Is the composition of blowing-up a blowing-up?

Short answer

(After Huneke, C. and Swanson, I. (2006). Integral closure of ideals, rings, and modules, exercise 5.10)

Take $R$ a noetherian ring, an ideal $I$ of $R$ and an element $x$ of $I$. Define $S$ to be the ring $R[\frac Ix]$. Let $J$ be an ideal of $S$ and $y$ an element of $J$.

There exists an ideal $K$ of $R$ and $z$ in $K$ such that $R[\frac Ix][\frac Jy] = R[\frac Kz]$.

Indeed, let $J = (y, a_1, \dotsc, a_n)$. We can assume that all the generators of $J$ are in $R$ since $S[\frac Jy] = S[\frac{fJ}{fy}]$ for all $f$ in $R$. Set $K = IJ$ and $z = xy$ and you're done.

How ever, it is not clear how things patch together when working with non affine varieties.

Long answer

Let $X_2 \to X_1 \stackrel{\epsilon}{\longrightarrow} X_0$ a sequence of blowups, with center $I_0\subset \mathcal O_{X_0}$ and $I_1\subset \mathcal O_{X_1}$. Let $E_0\subset \mathcal O_{X_1}$ the exceptional divisor of the first blowup. There exists a $p$ and an ideal sheaf $J\subset \mathcal O_{X_0}$ such that $J\cdot \mathcal O_{X_1} = E_0^p I_1$.

I claim that the morphism $X_2 \to X_0$ is the blowup with center $I_0 J$, and I'll check this by showing the universal property of the blowup.

Let $Z$ be a variety with a morphism $Z\to X$ such that $I_0 J \cdot \mathcal O_Z$ is invertible. In particular, the ideal sheaf $I_0\cdot O_Z$ is invertible, so $Z\to X_0$ factors through $X_1$. I don't know what is $I_1\cdot \mathcal O_Z$ but $E_0^p I_1 \cdot \mathcal O_Z$ is $J\cdot \mathcal O_Z$, which is invertible. So $I_1 \cdot \mathcal O_Z$ is also invertible, and thus $Z\to X_1$ factors through $X_2$. The uniqueness of the factorization is implied by the uniqueness of $Z\to X_2$ given $Z\to X_1$ and the one of $Z\to X_1$ given $Z\to X_0$.