1
$\begingroup$

I have been thinking about resolution of singularities of plane curves in terms of blow-up and integral closure and i am trying to see how the two approaches relate.

Let us consider the affine curve given by the equation $y^2 - x^2 - x^3 = 0$. Then its blow-up is a parabola, which is nonsingular. This is explained in Hartshorne example I.1.4.9 (see also Example I.4.9.1 in Hartshorne (blowing-up)) or in Fulton's Algebraic Curves page 84.

On the other hand, the integral closure of $k[x,y]/(y^2 - x^2 - x^3)$ is $k[t]$ (see https://mathoverflow.net/questions/1504/what-is-the-geometric-meaning-of-integral-closure or isolating locally the branches of a curve) and so the affine variety that corresponds to the integral closure is just the affine line.

Question: The curve that we obtain from the blow-up is geometrically related to the original curve, i.e. the original geometry of the curve is respected "as much as possible". On the other hand, the affine line that we obtain via the integral closure has completely eliminated the original geometry. So what is the point of considering the integral closure? I mean, we do want to "resolve" the singularity at $(0,0)$ but the integral closure seems to be doing this in very "crude" way.

$\endgroup$
3
  • $\begingroup$ The parabola is isomorphic to a line. So in what sense do you mean that the parabola is better? $\endgroup$
    – Amitai Yuval
    Mar 12, 2015 at 19:18
  • $\begingroup$ @AmitaiYuval: In the sense that the parabola is directly linked to the equations of the original curve. The affine line seems like an arbitrary result to me. $\endgroup$
    – Manos
    Mar 12, 2015 at 19:20
  • $\begingroup$ @Manos: Directly linked how? To make an analogy: it's almost as if one way you're getting the dihedral group of order 6 and the other you're getting $S_3$ and then objecting that they are different. Unless I'm missing something. $\endgroup$
    – RghtHndSd
    Mar 13, 2015 at 13:55

1 Answer 1

Reset to default
2
$\begingroup$

As you have said, the integral closure seems to be not very related to the original geometry, however usual when you take the normalization $p:Y\to X$ of a curve, then you also have a map, and further when you take the isomorphism from the normalization to the affine line you also have a map. If you take into account both the spaces and the maps, then you still have a direct link to the original equations.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .