C$ (which I still call $C$), and a regular morphism $\Phi\colon\widetilde X\to\overline X$ such that:


* $\Phi|\_{\widetilde X\setminus Y}\colon\widetilde X\setminus Y\quad\overset{\simeq}\to\quad\overline X\setminus C$
* $\Phi|\_{Y}\colon Y\to C$ is the $\varphi$ above.


I guess this is very classical, but I wasn't able to find it.


Thanks in advance! 



Blowing-up an ordinary double point, then contracting the exceptional locus to a curve
I guess the answer in **no** when $\deg(X) \geq 3$. 


In fact, in this case Griffiths showed that the two rulings $L\_1$ and $L\_2$ in $E$ are *numerically equivalent* in $\widetilde{X}$ (although not algebraically equivalent in general), so the corresponding $K\_{\widetilde{X}}$-negative rays $R\_1$ and $R\_2$ in the Mori cone of $\widetilde{X}$ coincide. 


It follows that the contraction of $R\_1$ is actually the contraction of the whole quadric, that is it is not possible to contract the two rulings separately. 


The situation is very subtle: in fact, the completed local ring $\widehat{\mathcal{O}}\_{X,o}$ is *not* factorial, but the fact that $L\_1$ is numerically equivalent to $L\_2$ implies that $\mathcal{O}\_{X,o}$ is factorial.


See [Debarre, Higher Dimensional Algebraic Geometry, p. 160] and the references given therein.


