## Is the strict transform of a smooth subvariety under the iterated blowup of a smooth varitiety still smooth?

Suppose $Y$ is a smooth variety, $X$ is a smooth hypersurface of $Y$, and $D_1, \ldots, D_n$ are smooth subvarieties of $Y$ that intersect transversely such that the sum of the codimensions of the $D_i$ is equal to the dimension of $Y$. Suppose further that $X$ and any combination of the $D_i$'s intersect transversely, EXCEPT that the intersection of $X$ with the intersection of all of the $D_i$'s intersect at a single point and for dimensional reasons do not intersect transversely. Is it still true that the proper transform of $X$ under the iterated blowup of $Y$ along the $D_i$'s is smooth?

As an example, set $Y = \mathbb{A}^6$ with coordinates $x_1,x_2,x_3,y_1,y_2,y_3$, set $X$ to be the linear hypersurface defined by $x_1+x_2+x_3 = y_1 + y_2 + y_3,$ and set $D_i = (x_i,y_i)$ for $i = 1,2,3.$ I keep getting equations along the lines of $y_i = u_ix_i$ (which are fine) along with $u_1x_1 + u_2x_2 + u_3x_3 = x_1+x_2+x_3$, which appears to be singular when $u_1 = u_2 = u_3 = 1$ and $x_1 = x_2 = x_3 = 0$.

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 What do you mean precisely by iterated blowup along $D_i$'s? The order may be important. – Sasha May 6 2011 at 5:59 Since everything is smooth and the $D_i$'s intersect transversely, the order of the blowups doesn't matter. I didn't mention that $X$ and $Y$ are also irreducible (or equivalently, connected), but that can also be assumed. – Victor May 6 2011 at 13:39

## 1 Answer

I think what you would like is true. Think of it this way: Forget about $Y$ and just consider the blow ups of $X$ along $X\cap D_i$. According to your assumptions in each step this should be a blow up of a smooth subvariety on a smooth variety so you get a smooth variety.

As for you example, I think you need more equations for the strict transform. That equation seems to include the exceptional divisor so you will necessarily get singular points where the different components intersect.

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