Let $X$ be a variety over an algebraically closed field with null characteristic. Let $C$ be a smooth subvariety of $X$ of dimension 1, and let $x$ be a point of $C$. We assume that $X$ is analytically irreducible at $x$.

I consider the normalization $X'$ of $X$, and more precisely the morphism $C'\to C$ where $C'$ is the pull back of $C$ in $X'$. I assume that $C'$ is reduced, so that $C'\to C$ is a branched covering. Let $x'$ the unique point of $C'$ lying over $x$ — this point is unique since we assume that $X$ is analytically irreducible at $x$.

It may happen that $C'$ is not smooth at $x'$. By blowing-up I want to reduce to the situation where $C'$ is smooth.

Let $Y$ be the blowing-up of $X$ at $x$ ; let $D$ be the strict transform of $C$ in $Y$ ; and $y$ the unique point of $D$ which lies over $x$. As before, consider $Y'$ the normalization of $Y$ and $D'$ the pull-back of $C$ in $Y'$. Note that $D$ is isomorphic to $C$ since $C$ is smooth.

Since $Y'$ is normal, the morphism $Y'\to X$ factors through $X'$, and thus the morphism $D'\to C$ factors through $C'$. At least, I would like to prove that the morphism $D'\to C'$ is not an isomorphism if $C'$ is not smooth, so that by repeating the blowup, the curve $D'$ would eventually be smooth. Ideally, I wish I could prove that $D'\to C'$ is the blow-up at $x'$ of $C'$.

  1. Does $D' \simeq C'$ implies that $C'$ is smooth ?

  2. More precisely, does the morphism $D'\to C'$ factors through the blow-up at $x'$ of $C'$ ?

  3. Even more precisely, is $D'\to C'$ the blow-up of $C'$ at $x'$ ?

  4. What if we had the following additional assumptions : $X$ is an hypersurface of a smooth variety of dimension $n+1$, the order of $X$ at $x$ is $n$, and $X$ is generically a normal crossing point or order $n$ along $C$ ?

Any idea, reference, or counter-example is welcome. I'm really stuck at it... Thanks in advance for your help.


Consider the variety $X$ defined by the quotient ring $A = k[x,y,t]/(tx^2-(t^2+y)y^2)$. Let $C$ be the smooth curve $V(x, y)$. The normalization $X'$ of $X$ is given by $A[\frac{t x}{y}]$, which is isomorphic to $$k[x,y,t,u]/(yu-tx, u^2-t(t^2+y)),$$ with $u\mapsto tx/y$. The pull back of $C$ in $X'$ is given by $A'/(x,y)A'$, which is $k[t,u]/(u^2-t^3)$. This curve is singular at the origin.

Consider know the blow-up $Y$ of $X$ at the origin. An affine chart containing the strict transform of $C$ is given by $B = A[\frac xt, \frac yt]$, which is isomorphic to $$k[x_1,y_1,t]/(x_1^2-(t+y_1)y_1^2),$$ with $x_1 \mapsto \frac xt$ and $y_1 \mapsto \frac yt$. The strict transform $D$ of $C$ is $V(x_1,y_1)$. The normalization of $Y$ is given by $B[\frac{x_1}{y_1}]$ which is isomorphic to $$k[x_1,y_1,t,v]/(y_1v-x_1, v^2-(t+y_1)).$$ And then, the pull back of $D$ is $Y$ is given by $$k[t,v]/(v^2-t),$$ which is the blow-up of $C'$ at the origin.

  • $\begingroup$ My guess is that this won't work but I don't have an example. Thus what follows is just a long comment... The blowup $Y \to X$ is the blowup of the maximal ideal $I_x$. At the level of the strict transform, it can be viewed as the blowup of the ideal of the origin in $C$. Since $C$ is smooth this does nothing, in other words $I_x \cdot O_C$ is a principal ideal. Certainly if $Z \to X'$ is the normalization of the blowup of $I_x \cdot O_{X'}$, then since $I_x \cdot O_{Y'}$ is principal and $Y'$ is normal, we have a map (from universal properties) $Y' \to Z$. $\endgroup$ – Karl Schwede Jul 31 '12 at 3:06
  • $\begingroup$ At first glance, this looks like it would usually be an isomorphism. But maybe I'm being dumb, I haven't tried it in your example. Anyway, this somewhat limits our options. $\begin{array}{r} \newline\end{array}$ Suppose that indeed $Z = Y'$. We have $I_x \cdot O_{C'} = (I_x \cdot O_{C}) \cdot O_{C'} = (I_x \cdot O_{X'}) \cdot O_{C'}$ is still principal locally, and so the blowup $W \to X'$ of $X'$ at $I_x \cdot O_{X'}$ does not change $C'$ at all. It could still be that the normalization $Y' \to W$ saves us? I don't see why off the top of my head. $\endgroup$ – Karl Schwede Jul 31 '12 at 3:07
  • $\begingroup$ In particular, I'm pretty confident that #3 above can't be right in general. Probably #2 isn't right either, I'll try to find some real examples tomorrow. $\endgroup$ – Karl Schwede Jul 31 '12 at 3:10
  • $\begingroup$ @Karl: Thank very much for your comments. Note that when I talk about the blowup of $x'$, I don't mean the blowup of $I_x \cdot O_{X'}$. This ideal is not reduced in general. However $I_{x'} = \sqrt{I_{x}\cdot O_{X'}}$. Sorry agree with your factorization $Y'\to Z$, and I agree as well that the blowup of $I_{x}\cdot O_{X'}$ does not change $C'$. So yes, all would rely on the normalization of $W$... I thought of using some fancy valuative criterion, but I did not manage. I invite you to have a look at my example, I think it is striking. If it is not clear enough, I'd be happy to complete it. $\endgroup$ – Lierre Jul 31 '12 at 7:54
  • $\begingroup$ Dear Lierre, It's precisely because of the difference between $I_x \cdot O_X$ and $I_{x'}$ that I think these are going to be different. Blowing up different ideals cutting out the same scheme can give completely different (non-comparable) blowups. For example, you probably already know this but in the plane, blowing up $(x,y)$ and $(x^2,y)$ give schemes that can't be factored through each other. In fact, distinct valuations are created in both cases. $\endgroup$ – Karl Schwede Jul 31 '12 at 14:03

The answer for the four question is no, as this example shows.

Consider $X$ the surface defined by $x^2-(t^3+y)y^2 = 0$. Let $A$ denote the associated ring. Then $X' = \operatorname{Spec} A[\frac xy]$. The curve $D$ is given by $u^2-t^3$, with $u = \frac xy$.

The blowup (an affine chart of) at the origin of $X$ is given by $\operatorname{Spec} A[\frac xt, \frac yt]$. Note that $$ A\left[\frac xt, \frac yt\right] \simeq k[x_0,y_0, t]/(x_0^2-(t^2+y)ty_0^2), $$ with $x_0 \to \frac xt$ and $y_0 \to \frac yt$. Let $B$ denote this ring. Then $B' = B[\frac{x_0}{y_0}] = B[\frac xy]$. The curve $D'$ is again $u^2-t^3$. So nothing changed...

Nothing, really ?

If we repeat the blowup, we obtain the surface $x^2 - t^3 y^2$. And the corresponding $D'$ is smooth. So repeated blowups seem to be necessary to obtain one blowup of $D$.

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