For a $\mathbb C P^2$ is known a result: if through the generic points $p_1,p_2,\dots p_n$ with multiplicities $m_1,m_2\dots, m_n$ correspondingly a degree $d$ irreducible reduced curve passes then $d^2\geq m_1^2+m_2^2+\dots+m_n^2-min (m_i)$. (see Lemma 1 in "Curves in $\mathbb P^2$ and symplectic packings" Geng Xu) The idea of the proof is the following: move slightly a point $p_i$ and apply Bezout theorem; formally he gave a construction of a new curve via deformation of the old one.

I was wondering if the same proof works in positive characteristic ? In the article it is written $\mathbb P^2$ without mentioning of characteristic, but the proof doesn't work : as you know, a derivation of a polynomial in positive characteristic can drastically change its degree. From the other hand it is very frequently when a result in complex geometry automatically translates to any characteristic by some logic reasons, it is the case here?

  • $\begingroup$ Please try to use at least one top-level tag, those with a two-letter prefix corresponding to the math arXiv categories. $\endgroup$
    – user9072
    Oct 6, 2013 at 18:16

1 Answer 1


Xu's idea can be reproduced in arbitrary characteristic, without derivations, with the extra hypothesis that the singularities are ordinary (ie, $m_i$ distinct tangent directions at $p_i$), or at least each point has one direction of multiplicity one in the tangent cone. Without an assumption like this, I don't know if it can be done. EDIT: It can't be done. See below. Assume the base field $k$ is infinite; then the argument goes as follows.

Let $d$ and $m_1, \dots, m_n$ be such that for general $p_1, \dots, p_n$ there is an irreducible curve of degree $d$ with multiplicity $m_i$ at $p_i$. Fix such a set of points, and fix affine coordinates so that $p_1=(0,0)$ and $y=0$ is one of the tangent lines to the curve at $p_1$. Let $p(t)=(0,t)$ and for simplicity write $m=m_1$. By the assumption, for general $t\in k$ there is an irreducible curve of degree $d$ multiplicity $m$ at $p(t)$ and $m_i$ at $p_i$, $i=2,\dots, n$. Let $I\subset k[x,y]$ be the ideal of the points $p_2, \dots, p_n$ with their multiplicities, $I^e$ its extension to $k[x,y,t]$ and $J=I^e \cap (x,y-t)^{m}$. By hypothesis, there is an irreducible $F\in J$ of degree $d$ in $x$ and $y$.

Write $F=\sum_{i=0}^k t^iF_i$, where $F_i\in I$ are polynomials in $x,y$ of degree at most $d$, and $F_0$ has degree exactly $d$. Clearly $F\in J\subset(x,y,t)^m$ (a monomial ideal) and it is easy to see that this implies $F_1 \in (x,y)^{m-1}$. If we show that $F_1\not\in (F_0)$, then the rest of Xu's argument works.

Since the multiplicity of $F_0$ at $p$ is exactly $m$, one can write $F=\sum_{j=0}^m G_j x^{m-j}(y-t)^{j}$, where $G_j$ are polynomials in $x,y,t$ and at least $G_1$ has nonvanishing constant term (by the choice of tangent line $y=0$). Then the summand for $j=1$ contributes a term $cx^{m-1}t$ to $F_1$ which can not be cancelled with any other term. This shows that the multiplicity at $p$ of $F_1$ is exactly $m-1$ and in particular that $F_1$ is neither zero nor equal to a scalar multiple of $F_0$.

EDIT (added Oct 11): Now consider the family of irreducible curves $C_{a,b}:(x-a)^2(y-b)^2=x^5+y^5$ which have a 4-fold point at the point $(a,b)$. For any parameterized curve $(a(t),b(t))$ in the plane $(a,b)$, expanding $F_t=(x-a(t))^2(y-b(t))^2-x^5-y^5=\sum_{i=0}^k t^iF_i$ gives $F_1=0$ in characteristic 2 and so the intersection-theoretic argument does not work (even after a linear change of variables, or for a germ of curve parameterized by two power series, the argument is the same).

Xu's argument is local in nature, and it has been used in other settings by many authors, always in characteristic zero (notably Lazarsfeld, see 5.2.3 in "Positivity in Algebraic Geometry I" and references there). I am quite sure that families of curves like the one above can be used to obtain counterexamples in positive characteristic on suitable surfaces. On the plane, however, Xu's conclusion is likely to be true, or at least counterexamples will be hard to construct (otherwise Nagata's conjecture would fail in positive characteristic, which is unknown and -as far as I am concerned- unexpected). In any case, Xu's argument can't be pushed to positive characteristic without some extra assumption on the singularities as above.

  • $\begingroup$ thank you! By the way, do you know something to read about Nagata's conjecture and related areas in positive characteristic? I can prove some estimations via tropical geometry but I can not find anything which is already done in order to compare. $\endgroup$ Oct 11, 2013 at 19:18
  • $\begingroup$ for example in "Computing limit linear series with infinitesimal methods" Laurent Evain proved some deformation theorems (as I understand, for any char) but then switched to char=0. Surveys also frequently say nothing at all about characteristic $\endgroup$ Oct 11, 2013 at 19:23
  • $\begingroup$ Yes, most work on this area will assume char=0 for simplicity and not even check whether the results hold in positive characteristic. Évain's result that (uniform multiplicity) SHGH holds for squares has now been proved in three ways, all of them assuming characteristic zero. Even in my own work, the only bounds which I can tell for sure hold in positive characteristic are those in my 2001 paper dx.doi.org/10.1016/S0022-4049(99)00116-4 . In arXiv.math/0602213 I did point out what is ok in all characteristics and what is only char=0, but that may not be what you are after. $\endgroup$
    – quim
    Oct 12, 2013 at 14:10
  • $\begingroup$ By the way, I am greatly interested in estimations obtained via tropical geometry, even if they were currently not the best. $\endgroup$
    – quim
    Oct 12, 2013 at 14:11
  • 1
    $\begingroup$ Now that I think of it, toric degenerations as those used by Ciliberto-Miranda and (implicitly) Yang, Dumnicki, Eckl, ... should work in char=p as well. Only when Ciliberto-Miranda need a tansversality lemma (for matching conditions) char=0 is probably necessary (one should check). So (grosso modo) their results on Seshadri constants are ok in arbitrary characteristic, but some results on SHGH are only char=0, and also our results on "good rays" (passing from the large inequality in Nagata-like bounds to the strict one). $\endgroup$
    – quim
    Oct 12, 2013 at 20:50

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