User quim - MathOverflow

Answer by quim for Smooth curves in a linear system satisfying certain conditions.

2013-02-04T10:39:43Z

$\xi$ has to be "curvilinear", i.e., contained in some smooth curve of $S$, or equivalently, has to be of multiplicity one. It is clear that this is necessary.

Let me now show that it is sufficient. The simplest case is when $\xi$ is reduced. Then you consider the blow up $\pi_\xi:\tilde S\rightarrow S$ at all points in $\xi$, and notice that $|L\otimes I_\xi|\cong |\pi^*L-E_\xi|$. General curves in $|\pi^*L-E_\xi|$ are smooth and intersect $E$ transversely, because $\pi^*L-E_\xi$ is globally generated, so you are done.

Now assume the curvilinear scheme $\xi$ is nonreduced but irreducible of length $k$ supported at some point $p=p_1$, and consider the blow up $\pi_1: S_1\rightarrow S$ at $p$, with exceptional divisor $E_1$. Then $\pi_1^*I_\xi=I_{E_1}\cdot I_{\xi_1}\cong I_{\xi_1}\otimes \mathcal O_{S_1}(-E_1)$, where $\xi_1$ is a 0-dimensional curvilinear subscheme of $S_1$, irreducible of length $k-1$, supported at some point $p_2\in E_1$. More important, $I_\xi=\pi_{1*}(I_{\xi_1}\otimes \mathcal O_{S_1}(-E_1))$, so $I_{\xi_1}\otimes \mathcal O_{S_1}(-E_1)$ is globally generated. Blow up $p_2$ to define $p_3$, etc, till you blow up $p_k$; denote $\pi:\tilde S \rightarrow S$ the composition of the $k$ blowups, and let $E_i$ be the pullback on $\tilde S$ of the exceptional divisor above $p_i$. Now $I_\xi=\pi_*(\mathcal O_{\tilde S}(-E_1-\dots-E_k))$ with $\mathcal O_{\tilde S}(\pi^*L-E_1-\dots-E_k)$ base point free. A general member of $|\pi^*L-E_1-\dots-E_k|$ is smooth, intersects $E_k$ transversely at a point different from the singular point of $E_1$, so its image on $S$ is nonsingular.

The above analysis can be carried over for each point on the support of $\xi$.

A local version of this, including the description of what happens for non-curvilinear subschemes (the singularity type of a general member of $|L\otimes I_\xi|$ is determined by the "resolution" of $\xi$), can be found (as "Bertini theorem") on Casas-Alvero's book on singularities of plane curves.

Answer by quim for Set of plane curves which intersect a fixed curve with given multiplicity

2013-01-29T09:47:04Z

If $C$ is unibranch at $p$, $V_n$ is always a vector space. If $C$ is not unibranch, there exist $n$ such that $V_n$ is not a vector space (as noted by the previous answers).

To see the first claim, consider the normalization $\eta:\tilde C \rightarrow C$ of the curve. $C$ is unibranch at $p$ if $\eta^{-1}(p)$ is a single point $q$. Let $x,y$ be affine coordinates in a neighborhood of $p$, and $t$ a parameter of $\tilde C$ at $q$. Consider the induced map $\eta^*:\mathbb{C}[x,y]\rightarrow \mathbb{C}[[t]]$ and the ideal $I_n=(t^n)\subset \mathbb{C}[[t]]$. Then $V_n$ can be identified with $(\eta^*)^{-1}(I_n)\cap \mathbb{C}[x,y]_{\le d-1},$ where $\mathbb{C}[x,y]_{\le d-1}$ denotes the set of polynomials of degree at most $d-1$.

Answer by quim for Irreducible divisors containing an arbitrary closed set

2013-01-17T11:50:04Z

EDIT 18/1 to make it clearer and somehow address the singular case:

First assume $X$ nonsingular. Denote $|mA-V|$ the linear system of divisors in $|mA|$ containing $V$. For $m\gg 0$, the base locus of $|mA-V|$ is exactly $V$ (blowing up the components of $V$ you can transform $|mA-V|$ into a base-point free system). So by Bertini (in characteristic zero), either you have an irreducible $A'$ or the image of $X$ by the corresponding map $f$ to projective space is ${\mathbb P}^1$. As diverietti says in the comments, the exact sequence in cohomology of $0 \rightarrow \mathcal{I}_V(mA) \rightarrow \mathcal{O}_X(mA) \rightarrow \mathcal{O}_V(mA)\rightarrow 0$ and Serre vanishing show that $\dim |mA-V|$ grows like $m^{\dim X}$. This being larger than 2 does not guarantee that the image of $f$ is not $\mathbb{P}^1$ (you could have $H^0(\mathcal{I}_V(mA))=H^0(f^*(\mathcal{O}_{\mathbb{P}^1}(k))$) but if that were the case, the rate of growth of $\dim |tmA-V|$ shows that for $t \gg 0$, $H^0(\mathcal{I}_V(tmA))$ strictly contains $H^0(f^*(\mathcal{O}_{\mathbb{P}^1}(tk))$ and so for some $t$ the system is not composed with a pencil.

In the singular case, let $\pi:Y \rightarrow X$ be a resolution of singularities such that $E=\pi^{-1}(\operatorname{Sing}(X))$ is a divisor, and assume that (*) for each component $V_i$ of $V$ there is an irreducible $\tilde V_i\subset Y$ of codimension at least 2 with $\pi(\tilde V_i)=V_i$. Let $\tilde A$ be an ample divisor of the form $\pi^*(kA)-D$ where $D$ is some divisor supported at $E$. The proof for the nonsingular case gives an irreducible divisor in $|m\tilde A-\tilde V|$ whose image in $X$ is irreducible and lies in $|kmA-V|$.

(*) is always true if no component of $V$ is contained in the singular locus, and I guess it is also true for components contained in the singular locus but don't know of a quick proof. On the other hand, Olivier's proof does not need characteristic zero, so unless someone asks, I won't try to polish this one further.

Answer by quim for $d$ points on a curve which are in the base locus of a pencil of planes

2012-12-15T13:22:11Z

A slightly different way to prove Francesco+auniket's result is as follows. First, given two different planes in the family, all $d$ points must lie in their intersection, which is a line $L$.

Second, every plane intersecting $C$ properly does so in $d$ points (counting multiplicities). So for every plane $H$ through $L$ intersecting $C$ properly, $H\cap C \subset L$. If there is any such plane, then general planes through $L$ intersect $C$ properly. Let $H_1, \dots, H_m$ be the (therefore finitely many) planes through $L$ which meet $C$ nonproperly. Then $C \subset \bigcup H_i$, each $H_i$ contains a component $C_i$ of $C$ of degree $d_i$ which goes through $d_i$ of the $d$ points. If there is no such plane on the other hand, then $L$ is a component of $C$.

Actually, this is projecting from $L$ rather than projecting from a point of $L$.

Are irreducible components of a flat family flat?

2012-11-22T13:50:29Z

Let $f:X\rightarrow Y$ be a flat morphism of schemes of finite type over a field $k$, and assume $Y$ is irreducible. Let $X_1, \dots, X_n$ be the scheme-theoretic irreducible components of $X$ (i.e., including embedded components).

Is it true that each $X_i$ is flat over $Y$?
If there are counterexamples to flatness of the $X_i$, is it true at least that each of them has equidimensional fibers?

Irreducible "family" of relative effective divisors of a smooth morphism

2012-07-11T06:34:21Z

Let $\pi:X\rightarrow Y$ be a smooth proper (assume projective if needed) morphism of schemes with $Y$ locally noetherian, and let $Z\subset X$ be an irreducible integral closed subscheme containing no fiber of $\pi$.

Is the locus $Pic_\pi(Z)=\{y\in Y:Z_y \text{ is Cartier in }X_y\}$ closed in $Y$?
If not, what extra hypotheses would make it closed?

As $\pi$ is smooth, Cartier and Weil divisors on fibers are the same, and as it is proper, the dimension of fibers is semicontinuous, so the issue is actually about components of smaller dimension in the fibers. Thus I'd drop the hypothesis on not containing fibers and replace it by the hypotheses that $Z$ dominates $Y$; then the question would be:

Is the locus of $y$ such that $Z_y$ has a (possibly embedded) component of codimension $\ge 2$ in $X_y$ open in $Y$?

I have the feeling that this is related to Zariski's main theorem, although in the first formulation it seems closer to asking whether the subscheme of relative effective divisors is closed in the Hilbert scheme when $\pi$ is smooth. But I can't pin it down.

(In my situation, $Z$ is actually of codimension 2, and everything is over the complex field, but I don't think this is necessary).

Answer by quim for About Newton Polygon to find a solution

2012-05-29T21:11:09Z

The Newton polygon of f alone does not determine the solutions nor the number of branches. (Example: $f=x^2-y^2$ and $f=(x-y)^2-y^3$ have the same Newton polygon.) However, the Newton-Puiseux theorem shows that by the Newton algorithm one can find all solutions and in particular the number of branches, and this algorithm is based on computing the Newton polygons of a sequence of polynomials starting from f and determined by changes of variables as in John Mangual's answer. This is explained in detail in Casas-Alvero's book "Singularities of plane curves" and also in Brieskorn-Knörrer "Plane Algebraic Curves".

Answer by quim for What is the original statement of Jung-Abhyankar theorem?

2012-03-30T10:46:46Z

The original papers are accessible online:

H. W. E. Jung, Darstellung der Funktionen eines algebraischen Körpers zweier unabhängigen Veränderlichen x, y in der Umgebung einer Stelle x = a, y = b. Journal für die reine und angewandte Mathematik 133, 289-314 (1908)
S. S. Abhyankar, On the ramification of algebraic functions. Amer. J. Math. 77 (1955), 575–592.

