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Let $X$ be a Toric Variety and let $x\in X$ be a point (not necessarily smooth). Then the blow up $Bl_{x}X$ of $X$ in $x$ is Toric. Let $Y:=WBl_{x}X$ be the weighted blow up of $X$ in $x$ with weights $a_{1},...,a_{n}$.

Is $Y$ Toric ?

In particular, is a weighted blow up of a weighted projective space Toric ?

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For the blow up to be toric $x$ must be torus invariant – Jesus Martinez Garcia Aug 4 '11 at 18:19
If you are blowing up a torus invariant ideal, the resulting object should still be toric. – Karl Schwede Aug 4 '11 at 18:53
up vote 2 down vote accepted

(Essentially reposting Jesus Martinez Garcia and Karl Schwede's comments as an answer)

If you are blowing up a torus-invariant sheaf of ideals, then the Rees algebra (the thing you take Proj of to get the blow-up) has grading by characters of the torus, so the blowup has an action of the torus, so it is toric (since it also contains a dense copy of the torus that acts on it). Thus, if your weighted blow-up is weighted "along the coordinate hyperplanes", it will be toric.

If you blow up a non-invariant sheaf of ideals, you do not get something toric in general. If you blow up a non-invariant point of $\mathbb A^2$, then the blow-up map already cannot be toric, even though both varieties are toric. Blowing up two different points of $\mathbb A^2$ gives a total space which has no toric structure at all: the two (-1)-curves must be torus-invariant and blowing them down gives a toric map to $\mathbb A^2$, but any toric blow-up of $\mathbb A^2$ must have the exceptional locus lying over a single point.

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A caveat: the blow-up of a normal toric variety at a torus-invariant ideal need not be normal. For example, take the blow-up of $\langle x^2, y^2 \rangle$ in the plane is non-normal. So, the answer also depends on what you take to be a toric variety. – Dustin Cartwright Aug 5 '11 at 13:26

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