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Let $X$ be a noetherian scheme and let $Y$ be a closed subscheme of $X$. What relation is there between $\mathrm{Bl} _ {Y}(X)$ and $\mathrm{Bl} _{ Y _{\mathrm{red}}}(X)$ ?

Thanks.

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2 Answers

up vote 11 down vote accepted

There is no map from one blow up to the other, and definitely not an isomorphism. Please see my comments to J.C. Ottems answer.

However, if you replace radical by integral closure, then everything is fine.

Here's what I mean, if $I$ is an ideal and $J$ is its integral closure, then you always have an everywhere defined map

$$Bl_J X \to Bl_I X.$$

This need not be an isomorphism, indeed the integral closure of $(x^2, y^2)$ is $(x^2, xy, y^2)$. The blow up of the latter ideal is the normalization of the blow up of the former.

The other way you can get a map is if $J = \sqrt I$, and also if we can write $I = J \cdot \mathfrak{a}$ for some other ideal $\mathfrak{a}$. Then the blow up of $I$ is always the blow up of $\mathfrak{a}$ pulled back to $Bl_J X$.

In general, you should expect no relation between the blow up of two ideals with the same radical unless there is some integral closure relation between them and/or one ideal is the product of the other (and something else).

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In general they can be very different. For example take the subscheme $Y$ of $\mathbb{A}^2$ given by the ideal $(x^2,y)$. Here the blow up is covered by the two open subsets

$$U = \mbox{Spec} k[x, y][t]/(y − x^2t),\qquad V = \mbox{Spec} k[x, y][s]/(ys − x^2)$$

In particular the blow up of $Y$ is singular, whereas the blow-up of $\mathbb{A}^2$ at a point is not.

In general, even if you assume that both blow-ups are smooth, all sorts of things can happen depending on how complicated the ideal sheaf is. For example the blow-ups can have a different number of exceptional divisors and not even be related by a finite map. Even worse, every birational morphism $X'\to X$ is the blow-up of $X$ along some ideal sheaf.

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Thanks for the example. But at least, there is a natural map from one to another? –  gio Mar 7 '12 at 21:25
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I believe there is no map from one to the other. In the example J.C. Ottem gives, the blow up of $(x^2,y)$ can be obtained as follows. Blow up $(x,y)$, then blow up another point on that first blowup (the origin on one of the usual charts), and then contract the first exceptional curve. There's no map between $Bl_{(x^2,y)}X$ and $Bl_(x,y) X$, at least no map over $X$.$$\text{ }$$ Just because you have an inclusion of Rees algebras, does not mean that there is an everywhere defined map of the blow-ups. In the given example, one of the points of the overring contracts to an irrelevant ideal. –  Karl Schwede Mar 8 '12 at 4:39
    
You are right, Karl. Thanks. –  J.C. Ottem Mar 8 '12 at 8:06
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