quim
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May 9 |
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regularity of finite flat branched covers My comments don't make sense any more, so I'll delete them in a moment. |
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May 8 |
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regularity of finite flat branched covers @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. |
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Apr 29 |
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Hilbert function of (weighted) projectivized tangent cones 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. |
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Apr 16 |
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Tropicalization of the Grassmannian I don't know the answer, but why not contact Filip Cools? He proved the equality when m=3, right? |
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Apr 16 |
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Algebraic machinery for algebraic geometry 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. |
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Apr 4 |
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Existence of smooth surfaces containing a curve 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. |
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Apr 3 |
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Existence of smooth surfaces containing a curve 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. |
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Apr 3 |
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Existence of smooth surfaces containing a curve Actually, the complement to $U_p$ is a linear subspace $V_p\subset |I_C(d)|$ which has codimension 2 if $C$ is reduced, but codimension 1 if it is nonreduced. Since $I_C(d)$ is globally generated, the $V_p$ are all distinct. Therefore $\bigcup_p V_p=|I_C(d)|$ if and only if $C$ is reduced. |
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Apr 3 |
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Existence of smooth surfaces containing a curve Rather, the other point must be $q\in C$ ;) |
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Apr 3 |
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Existence of smooth surfaces containing a curve I don't follow either. In Jason's example, the singularities of $R$ lie on the base locus (and for every $d>1$, the base locus of $\mathcal{I}_C(d)$ is just $C$). But these singularities move with $R$. That is, for every $p\in C$ there is indeed a surface smooth at $p$ and containing $C$, but it must be singular at some other point $p\in C$. |
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Mar 9 |
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“Degree” of an algebraic variety So the degree of a point in $\mathbb{A}^n$ would be $n$? |
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Mar 6 |
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Crepant Birational Map on the Blow-up @Andrew: doesn't $f$ restrict to an isomorphism between the obvious open subsets isomorphic to $\mathbb{A}^n$? |
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Feb 4 |
accepted | Smooth curves in a linear system satisfying certain conditions. |
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Feb 4 |
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Smooth curves in a linear system satisfying certain conditions. For each point $p$ of the support of $\xi$ (or, for each irreducible component of $\xi$) the multiplicity of $\xi$ at $p$ is the maximal integer $m$ such that the stalk at $p$ of $I_\xi$ is contained in the $m$-th power of the maximal ideal. What I mean is that all components have to be of multiplicity 1. |
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Feb 4 |
answered | Smooth curves in a linear system satisfying certain conditions. |
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Feb 1 |
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bivariate polynomial Are you counting "distinct" roots, or you take multiplicities into account? |
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Jan 29 |
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Set of plane curves which intersect a fixed curve with given multiplicity And for degrees $\ge d$, the multiples of the equation of $C$ are themselves a linear subspace, which is the intersection of all $V_n$. |
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Jan 29 |
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Set of plane curves which intersect a fixed curve with given multiplicity Yes. If there are $k$ branches, the intersection multiplicity with $C$ is the sum of the intersections with each branch, so there would be a vector subspace for each partition of $n=n_1+\dots+n_k$. (In fact not all partitions need to be considered, because of the intersections of branches with one another). |
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Jan 29 |
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Set of plane curves which intersect a fixed curve with given multiplicity Besides, there is nothing special about degree d-1 forms for this question. |
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Jan 29 |
answered | Set of plane curves which intersect a fixed curve with given multiplicity |
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Jan 24 |
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Universal property of blowing down jstor.org/stable/1970602 Do you mean lemma 3.7 there? It is quite different from what the OP expected... |
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Jan 18 |
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Irreducible divisors containing an arbitrary closed set added 31 characters in body; edited body |
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Jan 18 |
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Irreducible divisors containing an arbitrary closed set added 1674 characters in body |
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Jan 17 |
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Irreducible divisors containing an arbitrary closed set Ok, I'll rewrite it more carefully asap (tonight possibly). About the reduction to the nonsingular case I am not 100% sure, I'll think of it. |
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Jan 17 |
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Irreducible divisors containing an arbitrary closed set added 38 characters in body; added 71 characters in body |
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Jan 17 |
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Irreducible divisors containing an arbitrary closed set Oh, sorry: I missed the "normal X" hypothesis, and assumed smooth. In that case, you can reduce this to Hartshorne's III,10.9 by blowing up (the components of) V. The proper transform of the linear system $|mA-V|$ is basepoint free. |
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Jan 17 |
answered | Irreducible divisors containing an arbitrary closed set |
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Dec 17 |
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Milnor number in terms of minimal resolution of an isolated singularity. "co"dimension of the "Jacobian" ideal? |
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Dec 15 |
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$d$ points on a curve which are in the base locus of a pencil of planes Added comment. |
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Dec 15 |
answered | $d$ points on a curve which are in the base locus of a pencil of planes |
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Dec 13 |
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$d$ points on a curve which are in the base locus of a pencil of planes Wow! actually this answer even fits the degenerate case in the picture (with m=1). |
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Dec 12 |
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$d$ points on a curve which are in the base locus of a pencil of planes Actually, there is another possibility: that the line L is a component of C, then there are no restrictions on the rest: C is composed of a line through the d points plus a curve of degree $d-1$. The argument "fails" in that the projection is not birational on C in that case. |
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Dec 11 |
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$d$ points on a curve which are in the base locus of a pencil of planes Even simpler: any two different planes in the family meet along a line. All d points must belong to it. |
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Dec 11 |
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Irreducible “family” of relative effective divisors of a smooth morphism In short, the trick of both Mumford and Grothendieck's proof is to use the valuative criterion for properness. Then one has to cope with a situation in which $X\rightarrow Y$ is smooth and Y regular, so X is regular too, and divisors are simpler to deal with. |
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Dec 11 |
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Irreducible “family” of relative effective divisors of a smooth morphism Finally found a reference that settles the flat "special case". In Grothendieck's FGA, last éxposé (VI), Theorème 2.1 (i). A maybe simpler explanation (which avoids part of the algebra but focuses on the surface case) appears in Mumford's Lectures on Curves..., in lecture 15, VIII. But I didn't understand why embedded components of fibers are impossible until I read Grothendieck's proof. |
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Dec 5 |
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What is the multiplicity of a Cartier divisor at a point？ In (3), the intersection product should be replaced by the local intersection multiplicity. Then, if X is smooth at P, the three definitions are equivalent. |
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Nov 29 |
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Are irreducible components of a flat family flat? @Tom Goodwillie: Thank you! I have been able to track down this example to EGA IV, Remarques 6.5.5(ii) and 12.1.2(i). On the other hand, the immediately preceding proposition 12.1.1.5 shows that my second question has positive answer |
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Nov 29 |
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Irreducible “family” of relative effective divisors of a smooth morphism @Qing Liu: It seems (by the last reference in your answer or, rather equivalently, reading Kleiman's exposition on relative effective divisors in his "FGA Explained" chapter) that if $Z\rightarrow Y$ is flat, then $Pic_\pi$ is open. So using the flattening stratification would yield constructibility. I also see that your proof attempts to be more elementary and avoid flattening stratification. OTOH, these results do not require $\pi$ smooth, and I am still wondering whether $Pic_\pi$ is closed in the case that $X\rightarrow Y$ is smooth, and $Z\rightarrow Y$ is flat? |
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Nov 23 |
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Are irreducible components of a flat family flat? By each component having equidimensional fibers I mean that $dim (X_i)_y$ is independent of $y$. Of course these dimensions will depend on $i$. |