Jung's paper is devoted exactly to this result, whereas Abhyankar's gives a more contextualized explanation (his was an unsuccessful attempt to pass to characteristic $p$); I think the version of Abhyankar-Jung in Abhyankar is Theorem 3 (but it might be worth studying the paper carefully).

In Jung's statement, which I reproduce here, the field $K$ is defined as $K=\mathbb{C}(x,y)[z]/(f)$ for some irreducible polynomial $f$ (which is implicitly assumed to involve all three variables).

Man kann Funktionenpaare $u, v$ des Körpers $K$ bestimmen derart, daß $x$ und $y$ gewönliche Potenzreihen von $u$, $v$ werden, die für $u=v=0$ verschwinden, während alle anderen Funktionen von $K$ entweder gewöhnliche Potenzreihen von $u, v$ werden, oder Quotienten solcher. Eine endliche Anzahl solcher Funtionenpaare und Entwicklungen genügt, die Funktionen von $K$ für die ganze Umgebung von $x=0, y=0$ darzustellen.

My translation:

It is possible to determine pairs of functions $u, v \in K,$ such that $x$ and $y$ become usual power series in $u, v$, vanishing for $u=v=0$, while every other function in $K$ is either a usual power series in $u,v$ or a quotient of such. A finite number of such pairs and series is enough to represent all functions of $K$ in a neighborhood of $x=0, y=0$.

This seems to be equivalent, in the formulation usual in more recent papers, to the following (I use $\mathbb{C}\{x\}$ to denote convergent power series):

Let $f\in\mathbb{C}\{x,y\}[z]$ be a monic irreducible Weierstrass polynomial having a discriminant of the form $x^\alpha y^\beta u$, with $\alpha, \beta$ nonnegative integers, and $u\in \mathbb{C}\{x,y\}$ a unit. Then there exist positive integers $n, m$ such that $f$ has all its roots in $\mathbb{C}\{x^{\frac{1}{n}},y^{\frac{1}{m}}\}$.

Abhyankar considers the case of $n$ variables over an algebraically closed field of characteristic zero.

Answer by quim for Calculating the local index of intersection of two algebraic curves.

2012-03-26T09:23:44Z

The index of intersection satisfies certain properties which are easier to apply than the two definitions you give. For instance, if the tangent cones at P (initial forms, if $P=(0,0)$) of $F_1$, $F_2$ have no common factors, the index of intersection is just the product of multiplicities at $P$. And the index of intersection of $F_1=0$, $F_2=0$ coincides with the index of intersection of $F_1=0$ and $F_2+G\cdot F_1=0$ for every $G$. You can find a list of the relevant properties, with examples on how to apply them, in the relevant section of Fulton's book on Algebraic Curves.

There are two other ways to compute this number. One is, as you say, resolve the singularities of the union $F_1\cdot F_2=0$; then the intersection index is the sum of the products of the multiplicities of (the strict transforms of) $F_i=0$ at all blown up points (this is due to Max Noether). The other is to parameterize all branches of one of the curves $F_1=0$ and substitute the parameterizations in the other equation $F_2=0$. If the parameterizations are minimal, the intersection index is the sum of the resulting orders. Both these ways are explained in Casas-Alvero's book on Singularities of Plane Curves.

Answer by quim for How can we find a surface with a given singularity?

2012-01-13T11:38:49Z

For plane curves, general sufficient conditions have been given by Shustin (Trans. AMS 356, 2004, 953–985) although for particular singularity types (such as A-singularities) sharper results are known (see J. Alg. 302, 2006, 37-54). For one single $A_m$ singularity, I think the best sufficient condition is due to Lossen, via explicit equations (EDIT: Comm. Algebra 27, 1999, 3263–3282). In general it is not enough that the linear system of plane curves of degree $d$ has dimension at least equal to the codimension of the singularity type (except for the case of $m$ nodes, when this is necessary and sufficient).

In higher dimension, less is known, but again I'd suggest to look at Shustin-Westenberger, J. London Math. Soc. 70, 609–624.

Answer by quim for Nagata's conjecture, Seshadri constant

2012-01-09T18:28:17Z

I assume $S$ is a projective smooth toric surface.

If $D_1, \dots, D_n$ are the irreducible toric divisors on $S$, then $-K_S=D_1+\dots+D_n$ is an anticanonical divisor. Thus blowing up any point on any of these divisors one obtains a smooth anticanonical rational surface $\tilde S$; such surfaces are very well known by work of Brian Harbourne. Maybe more simply, by Mori theory, since $-K_{\tilde S}=-\pi^*K_S-E$ is effective on $\tilde S$, the cone of curves is spanned by extremal rays $\mathbb{R}C_i$ with $-K_{\tilde S}\cdot C_i>0$ and components of $-K$; these curves are in our case the exceptional divisor and the birational transforms of a subset of $D_i$'s [EDIT: the previous (italicized) sentence was not correct because new extremal rays do appear with the blowup, but it is still true that these extremal rays can be controlled with Harbourne's results]. Thus to determine nefness on $\tilde S$ and so the Seshadri constant of any ample divisor at the given point is not difficult. In particular these Seshadri constants are rational, and they only depend on the ample class and the $D_i$(s) to which the point belongs. [EDIT: these conclusions are still correct].

