TITLE: which varieties can appear as exceptional divisors?
QUESTION [6 upvotes]: Let $\pi$ be the blow up of $\mathbb{A}^n$ at an ideal 
$I_{p}$ which is supported at a close point $p$; let $E$ be the associated exceptional divisor.  What are the possible varieties $Y$ that can appear as $E \cong Y$?
$\textbf{Partial Answers:}$ 
(1) If $I_p = (x_0, \ldots x_n)$, then $E \cong \mathbb{P}^n$. 
In this case $\pi$ is the blow up of $\mathbb{A}^n$ at a smoothing point.
(2) If $I_p = \left( x^{d/w_0}, \ldots, x_n^{d/w_n} \right)$ where $d=lcm(w_0, \ldots, w_n)$. Then  $E \cong \mathbb{P}(w_0, \ldots w_n)$. In this case $\pi$ is the weighted blow up with respect the weights $(w_0, \ldots w_n)$. 
(3) Kawakita article [K] proves that a divisorial contraction in dimension $n=3$ which contracts its exceptional divisor to a smooth point is a suitable weighted blow-up.
I am wondering if there are similar picture for other varieties? ( The toric case is clarified below by Lev Borisov's answer). I am particularly interested in an explicit description! This means: To obtain the variety $E \cong Y$ we must blow up the ideal $I(Y)_{p}$ which  generators are...
Particularly, is it possible to obtain a grassmannian this way?
Thanks!
[K] Divisorial contractions in dimension three which contract divisors to smooth points,Kawakita, Masayuki, Inventiones mathematicae, 145, 1, 105--119,2001

REPLY [4 votes]: Assuming that the blowup in question is toric, and exceptional divisor has one component, you are not going to get anything beyond these examples. After all, it would mean that you have a subdivision $\Sigma$ of a unimodular cone $C$ that gives $\mathbb A^n$. If the exceptional locus is an irreducible divisor, it means that there is exactly one new ray in $\Sigma$. Then it is just a matter of looking where this ray is, and you get your weighted projective space picture.
If you don't assume that the divisor has only one component, then you can probably get any projective toric variety as a component of the exceptional locus, as a subdivision
of $C$ can be as messy as you want it to be (and it would still correspond to an ideal under projectivity assumptions).