Is this divisorial contraction a blow-up?

Question

Let $C$ be a curve in a smooth $3$-fold $X$ with an ordinary node $p\in X$. Blow-up $p$ let $E$ be the exceptional divisor, and $\widetilde{C}$ the strict transform of $C$. Furthermore let $L$ be the line in $E$ through the two points $E\cap \widetilde{C}$.

Now, let us blow-up $\widetilde{C}$ (with exceptional divisor $E_{C}$) and then flop the strict transform of $L$. Let $Z$ be the resulting variety. The strict transform of $E$ in $Z$ is isomorphic to $\mathbb{P}^1\times\mathbb{P}^1$, and can be blow-down to another variety $W$. We get then a divisorial contraction $f:W\rightarrow X$ contracting the strict transform of $E_C$ to $C$.

Is $f$ the blow-up of $X$ along $C$ ?

To me it is clear that it must be the blow-up over any point $q\in C\setminus \{p\}$ but I am quite confused on what happens over $p$.

Answer 1

The variety $Y$ you constructed is terminal. It is singular at the point where $\mathbb{P}^1\times\mathbb{P}^1$ is contracted. However this singularity is normal.

Now, you can write $C$ as $\{y^2-x^3-x^2=z=0\}\subset\mathbb{A}^3$. The blow-up $Z$ of $\mathbb{A}^3$ along $Z$ is defined by $\{s(y^2-x^3-x^2)-tz=0\}\subset\mathbb{A}^3\times\mathbb{P}^1$. Note that $Z$ has a unique singular point, namely $((0,0,0),[1:0])$ which is an ordinary double point. Therefore, $Z$ is terminal.

Since there exists at most a unique terminal extraction of $C$ we get $Y\cong Z$.

Answer 2

This $f$ is the blowup of $C$. See qu. 22 of Kollár's Exercises in the birational geometry of algebraic varieties.

A subtle point is whether you assume that $C$ is irreducible or not.

If $C$ is irreducible then the divisor isomorphic to $\mathbb{P}^1\times\mathbb{P}^1$ is contracted to a $\mathbb{Q}$-factorial ordinary double point (i.e. locally analytically isomorphic to $xy=zt$). The fibre above $p$ is isomorphic to $\mathbb{P}^1$ (the flopped curve) and this singularity appears somewhere along it.

If $C$ is not irreducible then the two factors of $\mathbb{P}^1\times \mathbb{P}^1$ are not numerically equivalent and can be contracted independently of each other (with the order you contract them in corresponding to blowing up one branch of $C$ first and then the other). Contracting the whole $\mathbb{P}^1\times \mathbb{P}^1$ gives a variety that looks the same as before, except that the singular point is not $\mathbb{Q}$-factorial.