Blowing up at a general point of $S$ may give a surface which is anticanonical (if $-K$ is not fixed on $S$) or non-anticanonical (when $|-K|=\{-K\}$ ). In the first case, similar considerations would lead to the computation of the Seshadri constant. In the second case, I am afraid the problem can be difficult, and indeed related to the Nagata conjecture. The simplest interesting example would be the following: start with $\mathbb{P}^2$ as a toric surface and blow it up at the three three toric points. Now blow up the resulting surface at its six toric points. The result is a toric surface, the blow up of $\mathbb{P}^2$ at three clusters of three infinitely near points, where the Seshadri constant of your preferred ample divisor $L$ at a general point is presumably unknown. It might be irrational, if $L^2$ is not a square.

An analogon of the Nagata conjecture for toric surfaces (Seshadri constants at sets of $r\gg 0$ general points) can of course be stated, as particular cases of the conjecture stated by Lazarsfeld in 5.1.24 of "Positivity in Algebraic Geometry". I have nothing particularly relevant to say about that, except that it will probably be just as difficult as Nagata's original conjecture.

Is the radical of a homogeneous ideal homogeneous?

2011-11-22T17:26:32Z

Let $S$ be an $M$-graded $R$-algebra, where $M$ is some monoid, and $I\subset S$ an homogeneous ideal. The original, naïve, question, was: is it true that $\sqrt{I}$ is homogeneous? In this generality, the answer is no (see comments). However, if $M$ is a cancellative monoid with a total order, then the usual proof for $M=\mathbb{N}$ works, and indeed $\sqrt{I}$ is homogeneous. So:

Is there a natural class of monoids $M$ (larger, or different, from totally ordered cancellative monoids) such that in every $M$-graded algebra the radical of a homogeneous ideal is homogeneous?
The same question, for fixed $R$. In darij grinberg's example, it is relevant that the characteristic of the ring is 2. So, given a ring $R$, is there a natural class of monoids such that in every $M$-graded $R$-algebra the radical of a homogeneous ideal is homogeneous?

I am assuming everything in sight is commutative.

Answer by quim for Is there an upper bound and a lower bound on the contribution to the genus, for a singularity of codimension k?

2011-10-13T21:02:38Z

I assume you are talking about equisingular/topologically equivalent singularities (if you are talking about the analytical types, then the codimension is even higher). In that case, the relation can be computed from the embedded resolution of the singularity, as follows. This resolution consists in blowing up the point, and as long as the obtained curve plus the exceptional divisor does not have Normal Crossings, keep blowing up the points at which this fails. For each point $p$ that has to be blown up, let $m_p$ be the multiplicity of the curve at $p$; and let $f$ be the total number of non-satellite points to be blown up (a point is satellite if it is the intersection point of two exceptional components of previous blowups). Then $k=g+\sum m_p-f-1$ [Edit: because both $k$ and $g$ can be computed from the resolution; see Kleiman-Piene, Enumerating singular curves on surfaces, in "Algebraic geometry: Hirzebruch 70" and references therein. In Kleiman-Piene, $k$ is called "cod" and $g$ is called $\delta$.]

Edit: as soon as the singularity has multiplicity $\ge 4$, $\sum m_p-f-1\ge 4-1-1$ and your lower bound follows. If it has multiplicity 3 and at least one more point in the resolution has multiplicity 3, then $\sum m_p-f-1\ge 6-2-1$. All remaining types are A, D or E and hence covered by your argument so the upper bound as you claimed.

If you want a lower bound of the type $g\ge k- constant$ (I had not noticed this part of the question, sorry) then the formula tells you it is impossible. For instance, the ordinary singular point of multiplicity m has $k=g+m-2$ and $m$ can be arbitrarily large. On the other hand, since both $k$ and $g$ grow quadratically with $m$, you can surely get bounds of the form $g \ge k \times constant$ (and pick the constant arbitrarily close to 1 by restricting to large $k$). (If you are interested in the codimension of the equianalytic stratum, this argument does not work, but I still believe a lower bound of this sort may exist).

Answer by quim for Closure of singular points

2011-10-11T15:02:13Z

Almost everything in this answer has already been said by qui-vadis or in the comments, but now I'll translate it to your notation. I'll write $f(x,y,t)$ for $f$, and $f(x,y,0)$ for the limit $f$.

