3
$\begingroup$

I am reading a text by Prof. Szpiro Tata lectures on equations defining space curves. In the proof of Proposition $1.2$ on page $12$ he gives explicit description of the defining equations of a local complete intersection curve, say $C$ in $\mathbb{P}^3$. In the first step he produces a polynomial $F_1 \in H^0(\mathcal{I}_C(n))$, for some integer $n$, such that $C$ is an effective Cartier divisor in the surface defined by $F_1$. The question is: From the existence of the polynomial $F_1$ (given on page $13$) it is not clear if $n$ is some number or could be any "large enough" number. In particular, for any $n$ large enough, can we find a hypersurface in $\mathbb{P}^3$ of degree $n$ containing $C$ as an effective Cartier divisor?

The reason for asking this question is: I would have thought that the answer to the last question should be false if $C$ is not reduced since the adjunction formula (which exists for an effective Cartier divisor on any hypersurface in $\mathbb{P}^3$, a Gorenstein scheme) tells us that the arithmetic genus of a non-reduced curve/effective Cartier divisor depends on the degree of the hypersurface containing it as a Cartier divisor.

$\endgroup$
4
  • $\begingroup$ $n$ is any large enough number, as Szpiro's proof shows. I do not understand why you think this gives a contradiction. $\endgroup$ – abx Mar 20 '15 at 17:04
  • $\begingroup$ @abx: I will explain my confusion using this example: Let $X, Y$ be two smooth surfaces in $\mathbb{P}^3$ of degree $d_1, d_2$ respectively containing a line, say $l$. Denote by $2l_X$ (resp. $2l_Y$) the cartier divisors on $X$ (resp. $Y$). Using the adjunction formula, see that the arithmetic genus of $2l_X$ is $1-d_1$ and of $2l_Y$ is $1-d_2$. So, they are definitely not the same curve. If I can embed $C$ into any large enough surface as a Cartier divisor, I will face a similar problem with the arithmetic genus. Am I wrong? $\endgroup$ – Chen Mar 20 '15 at 17:11
  • $\begingroup$ Now I see your point. This is indeed disturbing. $\endgroup$ – abx Mar 20 '15 at 18:31
  • $\begingroup$ I think although $C$ is a divisor in the surface defined by $F_1$, the associated reduced scheme $C_{\mathrm{red}}$ or its irreducible components need not be cartier divisors on this surface. $\endgroup$ – Chen Mar 22 '15 at 16:12
1
$\begingroup$

Let $\ell$ be the line in $\mathbb{P}^{3}$ cut out by the ideal $I=(x,y).$ It is true that if you look at the subscheme $\overline{\ell}$ of $\mathbb{P}^{3}$ cut out by $I^2$ (which is saturated) and its induced subschemes of smooth surfaces of different degrees containing $\ell,$ you will get nonreduced curves of different arithmetic genera. However, $\overline{\ell} \subseteq \mathbb{P}^{3}$ is not a local complete intersection.

EDIT 1: The number $n$ in Szpiro's proof is a means to the end of finding 4 generators for the homogeneous ideal of a given equidimensional lci curve in $\mathbb{P}^{3}.$ Picking a different $n$ may result in a different Cartier divisor on a different surface, and down the line this may result in a different collection of 4 generators, but this is OK if we fix $n$ once and for all in the "Existence of $F_1$" step.

EDIT 2: What is more problematic is the discussion at the top of p.14 in which the number $n$ is repurposed. However, I think this can be fixed if instead of $n$ we use $n+n'$ for some sufficiently large $n'.$

$\endgroup$
8
  • $\begingroup$ @Mustopa: Any Cartier divisor on a hypersurface in $\mathbb{P}^3$ is a local complete intersection curve in $\mathbb{P}^3$, by definition. Also note that in my notation, $I_{2l_X}$ or $I_{2l_Y}$ is not $I^2$, but $I^2+F_X$ and $I^2+F_Y$, respectively, where $F_X$ (resp. $F_Y$) are the equations defining $X$ (resp. $Y$). $I^2$ need not contain $F_X$ or $F_Y$. Am I missing something? $\endgroup$ – Chen Mar 20 '15 at 17:49
  • $\begingroup$ You are right. However, $\overline{\ell}$ is not a subscheme of any smooth surface in $\mathbb{P}^{3}$; the $2\ell$ Cartier divisors you mention are intersections of $\overline{\ell}$ with smooth surfaces. The input of the proposition you describe is a local complete intersection curve in $\mathbb{P}^{3}$--when you speak of the surfaces $X$ and $Y,$ you are introducing something extra. $\endgroup$ – Yusuf Mustopa Mar 20 '15 at 17:52
  • $\begingroup$ @Mustopa: No. A local complete intersection curve already contains the data of the surface in which it is a cartier divisor. There is nothing extra, if I understand correctly. $\endgroup$ – Chen Mar 20 '15 at 17:55
  • $\begingroup$ Being a local complete intersection is an intrinsic property of a scheme. In general, there is no unique surface in which the curve is a Cartier divisor. $\endgroup$ – Yusuf Mustopa Mar 20 '15 at 18:00
  • $\begingroup$ I agree, but as you see that $2l_X$ is a Cartier divisor on $X$, hence a lci in $\mathbb{P}^3$ but is different from the Cartier divisor $2l_Y$, which is also lci in $\mathbb{P}^3$. $\endgroup$ – Chen Mar 20 '15 at 18:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.