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Let $V_1$ and $V_2$ be two distinct smooth subvarieties of the smooth variety $X$ which are regularly embedded. I would like to find a reasonable criteria which guaranties the smoothness of the Blow-up $\widetilde{X}$ of $X$ along the union of $V_1$ and $V_2$.

For example, is $\widetilde{X}$ smooth if $V_1$ and $V_2$ meet transversally?

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Take two lines in $\mathbb{P}^3$. Then the blow up is smooth if and only if the lines do not intersect. –  J.C. Ottem Nov 19 '10 at 13:50
    
In the example of two lines in a threefold these subvarities don't intersect transversally and the blow-up is not expected to be smooth. I would like to see a non-trivial example or a useful criteria for the smoothness of the blow-up. –  Passenger Nov 19 '10 at 14:14
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It seems to me that with more or less the same proof you can show that if a subvariety $V$ of a smooth variety $X$ is complete intersection, then the blow-up of $X$ along the (reduced) ideal of $V$ is smooth if and only if $V$ is smooth –  Francesco Polizzi Nov 19 '10 at 14:36
    
Francesco, do you mean a subvariety of codimension more than one? –  roy smith Nov 19 '10 at 14:51
    
@roy Yes, of course I was thinking in codimension at least 2. Thank you! –  Francesco Polizzi Nov 19 '10 at 16:50
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3 Answers

Appearently, the blow-up IS smooth if $V_1$ and $V_2$ intersect transversally. In this case we have that $$ Bl_{V_1 \cup V_2} X = Bl_{\bar{V_1}}Bl_{V_2}X =Bl_{\bar{V_2}}Bl_{V_1}X $$where $\bar{V_i}$ denotes the proper transform of $V_i$. This is essentially Proposition 2.9 in Kiem and Moon's article http://arxiv.org/abs/1002.2461.

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In general the answer is no.

Take $X=\mathbb{A}^3$ with coordinates $x,y,z$ and let $V_1$ and $V_2$ be two lines meeting in one point, for instance

$V_1:=\{x=y=0\}, \quad V_2:= \{x=z=0\}$.

Then the ideal of $V_1 \cup V_2$ is $I=(x, yz)$ and the equation of the blow-up $\widetilde{X}$ of $X$ along $I$ are given in $X \times \mathbb{P}^1$ by

$\lambda x - \mu yz=0$,

where $[\lambda : \mu]$ are homogeneous coordinates in $\mathbb{P}^1$. In the chart $\mu=1$ the blow-up is then given by

$\textrm{Spec }k[x,y,z, \lambda]/(\lambda x - yz)$,

hence it has an isolated singularity at the origin.

The other chart $\lambda=1$ is instead smooth, so this is actually the unique singular point of $\widetilde{X}$.

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If I'm not completely mistaken the blow-up here is still Cohen-Macaulay. Is there a counterexample where the blow-up is not CM? –  J.C. Ottem Nov 19 '10 at 15:28
    
In this example the blow-up has a unique isolated hypersurface singularity, in particular it is Gorenstein (hence Cohen-Macauley). Any birational morphism $f \colon Y \to X$ can be obtained as the blow up of some sheaf of ideals $I$ in $X$, so a priori everything can happen. However, the ideal can be horrible. Actually, I do not know if there exists any example with non-CM blow-up in the situation proposed in the question... –  Francesco Polizzi Nov 19 '10 at 17:12
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This example shows that maybe we do not want to call V_1 and V_2 transversal. –  ABC Nov 29 '10 at 13:41
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Let $V$ be a finite-dimensional complex vector space, let $A$ be a subspace, and let $\alpha:V\to V/A$ be the projection. The blow-up of $PV$ along $PA$ can then be identified with

$\{(L,M)\in PV\times P(V/A): L\leq \alpha^{-1}(M)\}$

Now suppose we have another subspace $B$. I think that the blow-up along $PA\cup PB$ is just the fibre product of the blow-ups along $PA$ and $PB$, namely

$X=\{(L,M,N)\in PV\times P(V/A)\times P(V/B) : L\leq \alpha^{-1}(M)\cap\beta^{-1}(N)\}$

We have $PA\cap PB=P(A\cap B)$, and this intersection is transverse iff $A+B=V$. If so, then the map $(\alpha,\beta):V\to V/A\times V/B$ is surjective, so the spaces $\alpha^{-1}(M)\times\beta^{-1}(N)=(\alpha,\beta)^{-1}(M\times N)$ form a vector bundle over $P(V/A)\times P(V/B)$, whose associated projective bundle is $X$; this shows that $X$ is smooth. Thus, the question has an affirmative answer at least for linear subspaces of projective space.

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