First remark that $u^4(u-t)^2$ divides $f(L_1u,u^2,t)$ so $u^6$ divides $f(L_1u,u^2,0)$ (this is qui-vadis' local Bézout). Expanding this and passing to the limit, $$\frac{L_1^5}{5!}f_{50}+\frac{L_1^3}{3!}f_{31}+\frac{L_1}{2}f_{12}=0.$$

Next oberve that the limit vanishings of $f_{40}$, $f_{21}$ and $f_{02}$, together with $f_{40}(L_1t,t^2,t)f_{02}(L_1t,t^2,t)−3f^2_{21}(L_1t,t^2,t)=0$, give $$Q:=f_{t40}(0,0,0)f_{t02}(0,0,0)−3f^2_{t21}(0,0,0)=0,$$ i.e., the limit of $f_t=\partial f/ \partial t$ also has an $A_4$ at least. Now using $f_x=0, f_y=0$ and the vanishing (in the limit) of $f_{ij}$ for $i+2j\le 4$ which you already know, it is possible to write the unknowns $f_{t40}(0,0,0)$, $f_{t02}(0,0,0)$, $f_{t21}(0,0,0)$ in terms of $f_{50}$, $f_{31}$ and $f_{12}$. Substitute in $Q$, and the resulting equation is exactly what you were looking for.

Answer by quim for Closure of singular points

2011-10-04T16:27:08Z

Assuming you know that a node colliding transversely with an ordinary cusp gives as a limit singularity a $D_5$ singularity, the answer is easier.

Blow up the point $(0,0)$ (ie, take y=xz, $f_t$ becomes divisible by $x^2$) and the family of proper transforms $f_t(x,xz)/x^2$ of your $f_t$'s have exactly an ordinary cusp and a node approaching transversely. The limit curve has at least a $D_5$, ie, it has intersection multiplicity 3 with the exceptional. The proper transform of a point of multiplicity 2 cannot intersect with multiplicity 3, so x (the equation of the exceptional divisor) divides $f_0(x,xz)/x^2$ at least once (it is the smooth branch of the $D_5$), and the quotient (which is the actual strict transform of the limit curve, $f_0(x,xz)/x^3$) has at least an $A_2$ (ie an ordinary cusp). So the limit curve has a $D_7$ as you say.

Answer by quim for What is so "plactic" about the plactic monoid?

2011-09-26T13:54:21Z

This is wild speculation, stemming only from the italian abstract to Lascoux & Schützenberger. There, "monoide a placche" is used alongside the parallel construction "varietà a bandiere" (flag varieties). At the end of the "préface," "la cohomologie des variétés drapeaux sur les corps finis" appears (again, drapeau=flag) as one of the connections worth mentioning. It may be plausible to think of a "plaque" as a "rigid flag", or a "discrete flag". Which then prompts the question about the origin of that name...

Answer by quim for sequence of sheaves for studying intersection

2011-09-26T09:40:00Z

is ok. For 2: I don't think you can prove the existence of flexes this way, but assuming P is a flex, then unicity of the line does follow from the iso. BTW, the unique "conic" that intersects B in 6P (P a flex) is the tangent line, doubled.

Answer by quim for Anticanonical divisor of the blow up of P^2 in 9 points

2009-11-19T16:05:41Z

Your nine points must be [EDIT: very] general [see MP's answer], otherwise it can be semiample.

The only effective anticanonical divisor is then (the strict transform of) the cubic C through the nine points. Since there is no other cubic curve cutting C in your nine points, the restriction of -K_S to C is a noneffective divisor of degree 0 (C has genus 1). So the restriction of -mK_S is also noneffective for all m (the points are [very] general in C! [~~I guess~~] Torsion points can make a difference) which means the only effective divisor in -mK_S is mC. Thus -mK_S is never base point free.

Space of sections

2011-09-23T09:18:31Z

If S is a noetherian scheme and π : Z → X a morphism of S-schemes, where X is proper over S and Z is quasi-projective over S, then the set-valued contravariant functor $\Pi_{Z/X/S}$ on locally noetherian S-schemes, which associates to any T the set of all sections of $π_T : Z_T → X_T$, is representable by an open subscheme of $Hilb_{Z/S}$. This is an exercise in Nitsure, "Construction of Hilbert and Quot Schemes", and the proof is similar to the construction of the scheme of morphisms "Mor". My questions are:

What is known about this scheme? It should be locally noetherian and quasiprojective, right? Can we say anything more?
Can this be generalized somewhat? For example, can we dispense with the "X proper" assumption?
Is there an analogous statement in the analytic setting? By wich I mean, assuming π : Z → X a morphism of complex varieties, where X is compact and Z quasi-projective, for instance.

References would be nice, even to FGA if that is the right place to look.

