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I arrived at the following phrase at a material that I'm reading:

Let $\pi :N'\rightarrow N$ be the blow-up of center $P$. For a given $a\in\mathcal{O}$ and $P'\in\pi^{-1}(P)$, the strict transform of $a$ in $P'$ is the ideal $str(a;P')$ of $\mathcal{O}_{N',P'}$ generated by $f^{-\upsilon_P (a)}\cdot(a\circ\pi)$ where $f=0$ is a reduced equation of $\pi^{-1}(P)$ in $P'$.

Some context:

  1. $\mathcal{O}$ is the ring of germs of holomorphic functions at $P\in N$, $N$ is a bidimensional analytic manifold with a foliation $\mathcal{F}$, and $\mathcal{M}$ is the maximal ideal of $\mathcal{O}$.
  2. $\upsilon_P (a)=max\{t:g\in \mathcal{M}^t\}$, for $g\in\mathcal{O}$, that is, $\upsilon_P (a)$ is the multiplicity of the zero of $g$ at $P$.

My doubts are about:

  • What is the meaning of $f^{-\upsilon_P (a)}$?
  • What is the meaning of a reduced equation as mentioned above?

I don't know if I gave enough data to make this understandable, so just ask if you need more context. Any explanation is very, very welcome. Thanks in advance!

Matt
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Marra
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2 Answers2

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This is how I like to think about proper transform. This is somewhat loosely speaking be warned.

Say you are blowing up $Z\subset X$, to get a space $\pi:\hat X\to X$. For a point, say $p$, in the base space, if $p\notin Z$ there is a unique preimage $\hat p$ in $\hat X$ which 'lives above' $p$. On the other hand, if $p\in Z$, then living above $p$ a big chunk of the exceptional divisor, namely the fiber over $p$.

Now suppose you are blowing up $p\in \mathbb C^2$, then the preimage of $p$ is an entire $\mathbb P^1$. On the other hand, say we blowup a line $\mathbb P^1 \cong Z\subset\mathbb P^3$, the entire exceptional divisor is $E\cong\mathbb P^1\times \mathbb P^1$, and the fiber over any point $p$ sees one fiber in $E$.

Great, now in the latter situation, what if you had a curve $C$ passing through $Z$, say intersecting it at a single point. If you just take the set theoretic inverse image you get a curve with a $\mathbb P^1$ attached to it in $\hat X$. This is called the total transform and isnt really what we want a lot of the time.

For this reason, you can follow the following mantra to get the 'right' curve $\hat C$ living above $C$ in $\hat X$. Take $\pi^{-1}(C)$, and 'throw out the exceptional stuff', i.e. set theoretically remove the intersection with the exceptional divisor. Now take closure. This is the proper transform in this case, and you get more what you expect. Instead of this exceptional fiber attached to $\hat C$.

Working out a few examples, like the one above can be quite helpful. You can do this in coordinates even! Take your favourite line inside $\mathbb P^3$ and blow it up, check what the inverse image is, try to figure out what the proper transform looks like.

Dhruv Ranganathan
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I think the idea is that when $a$ vanishes at $P$ with multiplicity $v_P(a),$ the pullback $\pi^*(a)$ must contain a factor of at least $v_P(a)$ times the equation of the exceptional divisor. The strict transform is basically the inverse image of the vanishing of $a,$ but without the exceptional divisor. So, we get it by ignoring the factor, i.e., we divide the pullback equation $\pi^*(a)=a\circ\pi$ by $f^{v_P(a)}$.

For an algebraic example, to help get your bearings, consider the blowup of the affine plane $N=\mathbb A^2$($=\mathbb C^2$) at the origin, and the variety cut out by $a=y^2-x.$ In the chart of the blowup with coordinates $x,y/x$, the pullback of $a$ becomes $\pi^*(a)=(y/x\cdot x)^2-x=x((y/x)^2\cdot x-1)$ which has exceptional factor $f=x$ ($y^2-x$ vanishes to order one at the origin). Getting rid of this factor, our strict transform is given by the vanishing of $(y/x)^2\cdot x-1.$ In the other chart, with coordinates $x/y,y,$ the pullback is $\pi^*(a)=y^2-x/y\cdot y=y(y-x/y),$ and again we can factor out $f,$ which locally in this chart is $f=y,$ to get the strict transform $V(y-x/y).$ The analytic situation is very similar.

The meaning of the reduced equation is easy to understand via a different example. Take $N=\mathbb A^2$ as before, but now $a=y^2-x^3.$ Then $a$ vanishes to order two at the origin.

In the chart with coordinates $x,y/x,$ we have $\pi^*(a)=(y/x\cdot x)^2-x^3=x^2((y/x)^2-x),$ so in this case the reduced equation $f$ is $x,$ and we can factor out $f^{v_P(a)}=x^2.$ Again in the second chart we will factor out $y^2,$ and the reduced equation for the exceptional divisor is $f=y.$

We could have taken $f=x^2$ (in the first chart), which would still determine the same underlying vanishing set as $f=x,$ but we want to keep track of multiplicities and order of vanishing, so it makes sense to use $f=x.$

Andrew
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  • What is $\mathbb{A}$? – OhMyGod May 09 '13 at 01:12
  • Dear @OhMyGod, $\mathbb A^2$ denotes the affine plane, which is just $\mathbb C^2$ when we work over the complex numbers. – Andrew May 09 '13 at 01:16
  • If I understood it correctly, the stric transform are the points $x\in N'-E$, $E$ is the exceptional divisor, such that $a\circ \pi (x)=0$. Is that correct? – Marra May 09 '13 at 01:48
  • Dear @GustavoMarra, that is almost correct, the only thing is you will need to take the closure of that set (i.e. we might add back some points of the exceptional divisor if they are in the closure of the set of points you describe). – Andrew May 09 '13 at 03:32
  • Do you mean $V'=\overline{\pi^{-1}(V-{ 0} )}$, where V is the set $a=0$ ? – Marra May 09 '13 at 03:38
  • Dear @GustavoMarra, yes, that's what I mean. – Andrew May 09 '13 at 03:42