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.

  • $\begingroup$ $n$ 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, 2015 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, 2015 at 17:11
  • $\begingroup$ Now I see your point. This is indeed disturbing. $\endgroup$
    – abx
    Mar 20, 2015 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, 2015 at 16:12

1 Answer 1


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 $n'.$

  • $\begingroup$ @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, 2015 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. $\endgroup$ Mar 20, 2015 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, 2015 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. $\endgroup$ Mar 20, 2015 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, 2015 at 18:02

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