Answer by quim for Upper semicontinuity of multiplicities for finite morphisms between varieties

2011-09-22T08:16:43Z

M. Lejeune-Jalabert and B. Teissier. Normal cones and sheaves of relative jets. Compositio Math., 28:305–331, 1974

Actually the result is much more general, not just for finite morphisms.

Answer by quim for (Anti)Canonical divisor of a blow up

2011-09-13T08:33:11Z

If p:Y->X is the blowup of the surface X at a point x, and E is the exceptional divisor, then $K_Y=p^*K_X+E$. Hence the formula for the blowup of P^2 at 9 points. In the $P^1 \times P^1$ case, the canonical divisor is $-2(H_x+H_y)$, and you can compute the canonical divisor on the blowup easily. You can find all this in any book on algebraic surfaces, I bet.

Answer by quim for How many points determine an algebraic surface ?

2011-06-01T05:47:16Z

The answer has essentially been given by J. C. Ottem in a comment. I just put it here (with a couple details) so that the question is "answered".

The space of degree $d$ polynomials in n+1 variables has dimension $\binom{n+d}{d}$ (coefficient count), so hypersurfaces of degree d in $\mathbb{P}^n$ are parameterized by a projective space of dimension $N:=\binom{n+d}{d}-1$. Asking that the hypersurface goes through a given point is a linear equation on the coefficients of polynomials.

If the base field (or domain) is infinite, then the conditions imposed by general points are independent. You prove this by induction on the number of points: assume that hypersurfaces of degree d in $\mathbb{P}^n$ through $k-1 < N-1$ general points are parameterized by $\mathbb{P}^{N-k+1}$. EDIT: Choose one of these hypersurfaces $X$; it does not contain all of the points in $\mathbb{P}^n$, as the base field is infinite. So if the $k$-th point is out of $X$ (which we can assume as the points are general) the parameter space of hypersurfaces containing all $k$ points is strictly conained in $\mathbb{P}^{N-k+1}$, which means the last condition is linearly independent and defines a hyperplane $\mathbb{P}^{N-k}$.

So $N$ is the number of general points that uniquely determine a hypersurface of degree $d$.

Answer by quim for resolution of singular points on curve

2011-05-31T15:05:25Z

It won't be possible in general to get a single equation, because the curve does not necessarily admit a nonsingular plane model (globally). It is possible to get equations using Computer Algebra Systems, for instance Singular does it as explained here.

Concerning blowups: if the curve has a singular point which is not rational, then all of its conjugate points will be singular too, and they'll form a scheme that is defined over $k$. I never had to deal with such examples, but if $k$ is perfect I suppose that the blowup centered at that scheme will simplify the singularity and, iterating, eventually resolve it.

Answer by quim for de Rham cohomology vs. iterated tangent bundles?

2011-05-24T13:13:56Z

EDIT: I think an answer to your first question is explained in the papers:

P.-A. Meyer, Qu'est ce qu'une différentielle d'ordre $n$, Exposition. Math. 7 (1989), 249–264.
Laksov, Dan; Thorup, Anders, These are the differentials of order $n$. Trans. Amer. Math. Soc. 351 (1999), no. 4, 1293–1353. Freely available online.

Quote from the second:

The higher order differentials were part of the folklore of mathematics up to the end of the previous century, and formulas like $d^2f=f'_xd^2x+f'_yd^2y+{f''_{x^2}}dx^2+2{f''_{xy}}dxdy+{f''_{y^2}}dy^2$ can be found in most classical calculus books. (...) the higher order differentials vanished (...) because the extensive user of exterior differentials led mathematicians to believe that $d^2$ should always be zero.

(...)

Let $C_n:=\mathcal{C}^\infty(T^nX)$. The differential of $\mathcal{C}^\infty$- functions on $T^nX$ can be viewed as a k-linear map $d:C_n\rightarrow C_{n+1}$, and we obtain a sequence of linear maps...

Then $\Omega^n$ is defined as a suitable submodule of $C_n$, and there is a product $\Omega^p \otimes \Omega^n \rightarrow \Omega^{p+n}$ such that $d(\omega\cdot\pi)=d\omega\cdot\pi + \omega\cdot d\pi$ .

Functions defined as infinite products

2011-05-16T16:47:45Z

Are there standard references on infinite products of rational functions and their convergence properties? I'd appreciate information on finite products too!

The original motivation for this is the (finite) product $f(n)=\prod_{i=1}^{n-1}(1-\frac{i}{i^2+n})$ that I had to bound some time ago. Applying some calculus (logarithm to convert into sum, relate to a series, bound with an integral) I could show that $f(n)>\frac{1}{\sqrt{n}}-\frac\pi{8n}$ if $n>9$ (and actually $f(n)\sim \frac{1}{\sqrt{n}}-\frac\pi{8n}$ for big $n$) but I was left with the question whether there is a closed form for "my" finite product, or for the corresponding infinite product $\phi(n)=\prod_{i\ge 1}(1-\frac{i}{i^2+n})$(Any information on it would also make my day).

EDIT May 17: the infinite product $\phi(n)$ is zero (see Robert Israel's answer). Nevertheless, the square of $f(n)$ is $(1/n)\prod_{i=1}^{n-1}(1-(\frac{i}{i^2+n})^2)$, and it is still possible that the infinite product $\prod_{i\ge1}(1-(\frac{i}{i^2+n})^2)$ converges.

So, is there a place to look for techniques to deal with such products if the need arises?

Answer by quim for Algebraic plane curves and their tangent curves

2011-05-07T09:36:08Z

The answer is clearly yes for d=2,3,4 and 6. I am skeptical about larger degrees.

The family of plane curves of degree (at most) d-1 has dimension (d-1)(d+2)/2. Imposing d(d-1)/2 tangency points with the given curve determines a family of curves of degree d-1 tangent to C of dimension at least d-1. For large d, the family of nonreduced solutions will have dimension bigger than d-1, so this gives no information about existence of reduced solutions. However, for d=2,4 and 6 there must be reduced solutions because the nonreduced families of solutions have dimension 0, 2 and 4 respectively.

Answer by quim for Singular locus of the punctual Hilbert scheme

2011-05-02T16:59:21Z

It follows from Briançon (Inventiones Math 41 (1977), no. 1, 45–89) Theorem III.3.1 that the codimension is 2 for n=3, and 1 for n>3.

Answer by quim for What are some correct results discovered with incorrect (or no) proofs?

2011-04-05T16:55:00Z

According to M. Meo, Cauchy's proof of Cauchy's theorem (existence of elements of order a given prime p in every finite group of order divisible by a p) is wrong.

Cauchy works with subgroups of $S_n$, and his proof depends on the construction of what we now call a Sylow subgroup of $S_n$. This subgroup is obtained as a semidirect product, which Cauchy seems to say is actually a direct product (which would be abelian). I am not completely sure whether Cauchy was really wrong, or he did know what was going on, and simply lacked the appropriate language. In any case, would be an example of Lack of foundations.

Answer by quim for General hyperplane sections and projection from a point

2011-02-16T11:26:23Z

EDITED to match clarifications in the question and in Sándor's answer.

The question is equivalent to asking whether the tangent cone at $x$ of the hyperplane section coincides with the hyperplane section of the tangent cone:

Blow up $x$, and denote $\tilde{\mathbb{P}}^n$, $\tilde X$, $\tilde H$ the resulting variety and the proper transforms of $X$ and $H$. Now $\phi$ extends to $\tilde \phi$, defined on the whole $\tilde{\mathbb{P}}^n$, and for every subvariety $Z\subset \mathbb{P}^n$, $\overline{\phi(Z\setminus {x})}=\tilde\phi(\tilde Z)$. Thus, $\overline{\phi(X)} \cap \overline{\phi(H)}= \tilde\phi(\tilde X) \cap \tilde\phi(\tilde H)=\tilde\phi(\tilde X\cap \tilde H)$, because $\tilde H=\tilde \phi^{-1}(\tilde \phi(\tilde H))$ as in Sándor's computation. On the other hand, $\overline{\phi(X \cap H)}=\tilde \phi(\widetilde{X\cap H})$.

So, whenever $\tilde X\cap \tilde H=\widetilde{X\cap H}$, your statement holds. Now we can work componentwise: assume $X$ is irreducible. It is clear that $\widetilde{X\cap H}=\tilde X \cap \tilde H$ unless $\tilde X \cap \tilde H$ has a component contained in the exceptional divisor $E\cong \mathbb{P}^{n-1}$. Since $\tilde X$ meets the exceptional divisor properly, the only case when $\tilde X \cap \tilde H$ can have a component contained in the exceptional divisor arises if $\tilde H$ does not meet $\tilde X \cap E$ properly. But if $H$ is general, so is $\tilde H \cap E=L$, and then it does meet $\tilde X \cap E$ properly!

In your example with non-general $H$, $\tilde X \cap E$ is a point which gives trouble because your $\tilde H$ happens to contain it.

Answer by quim for About direct image of ideal sheaves

2010-12-15T14:52:53Z

The answer is no.

Consider $\mu=\mu_z \circ \mu_y \circ \mu_x$ the blow up of a smooth surface at three points $x$, $y$, $z$, as follows: $x\in X$ is arbitrary, $y\in E_x:=\mu_x^{-1}(x)$, where $\mu_x$ is the blowup of $X$ centered at $x$, and $z$ is the "satellite" point that appears after blowing up $y$, ie, $z=\tilde E_x\cap E_y$, where $E_y:=\mu_y^{-1}(y)$ is the exceptional of the second blowup $\mu_y$ and $\tilde E_x$ is the strict (birational) transform of $E_x$.

Let $d(\tilde E_x)=d(\tilde E_y)=0$, $d(E_z)=1$, where again $\tilde E_x$ and $\tilde E_y$ denote the strict transforms of the first and second exceptional divisors (but now on $X'$, ie, after blowing up the third). Then in your notation $I_1=I_2=\mathfrak{m}_x$ is the maximal ideal of $x$.

An easy way to see it is that the pullback of $E_x$ to $X'$ is precisely $(\mu_z \circ \mu_y)^*E_x=\tilde E_x + \tilde E_y +2 E_z$.

Comment by quim

2013-05-30T14:29:16Z

If each variable can be written as a rational function of x, then the curve has genus zero. If not, then what does "every other variable has function of x the variable as it's solution" mean?

Comment by quim

2013-05-23T20:01:14Z

I would bet this is feasible (by degenerations, again) but I suspect it takes some work. It could be that they are not "maximal rank" curves, but one can compute their regularity, and work out low degree cases. Are you interested in some particular application?

Comment by quim

2013-05-23T19:49:21Z

And, answering the circles case: there are four intersection points, but two of them are complex conjugate. This, however, is basic, more appropriate at math.stackexchange than here.

Comment by quim

2013-05-23T17:59:57Z

In the BKK theorem, N is also the dimension of the ambient space.

Comment by quim

2013-05-23T08:58:07Z

I don't think much mathematical background is needed. The $\Delta_i$ is just the convex hull of the set of exponents of the terms involved in the $p_i$, inside $\mathbb{R}^2$, so in your case they are both equal to the triangle with vertices (2,3), (1,2), (0,0). The coefficients don't matter. Vol denotes "mixed volume" but I think that when all $\Delta_i$ are equal, this is just ordinary volume. Have a look at arXiv:0812.4688. BTW, just as Bezout, this only gives an upper bound on the number of solutions, which is the exact number if the coefficients are "general enough".

Comment by quim

2013-05-22T20:08:41Z

Since you fix the terms, the Bernstein-Kushnirenko theorem (or Bernstein-Khovanskii-Kushnirenko) would be more precise than Bezout's theorem.

Comment by quim

2013-05-09T14:47:58Z

My comments don't make sense any more, so I'll delete them in a moment.

Comment by quim

2013-05-08T14:41:07Z

@Francesco: We should be able to see that the schematic branch locus has multiplicity two by looking at the ring map $\mathbb{C}[x]\rightarrow\mathbb{C}[x,y]/(x^2-y^2)$. So Joël's question is pertinent: what is the definition of branch locus? If this is not a counterexample for the OP's definition of branched along $D$ (because it is actually branched along $2D$, which is not regular) then maybe your deleted (why?) answer was the right one.

Comment by quim

2013-05-01T20:26:15Z

If I understand it right then, the same case of two coplanar lines, but with d=3, would be a counterexample. Because a general plane through the third line cuts an irreducilbe conic on the cubic surface, which is linearly equivalent to the sum of the two given lines.

Comment by quim

2013-05-01T16:49:02Z

"We assume that the classes of the curves in the Neron-Severi group of X spans a rank r lattice not containing the hyperplane section of X." This seems a very very restrictive case, if it can occur at all (for a general X). Can you give some more context, or an example?

Comment by quim

2013-04-29T08:37:54Z

You can indeed weigh differently different variables locally, and get a tangent cone which lives in $\mathbb{P}(a_0,\dots,a_n)$. Whether this is useful or not will depend on the local nature of the singular variety, and not on whether it is globally embedded in some weighted projective space. Consider the tangent cone at the origin of the affine plane curve given by $F=(y^2-x^3)(y^2+x^3)+x^5+y^5$, with weights (2,3). It has two components, corresponding to the two branches of F=0. But if you do that to an ordinary singularity, you see nothing.

Comment by quim

2013-04-16T08:42:15Z

I don't know the answer, but why not contact Filip Cools? He proved the equality when m=3, right?

Comment by quim

2013-04-16T08:32:29Z

I sympathise with your heresy, so +1. That said, people differ as to their need to understand about the engine. Everybody can drive safely knowing very little, but to be comfortable and happy with it is another thing. It does not seem advisable to start doing algebraic geometry without at least some familiarity with the content of Atiyah-Macdonald, and if the OP feels this is not enough for them, going for more won't hurt. Note that also Sándor's thoughtful answer suggests giving lower priority to CA as one goes deeper into AG.

Comment by quim

2013-04-04T11:20:21Z

Actually, the correct statement should be that $V_p$ has codimension 1 exactl when $C$ is singular at $p$. Sorry. The conclusion that follows is the same.

Comment by quim

2013-04-03T12:56:28Z

Oops, the sentence in the previous comment should be, $\bigcup_{p\in C} V_p=|I_C(d)|$ if and only if $C$ is *non*reduced. And, to be more precise, $V_p$ has codimension 1 exactly when $C$ is nonreduced at $p$. So the general surface is singular if and only if $C$ has a nonreduced component.